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| Q&A
for 14A-1 |
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When will we be able to check our grades online with myucla.edu?
Hi David, Thursday December 22.
one last question.... for the Fall 2002 final question 5, why is the ground state electron configuration for O2+ have 12 electrons total? I thought there were only 11 electrons to fill in the MO diagram.
It is supposed to be 11, what is written is for O2 not O2+. Maryam
hi! I have a question about Q5 part b on the Summer 2005 Exam 2. Why did he round off the sigfigs so early? I thought you should do them last... and if you do, you get a [OH-]= 9.0 x 10^-6, pOH= 5.04, aqnd thus pH=8.96. So should you round them as soon as the initial moles of NH3 and produced moles of NH4+ ? Thank you!
You are correct, you should round off at the end. The answer to this question has been taken from the solution manual and they are not always careful with the sig. figs. Maryam
For Polyprotic Acids and Bases, our book and Dr. Lavelle said to just use Ka1 in our calculations. However, in lecture notes, it says to use Ka1 x Ka2 = Koverall . then you are to use k overall for the calculations...what are we to do?
You mean polyprotic acids. There is no such thing as polyprotic bases, since proton is acidic, anyhow. Just use Ka1. Maryam
Can you explain of "point C" in titrations, and then the difference between that and the stoichiometric point? Thanks!
The stoic. pt. is when you have added the same number of moles of acid/base from the burett as the initial number of moles of the base/acid in the flask, to the flask. So at the stoic. pt. all of the base/acid in the flask has reacted with the added acid/base. Half-way to the stoic. pt. is when you have added half of the volume of the acid/base in burett that you need to reach the stoic. pt. to the flask. So at this point half of the base/acid has reacted with the added acid/base. Half-way to the stoic. pt. is only interesting when we are titrating a weak base/acid with a strong acid/base, since pKa of a weak acid = pH at half-way (point C) and pKb of a weak base = pOH at half way, but remember that since on the titration graph we record pH not pOH, then for the case of a weak base you need to convert between pH and pOH. Maryam
I was just wondering if in the final we are going to be given a table of values for kb, and ka for the most important acids/bases, or are we supposed to memorize them all?
You don't need to memorize any K values. If the values are needed, they will be given. If the reason that you are asking this question is that to see in order to be able to compare acidity or basisity strengths, then if by just looking at the structures you will be able to compare their strengths, then it won't be given. But if it is for a question that you need if for calculation of for example pH, it will be given. Maryam
When given somrthing like CH3CHCHCOH, and asked to find the hybridization for the carbons, can we just write "sp2" or are we supposed to write out each bond such as "sigma H1s-C2sp2", "sigma C2sp2-C2sp2", "pi C2sp2-C2sp2"?
If it is asking about bonds, then you need to write, for example, sigma H1s-C2sp2 and so on. But if they are asking about hybridization then you should write sp2 (2sp2), sp3( 2sp3), and so on. Maryam
in hw problem 10.35 part c, the answer manual says HCLO2 is a strong acid than HBrO2. the chapter says that the strength increases down a group though. shouldnt the answer be HBrO2?
The strength of binary acids like HCl, HBr, ... (hydrogen and a nonmetal) increases down a group. Between HClO2 and HBrO2, HClO2 is stronger because Cl is more electronegative and it can stabilize the conj base formed upon deprotonation more. Read 10.9 and 10.10. Maryam
For XeF2 and I3 (minus), even though the electron arrangement is trigonal bypyramidal, how come the bond angle is 180 degree?
Because in trigonal bypyramidal there are three sets of angles: 120, 90, and 180. Since there are three lone pairs on these (molecule and anion), the other two sets of bond angles don't apply, so the only one remaining is 180. It is not less than 180, because the lone pairs are perpendicular to the two bonds. Remember when you have a trigonal bypyramidal electron arrangement and there are some lone pairs, they will be positioned at the equatorial position. Maryam
For a question such as Winter 2001 Final Q8, when they give you the normal pH of 7.4, are we to use the fact that there is only one significant digit in the concentration used to derive that pH? and therefore, arriving at an answer such as 2E-2 instead of .0222?
What you are saying is correct. But sometimes for questions like this Dr. Lavelle doesn't asign any points for the sig figs. But in order to be on the safe side, show what you have obtained, first, and then apply the correct number of sig figs to that answer. Maryam
Are we allowed to use graphing calculators on the final?
No you are not. Maryam
Maybe this was already mentioned, but will we need to memorize the table of color change indicators in Ch. 11 for the final, if we're asked to give a proper one for a titration? Thanks!
I don't remember that Dr. Lavelle mentioned anything about knowing the color changes for the indicators that you choose to memorize. If you want to be 100% prepared go ahead and take a look at the color changes for the 3 indicators that you have chosen to memorize. But I doubt that he will ask about the colors. Maryam
For a Nuetralization reaction forming a salt and water, do we always use the double arrows in writing the reaction, or do we use a single arrow when there is a strong acid or base?
For a neutralization reaction you always use a single arrow, except if it is a reaction between a weak acid and a weak base. But once at least one of them (acid or base) is strong, then you need to use a single arrow. Maryam
How do lone pair electrons play a role in bond length/strength?
Lone pairs play a role in bond strength, lone on neighboring atoms cause repulsion and makes the bond weaker. For example the bond in F2 is weaker than the bond in H2. Maryam
How is N2O not polar? If N-N-O, wouldn't the dipole moments both be going to the electronegative O and therefore, no "canceling" would be going on... right?
That is correct, just being linear is not enough for being nonpolar, but due to its structure (having a formal negative charge on N), it is slightly polar. Maryam
How is N2O not polar? If N-N-O, wouldn't the dipole moments both be going to the electronegative O and therefore, no "canceling" would be going on... right?
That is correct, just being linear is not enough for being nonpolar, but due to its structure (having a formal negative charge on N), it is slightly polar. Maryam
If we are drawing our molecular orbitals and we have 1 electron in our 2pi(px) and that is our highest occupied molecular orbital (HOMO), could/would our lowest unoccupied molecular orbital (LUMO) be the 2pi(py) orbital? Is it possible for the HOMO and LUMO to be at the same energy level? Thanks again!!
For a question like this (for the scope of this class), you need to show that, for your example, pi2(p) is HOMO and LUMO the will be the orbital with a higher energy level after HOMO, so it will be pi2p*. The reason that we are interested in the HOMO and LUMO is because of transitions between the HOMO and LUMO energy levels. So a an empty degenerate orbital with the HOMO is not LUMO. Maryam
winter 2004 question 2A, how does electron diffraction by a crystal support de Broglie's hypothesis of the wave nature of matter?
De Broglie's relation shows wave nature of particles (like electrons). Diffraction patterns are produced by waves. So electron diffration by a crystal supports de Broglie's hypothesis of the wave nature of matter. Maryam
For Winter 2001 Final Exam question 1, how do we know to use the equation 1/2mv^2 for e, and not for part h "wavelentgh emitted by...". What dictates when we should use what? Would we be able to use lambda=h/p for part e and say that because E=hc/lambda, E=cp? Are there certain cases where we must use one equation as opposed to another, and what makes us use such equation on this question?
For part e) since it is asking about the energy of an electron with a specific velocity, then what they mean is kinetic energy, so you need to use the kinetic energy equation. For part e) since it is asking about the wavelength of an electron, with a given velocity, so you know that de Broglie's relation gives the relation between the wavelength of a particle (like an electron) and its velocity. So depending on the question, you need to see what equation you need to use. Maryam
summer 2005 question 4D, how do we know that C2H5NH2 has a higher pH than NH3?
When you have alkyl substituents on N (CH3 group or C2H5 group, or longer chaines), since they are electron donating groups, they give more electron density to N and making it a stronger base, this is compared to having only hydrogens on N. Take a look at table 10.2. Maryam
Table 10.6 on p 382 says that for binary acids, there are two trends. Is there one of these trends that dominates the other? That is to say, what if we compare two anions that aren't necessarily from the same group or period? Thank you!
I suppose you mean binary acids (not anions). Then the electronegativity will dominate. As in H2S and HF, HF is a stronger acid compared to H2S, since F is more electronegative than S. Maryam
Are the sig figs for Q6 on Fall 2002 Final exam correct for the concentration of PCl5?
It should have 2 sig figs. Maryam
why is it that in 11.13, for c its ka3 and d is ka2? and do we have to look up the formula to figure out which is the acid and which is the conjugate base???
For c) it is KHPO4- (in this case this is the acid) and (KPO4)2- (conj base), so it deals with Ka3 of H3PO4 (the third deprotonation), for d) it is KHPO4- (conj base) and KH2PO4 (acid), so it deals with Ka2 of H3PO4 (the second deprotonation). If you are not familiar with the nomeclature of compounds, then take a look at fundamentals D, especially table D.1. Maryam
How do you tell if an oxide is basic, acidic, or especially amphoteric by just looking at it? For instance, if you were given As2O3, how would u figure that it is amphoteric?
Oxide of nonmetals are acidic, metal oxides are basic, and oxide of amphoteric elements are amphoteric. So you need to know which elements are metals, nonmetals, and amphters. If you look at the periodic table, you will see that metals and nonemetals are being separated by amphoters. Look at figure 10.5 p369, it shows the elements that are on the diagonal between metals and nonmetals (amphoters). Maryam
Winter 2001 final 1. f. What is a planar node? 1.h. For the uncertainty in position, why do they use h/2pi instead of h/4pi?
A planar node (or a nodal plane) is a plane that goes through an orbital and the electron density is zero on it. p orbitals have 1 planar node, and s orbitals do not have any planar nodes. For 1.h they should be using h/4pi. Maryam
when are our grades going to be posted online through my.ucla.edu??
They should be posted by the end of next week, if everything goes as planned. Maryam
This post is for my TA MATT, and concerning quiz four that i took but was not recorded. I was wondering if you (Matt) will be on campus on wednesday so that i can bring you quiz 4. Thanks
Then you need to email Matt. I'm answering the VOH questions not Matt. Maryam
Dr. Lavelle, In the final exam for fall 2002, question 5a, shouldn't the sum of the electrons in the electron configuration be eleven as opposed to twelve, as it is in the answer key (because O2+ only has eleven electrons)? Thank you. ~Silvia
I've answer this question several times by now! Yes, you are right it should have a total of 11 electrons as opposed to 12. 12 is for O2 not O2+. Maryam
which homework problem did professor indicate that we dont have to do because the solution was wrong?
10.69. Maryam
Looking at homework question 10.59 How can you tell if a salt is acidic, basic, or neutral? Can you explain the last 4 parts of the question.
When looking at salts you need to consider the anion and the cation forming the salt, one by one. Look at each and see if it is neutral, acidic, or basic. Cations like group one(Li, Na, etc) and group 2 (Ca2+ and heavier ones) are neutral, since they don't react with water, they don't change the pH of a solution. Anions of strong acids like Cl-, Br-, etc. are neutral since they don't react with with water and they don't change the pH of the solution. On the other hand anions that are conjugate bases of weak acids are basic, like CH3COO-. Upon reaction with water they form a weak acid and OH-. Since they form OH- in water they are basic. Cations that are conjugate acids of weak bases are acidic, like NH4+. Upon reation with water they form a weak base and H3O+. Since they form H3O+ in water they are acidic. Small highly charged cations like Al3+ and Fe3+ are also acidic, they form H3O+ in water as well. Take a look at tables 10.7 and 10.8 for a complete list of cations and anions and their acidic, basic, and neutral character in water. As of example 10.59, then you can see that for KF, K+ is neutral (group one cation), F- is basic (it is conjugate base of a weak acid). So the salt is basic. For KBr, K+ is neutral (group one cation), and Br- is neutral (anion of a strong acid) so the salt is neutral. For AlCl3, Al3+ is acidic (small highly charged cation), Cl- neutral (anion of a strong acid), so the salt is acidic. For Cu(NO3)2, Cu2+ is acidic (small highly charged cation), NO3- is neutral (anion of a strong acid), so the salt is acidic. Maryam
Do strong acids or strong bases ever have a Ka or Kb value?
For very strong acid and bases Ka and Kb are so large that the reaction is almost 100% going to the right, so there is almost no equlibium, and no equilibrium constant(or a very large K value that in the book they just say strong). Maryam
Fall 2002 1A, they use kg as the unit of measurent for the mass. Isn't it normally in grams, so that the answer should be the the power of 10^-41 instead of 10^-38?
The SI unit for mass is kg not g. Maryam
Dr. Lavelle's notes for Ch. 10 say that "Many nonmetal oxides react with water to form acids." The notes give as an example the reaction of H2O with CO2 to form carbonic acid. My question is, is CO2 in aqueous solution considered an acid?
Yes, CO2 is a gas. When you dissolve it in water you will get H2CO3(aq) (carbonic acid). CO2 (g) + H2O (l) => H2CO3(aq) Maryam
what is the bond angle for bent molecular structures?
I have answered this before. Depending on the electron arrangement, for example H2O's electron arrangement is tetrahedral and has a bent shape. Its bond angle is less than 109.5 (105). SO2's electron arrangement is trigonal planar and has a bent structure. Its bond angle is less than 120. Maryam
in the fall 2002 final question 5, the answer corresponded to the electron congfifuration for O2 because it has 12 e-. the question asked the ground state of O2+. Which one are they asking for ? is this an error?
I've answered this question before. Yes it is an error. Maryam
I read somewhere online that a Bronsted acid is always a Lewis acid. Is this true? for instance, is H2SO4 both a bronsted acid and a lewis acid because its proton acts as an electron pair recipient?
Yes, Lewis definition is the most general definition for acids and bases. So any Bronsted acid/base is also a Lewis acid/base, but not every Lewis acid/base is a Bronsted acid/base. Maryam
I read somewhere online that a Bronsted acid is always a Lewis acid. Is this true? for instance, is H2SO4 both a bronsted acid and a lewis acid because its proton acts as an electron pair recipient?
Yes, Lewis definition is the most general definition for acids and bases. So any Bronsted acid/base is also a Lewis acid/base, but not every Lewis acid/base is a Bronsted acid/base.
If I am not mistaken, O2+ has 11 valence electrons. How is it then, that for number 5a on Fall 2002 Final there are 12 electrons represented by the electron configuration instead of 11. I would not have squared the last term written (pi, 2p *) Thank you!
I have answered this question before. It should be (pi 2p*)1. What is written is for O2 not O2+. Maryam
For problem 10.69, the solutions manual has the pH of the acid as 1/2 the pKa1 + the pKa2. Can this formula be used for all acids given two pKas?
This problem has been omitted. This formula is for pH of amphiprotic salts (which we did not cover). Maryam
Would thymol blue be an acceptable indicator to use in question 10 on Winter 2001 Final?
Phenolphthalein is a better choice, but since Dr. Lavelle said that you need to memorize only one example of each indicator with an acidic, basic, and neutral pKln, then thymol blue will be acceptable too. Maryam
My tutor told me to approximate ovly if Ka was smaller than 10^-5 but in past finals, Lavelle approximates for 10^-4. Are we allowed to approximate for 10^-4 and smaller?
Yes, 10^-4 is small enough, for 10^-3 since is borderline, if you use approximation then you need to confirm it by checking to see if the value obtained for x is smaller than 5% of the value that you are doing the approximation against (the 5% rule). Maryam
Is the bond angles depend on the molecular shape or the electron arrangement? Thank you
Bond angles depend on the electron arrangement. For example, CH4 has a tetrahedral electron arrangement and tetrahedral electron shape, bond angles are 109.5. H2O has a tetrahedral electron arrangement and a bent structure, bond angle is less than 109.5 (105, because of the two lone pairs). NH3 has a tetrahedral electron arrangement and a trigonal pyramidal shape, bond angle less than 109.5 (107, because of the one lone pair). Maryam
Hi Maryam, I noticed you've been answering a lot of the same questions over and over for the past week, so I just wanted to say thank you for answering all of our questions. You are doing an amazing job!! :)
Hi Jackie, Thank you! That's very nice of you. Now I feel I have more energy to answer the questions. Maryam
Hi. For question 3A in the Fall 1 2001 midterm, the question asks for the wavelength after the transition from n=4 to n=2. The solution shows that the change in energy was E4-E2, but I thought when calculating the change in energy, we were supposed to do Efinal-Einitial. In this case, wouldn't it be E2-E4 for the change in energy?
Yes you are correct. That anwer has been just coppied from the solution manual, that's why. But since the question is asking about the wavelengh, not change in energy the sign of change in energy is not important. But make sure on exam you have the correct order. Alway E_final -E_initial. Maryam
On question 11.35, why is only phenolpthalein listed as a suitable indicator? The pH is 8.88, so why wouldn't thymol blue (option c) also work if its pH range is 8.0-9.6? Also, Dr. Lavelle told us that we should memorize some of the indictators and their respective pH ranges. Could you clarify on this? Should we be able to give an indicator's name for any possible pH? Do we need to memorize all of the indicators? What would you personally recommend?
For 11.35 thymol blue is also suitable. This is also under the Textbook Errors. Dr. Lavelle said that you just need to memorize 3 indicators, one with pKln of in acidic pH, another with a neutral pKln, and one with basic pKln. So depending on pH of the stoic. pt. of the reaction of interest, you can name a suitable indicator. Maryam
do you use the Henderson-Hasselbalch equation in 11.3 because the molarities given were already at eliquibrium?? since for 11. 9 there is a change in volume and thus molarity concentrations, you have to use the ice box to find pH?
For 11.3, for part a) the book uses H-H equation, but for part b) it uses icebox, you can use H-H equation for part b) too. When you use H-H equation you will be using the initial concentrations, not the equilibrium concentrations. For 11.9 also after finding the initial concentrations of HCN and CN- which are 1.5 *10^-2 and 2.1* 10^-2, you can plug them in the H-H equation. You will be getting a similar answer. Maryam
Do we have to know the name of compouds for the final....? For instance in one past final it asked is cis-dichloroethene polar or nonpolar. Will we have to know what cis-dichloroethene is?
You will study organic chemistry in 14B, but it is good to know the name and structure of some simple organic compounds. Most likely he will give the formula, but in case be familiar with the ones covered in class or in the homework problems. Maryam
Do the indicators we have to memorize come from Dr. Lavelle's lecture notes, or can we choose them from the textbook's list?
Doesn't matter as long as you remember three different indicators having a neutral, an acidic, and a basic pKln. Maryam
do we need to know the stoichiometry of polyprotic acid titrations????
No that hasn't been covered. Maryam
I have a question on 10.43. If a larger Ka means a stronger acid or base corresponding to a smaller pKa, then why would aniline with the smallest pKa be the weakest?
The compounds in that example are all bases, so the larger the pKb the weaker the base (the smaller the Kb the weaker the base). But the numbers given are pKas of the conj. acids of these bases. So you could either convert pKa to pKb by using pKa + pKb = pKw = 14 and then comparing those numbers or you can just say that the weakest base has the strongest conjugate acid. So the conj. acid with the smallest pKa is the strongest conj. acid , and it correspondst to the weakest base. Maryam
I understand that the order of electron shell filling is: 1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5d,6p, etc. However, when we write an electron configuration, do we follow that same order?
Is you mentioned that is the order of filling electrons. But when it is filled the order is like this 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f, 5s, etc. Maryam
I think that problem 1d in the fall 2004 final contains an error. The 1.819 x 10 raised to the -4 should be 1.819 x 10 raised to the negative 3.
I have answered this questin before, yes you are right, it should be 10^-3. Maryam
Hi, I'm wondering how the sig fig. is going to be for the final. For past exams, -log (6.13 x10^-13) is 12.21. But according to lecture notes for logs, the corrrect sig fig. should be three digits after the decimal. Thanks.
I'm not sure which past exam you are reffering to. So there might be one of the given values that has two sig figs and that could be why they have two digits after the decimal point. Otherwise if the smallest number of sig figs in the values given that you used to do the calculation to get to 6.13 * 10^-13 was 3, then you need 3 digits after the decimal for the anwer to -log (6.13 x10^-13). Maryam
Hi, I asked a similar question before, but the answer I was given did not specifically pertain to the problem in the book. For question J.3 part (b), why isn't the complete ionic equation: (CH3)3N (aq) + H+ (aq) + NO3- (aq) yields NH(CH3)3+ (aq) + NO3- (aq)?
Hi, I'm not sure what you are asking but I try to just go over the anwer and if that is not what you were asking then try to clearify what is that you are asking. It is a reaction between a weak base and a strong acid, so for the complete ionic equation you will have (CH3)3N (aq) + H3O+ -> NH(CH3)3+ (aq) + H2O (l). Since HNO3 is a strong acid it 100% dissociates in water to H3O+ and NO3-. And Sice NO3- is the anion of a strong acid it is neutral and won't react with water, so you can omit it, it is an spectator ion. Since (CH3)3N (aq) is a weak base, you keep it in its molecular formula on the left had of the equation, so now you have (CH3)3N (aq) reacting with H3O+ (aq) which will yield to the conjutate acid of (CH3)3N (aq), that is (CH3)3NH+ (aq) and H2O (l). Maryam
For a question like 11.33 that asks for the volume of tirant needed to reach halfway to stoimetric point, or any volume in general, how are we to take into accound sigfigs? For example, halfway of 25 ml would be 12.5 ml which IS the answer. however, the M used in the equation only has TWO sigfigs... so the answer really should be 13 ml (but this wouldn't be accurate would it...) what is the rule?
As you can see this problem was given on a couple of finals and you can see for a problem like this sig figs were ignored, but to be on the safe side, you always want to have the correct number of sig. figs. So 12.5 has 3 sig fig, and you need two sig figs => 12 (since it ends in 5, read the sig fig rules, especially for numbers end in 5 to see when round up and to round down). Maryam
What is the bond angle for bent/angular shape? Is it less than 109.5 degrees or 120 degrees? I keep getting confused.
Depending on the electron arrangement, if the electron arrangement is tetrahedral then the bent shape will have a bond angle less than 109.5, example H20 (AX2E2). If the electron arrangement is trigonal planar then the bent shape will have a bond angle less than 120, example SO2 (AX2E). Maryam
You said the the H-F being a weaker acid than H-Cl rule pertains to acids in the same group... how about acids in different groups... What would be the stronger acid in this case H2S and HF and for H2O and HF ?
When you are looking across a row, for example between H2O and HF (the second row), then HF is a stronger acid compared to H2O. In this case bond polarities dominate. So the bond between the more electronegative atom and hydrogen is more polar and the acid is stronger. HF is a stronger acid than H2S. Maryam
What is in general the best strategy when asked to determine which acid/base is the strongest one out of a group of acids/bases, when the values of pKa/pKb are not given (just formulas)? Thank You.
You need to look at their molecular structure, and compare them together. Look at the factors that make an acid strong: a weak bond between the acidic H and the atom that is conected to, a stable conjugate base. These two factors show a stong acid. Read 10.9 and 10.10 for a complete description of acid strength. For bases look at the factors that make a base strong: having available lone pairs, with high electron density on them. Maryam
I apologize if this has already been covered, but I have a question about acid strengths. The explanation for 10.35 states that HCl is a stronger acid than HF because it has a weaker bond (even though F is more electronegative than Cl). But, in the third part of the problem it says that HClO2 is stronger than HBrO2 because Cl is more electronegative than Br. Does this rule regarding electronegative only pertain to non-binary acids?
When you are comparing the acidity strengh of binary acids within the same group, like HCl and HF (group 7 for this example) the bond strengh is the dominate factor, not the electronegativity, and the longer the bond the weaker the bond and the stronger the acid. For oxyoacids when you have different central atoms with the same number of oxygens, the electronegativity is the factor that dictates the acid strength, so the acid with the more electronegative central atom is the stronger one. Read 10.9 and 10.10 for a complete description of different situations. Maryam
Why is the manner of solving the initial pH for problem 11.29(a) and 11.31(a) different; While in 11.29(a) only the -log of the molarity of the analyte was applied to obtain the initial pOH (thus, pH), 11.31(a) required to apply the ice box method in order to obtain its initial pH. Both problems seem to be asking the same questio but how they apporoach them is different. Do these problems have something to do with the analyte and titrant; are the methods different because the analyte for 11.29 is a base while the analyte for 11.31 is an acid (or perhaps conjugate acid)?
The reason is that problem 11.29 is dealing with a strong base (NaOH), so no need for an icebox, since it is 100% dissociated in water to Na+ and OH-. So the concentration of OH- is 0.110 M. But for 11.31, CH3COOH is a weak acid, so you need to use an icebox to solve this problem. When dealing with a weak acid or a weak base you need to use an icebox. Maryam
On the Fall 2002 Final, question 5a, the answer says (ð2p*)², shouldnt it be (ð2p*)1 because it says O2+?
Yes it should be (pi2p*)1, what is written is for O2 not for O2+. Maryam
In my calculations, i tend to keep the answers in my calculator and then round off at the end according to the sig figs required. However, in both the solution manual's working out as well as the practice exams, it seems like they don't keep the values in their calculator but just round off according to the correct sig figs at every step. My answers always slightly vary.. for example for the value of the pka of ka= 1.8 10^-5 i get 4.74 while the book gets 4.75? Why is that.. and what way should i do it?
What you are doing is correct. Always round at the end. Solution manual is not careful with sig figs. And those questions on the practice exams you are reffering to are from the homework problems, and the answer has been just coppied from the manual. Maryam
1) Are we supposed to know everything (up to chapter 11.7 or problem #11.109) for the final? 2) When calculating pH for buffer solutions, can we always estimate our answers instead of using the quadratic formula like in the answers to 11.9?
1) If what you mean is that the final is cumulative, your answer is yes, but if you mean that if there were some parts of the book that were omitted then the anwer is also yes, check the weekly outlines (under Handouts) on VOH for a detailed list of what parts of the chapters you need to read and what homework problems are assinged. 2) When calculating pH for buffers you usually can estimate your answer instead of using the quadratic equation (when you are using an icebox, remember you could use H-H equation too when dealing with buffers). The reson is that since a buffer is either a weak acid and its conj. base or a weak base and its conj. acid, and Ka or Kb are small numbers, smaller than 10^-3, so change (x) will be small compared to the initial concentration. Maryam
I had a quesiton about sig figs when using henderson hasselbach equation- do the same rules for pH apply to it? Also, I had a question about chapter 11 #33 part b , its a weak base, NH3, strong acid-HCl, titration. How should I go about this problem?? Also, sometimes when dealing with very small numbers and very negative exponents, my calculator gives the answer like this: .000000008 -so I'm not able to see the rest of the digits and it affects my answer- is there any way to fix this? thanks
Yes, when using H-H equation rules of pH apply, after all although it is not shown in the equation, but they are taking the log of [H3O+] (to get pH) in derivation of the equation. For 11.33 part b, you should first calculate # of moles of base (NH3) that you have, then calculate # of moles of acid (HCl) that is being added, then subtract them to see how may moles of NH3 are remained after adding HCl. Since you are titrating a weak base with a strong acid, all the acid added will react with the same number of moles of NH3. As of your calculator, you could just multiple your answer by 100 or 1000 or 10000 or ... to be able to see the last digits, but don't forget to divide it back by the same number! Maryam
For number 19, if I didn't use the Henderson-Haselbach equation and followed the procedure in example 11.2 on page 411. Is the reason why we don't use icebox for the equation because we already found out the final concentrations?
You could either use the H-H equation or use an icebox to solve this problem. Remember you can use H-H equation when dealing with buffers and it is an approximation. But it will make your calculation faster and easier. And also when using H-H equation you don't need the equilibrium concentration, you will be dealing with initial concentrations. Initial concentration of CH3COOH and CH3COO- after adding 10.0 mL NaOH are 5*10^-3 and 1.77*10^-1 mol/L. Maryam
in the winter 2004 final exam, for the titration problem in Q9 and Q10, they round extremely early in problems (so the answer is off significantly...) ie they change .00225 mol NH3 to .0023 mol NH3 in the problem where we are to find the pH after addition of 15.0 ml of 0.10 M HCL... we aren't supposed to round until the end correct? (giving an answer of 8.95 for the pH instead of 9.01...right?)
Yes, you are not supposed to round until the end. For grading this question this fact was considered. The answer posted is just taken from the solution manual (they are not always careful with sigfigs). Maryam
Fall 2004, Midterm 1 , Why does MgS have the most covelent character?
Because among the rest MgS has the smallest difference between the electronegativity of its atoms. The closer the electronegativities the more covlent character it has. You don't need to memorize the electronegativities, but you need to know the trends. Maryam
Hi, for problem 61 in chapter 10, i don't understand why the solution manual drop the na; instead of nach3co2 + h20, it becomes ch3c02
Because Na+ is a neutral cation, as with all the group one cations and group two heavy cations (Ca2+) and after. They won't react with water. So you could just drop it, it is an spectator ion. The same goes with anions of strong acids, like Cl-, they are neutral too and won't react with water, so you can just drop them. Maryam
On page 420 of the text, it says "At the halfway point of teh titration, [HA] = [A-] ..." Why is this so? If anything, I thought this is what the stoichiometric point was supposed to be. And in general, what are the properties of this halfway point/what is its definition? Ok, thanks a LOT! This is my last question (maybe :) ).
Don't mix the two different base with each other. When we say at the stoic. pt. # of moles of acid is equal to the # of moles of base, what we mean is that # of moles of base/acid added to the flask from the burett is equal to the # of moles of the acid/base that was ininitially in the flask, so all of the acid/base in the flask have reacted with the added base/acid. But at half-way to the stoic. [HA] = [A-], this is the case when you are titrating a weak acid (HA) with a strong base like NaOH, and at half way concentration of the weak acid is equal to its congutate base (not NaOH, the base that you are adding). That is why at the half way pH = Pka and the volume of the base, in this case, that you need to add to get to the half way is half of what you need to get to the stoic. pt. Maryam
I'm confused with the concept of the stoichiometric point. In the notes, it says the stoich. pt. is when the moles of acid equal the moles of base. But why does this matter? I thought, the stoich. pt. was where [H30] = [OH], where the solution is nuetralized (except for the case of a salt with a weak acid/base anion/cation). And I can't see the moles of acid = moles of base meaning the same thing as [H30] = [OH].
Stoich. pt. it when [H30+] = [OH-] only when you are titrating a strong acid/base with a strong base/acid. But in general at the stoic.pt. # of moles of acid/base originally in the flask = # moles of base/acid added, so this definition also considers when you are titrating a weak acid/base with a strong base/acid. Because in this case # of moles of H3O+ and OH- are not the same. That is why pH is 7 at the stoic. pt. of titration of a strong acid and strong base, but this is not the case when you have a weak base or acid. Maryam
Hello, I don't understand how a buffer can be prepared. For example, in this equilibrium: NH3 (aq) + H20 <--> NH4 + OH How can you get equal amounts of NH3 and HH4? When you add the salt containing NH4, the equilibrium should just shift to the left to accommodate the change. Besides this, since NH3 is a weak base, NH4 should be a strong acid, so it really would not want to stay in its NH4 form. Help!
You are correct that after adding salt of a weak base to a weak base solution, the equilibrium will shift to the left to relieve the change and so if we add 0.1 M of NH4Cl to a 0.1 M NH3 solution, after equilibrium the concentrations won't be the same, but the ratio of NH3 to NH4 relatively stays the same (this is an approximation, as it is an approximation with H-H equation that assume that initial concentrations of A- and HA to be the same as equilibrium concetrations). In order to convince your self set up this problem: first calulate to see what the equilibrium concentration of NH3 and NH4+ are in a 0.1 M solution of NH3 (using an icebox). Then set up a new problem by adding 0.1 M of NH4+ to a 0.1 M of NH3,and calculate the equilibrium concentrations of NH3 and NH4+ you will see that the change(x) in comparison with 0.1 is negligible, because NH3 is a weak base and has a small Kb. So at the end you will see the for the sencond case (buffer) the equlibrium concentrations of NH3 and NH4+ stay approximatly unchanged: 0.1 M. For the second part of your question, you need to consider the equilibrium as a whole, not only look at NH4+, since NH3 is a weak base there isn't much OH- in the solution, so NH4+ will stay in this form. Just set up the equilibrium table and you will see. Maryam
for fall 2000 final, question 10, shouldn't the answer have two decimal places after the decimal, since the number of sf is two?
Yes, it should. Maryam
On Summer 2005 2nd exam Q5(d), why at half stoichiometric point pOH=pKb ? How would this be different if we titrated a weak acid with a strong base? Thank You.
See below, I have answered two similar questions, read them both. Maryam
On page 409 of the text, there is a line at the top of the page that says "Because the conjugate base of a weak acid is a weak base (specifically, H20 is a weaker base than the CH3CO2 ion), we can predict ... blah blah blah. My question is that I thought if you have a weak acid, your conjugate base would be a strong base? I mean if the ratio of weak acid to base is about 100:1, it sure seems like the conjugate base is pretty strong.
What you are saying is relative, when you have a weak acid, its conjugate base is stronger, compared to the conjugate base of a stronger acid. But a conjugate base of a weak acid is still weak compared to a real strong base like NaOH. For example CH3COO- is a lot weaker than NaOH. Maryam
Hi, For number 45 in chapter 10 hw, how do i know when to assume that x is so small that x^2/.29-x is equal to x^2/.29 or when to consider x? Because is the solution manual for part a, they assume that the denominator x is equal to 0, but for part b they didn't?
When K is smaller than 10^-3 you can use approximation, for larger K values you cann't approximate. When is doubt, use the 5% rule. Maryam
On question 5b from chapter 11, I get the texttbook's answer "ph=1.22" after using the Henderson-Hasselbalch equation. But, after carefully using an icebox and making an equilibrium question out of it, I get a much different answer of "pH=1.40." I believe that this is because the H-H equation is only an approximation. Would be marked off points if we wrote the H-H answer of 1.22 instead of the more accurate 1.40?
You are right H-H equation is an approximation. On exam we will accept both methods for calculating pH of a buffer (either using H-H equation or using equilbrium concentraion by using an icebox. Maryam
In the winter 2000 exam, on the last problem, it says that the pH is 9.2. Shouldn't it have one more significant figure because you need 2 sf's and for pH you count the sf's after the decimal place, so a pH of 9.2 only has one sf?
You are right. Maryam
I was doing # 1d on the Fall 2004 final. Is my math really fuzzy, or should [mol N2] be on the power of 10^-3?
You are correct. Maryam
Why does S2- have a larger radius than K+ ?
Because S2- has 18 electrons and 16 protons, but K+ has 18 electrons and 19 protons. So the nucleus of K+ has a stronger attraction over the electrons compared to S2-, making S2- larger. Maryam
In winter 2000 final, number 7c, why is the answer SO2?
Because the rest in water form OH-, they are basic, but SO2 is an oxide of a nonmetal which when disolved in H2O will form an acidic solution. SO2 + 2H2O => HSO3- + H3O+ Maryam
can you guide me through the actual steps showing why at 1/2 stoimetric point pH = Pka ? (problem Q8 Fall 2004 final)
Let me first clear this as I mentioned in the previous anwer, there are two situation that you might encounter when dealing with pH at half way to the stoiciometric point: 1)then you are titrating a weak base with a strong acid (Q8 Fall 2004), then at half way to the stoiciometric point pOH = pKb, so pH = 14 - pKb or you can say pH = pKa, but since you are dealing with a weak base then you need to calculate the pKa from: pKa = pKw/pKb. 2) then you are titration a weak acid with a strong base, then at half way to the stoiciometric point pH = pKa. How to get to this relation? Since at the half way to the stoichiometric point solution is acting as a buffer, you can use Henderson-Hasselbalch equation. And at this point [A-] = [HA], so log [A-]/[HA] will be zero and pH = pKa. Maryam
On Summer 2005 Second Midterm Q5 (c) , shouldn't there be two sigfigs in the answer?
Yes it should be. Maryam
For 11.25, how do we approach part b?
Oh, I see, they forgot to type parb b in the solution manual. That 3.82, pH that they are giving is the pH at half way to the stoichiometric point, so it is equal to pKa of the weak acid. This is b/c 26.0 mL is half of 52.0 mL, so it is the half way. Maryam
For the fall 2004 final, number 8B, I do not understand how they got the pH. I understand why they half the volume to get 12.5mL, but what do you do from there?
At half way to the stoichiometric point for titration of a weak base with a strong acid the pKb is equal to pOH, so pH = 14 - pKb (for 8B), and for titration of a weak acid with a strong acid the pKa is equal to pH. Maryam
Will we be responsible for isomers? Or for naming wierd molecules on the final?
Hi David, Nope.
Professor Lavelle -- In review on friday from 4-5 you said that we can tell something is a weak acid if its ka value is less than 10^-3, but what should the kb value need to be less than/greater than inorder for it to be a weak BASE.
Hi Paige, For a weak acid Ka is less than 10^-3. For a weak base Kb is less than 10^-3. Remember these are rough guidelines.
Is "100 W" counted as one ssig fig or 3?
If it is written as 100 is one but if it is written as 100. then it is 3. Maryam
Hi, is this final cumulative??
Yes, it is. Maryam
In the fall 2004 final, number 2D, dowe need to know how to write formulas like that? We never went over it in class and I have never heard of "diaquabisoxalato ironIV". thanks.
Transition metal complexes are now moved to 14b. Maryam
I have noticed that several of the past finals have asked questions about the coordination and oxidation numbers of molecules. I don't remember covering these this quarter. Will we be responsible for knowing them anyway, and if so, what are they or where are they found in our book? Thank you so much!
The syllabus has changed for chem 14a, you will study transition metal complexes in 14B. Maryam
When finding the pH given the molarity of a weak acid I get a pH that is slightly off from the book because I don't round my answer until the very end. In Exampl1 10.7, for example, the book rounds the concentration of H3O+ to 1.3 x 10^-3 and takes the -log of this value and gets a pH of 2.89. I took the -log of the unrounded number that was still in my calculator to get a pH of 2.87. should I be rounded the H3O+ concentration first?
You should round at the end. Solution manual it not always very careful with SFs. Maryam
Are we going to get the same equations and constants for the final as we did in the quizzes and midterm?
Take a look at the old finals on VOH, you will be getting a similar cover page. Maryam
Hi! According to the lecture notes (pg 5, set #8 & 9), for a neutralization reaction, "the product is always a salt + water". However, the book states that a neutralization reaction always produces a salt, but only results in water when the base is strong (pg F72). Which of these is true? Also, in the net ionic equation for part (b) of question J.3, why is water included as one of the products? Aren't we dealing with a weak base here? Thanks!
Hi, Since a strong base like NaOH has an hydroxide group then that and a proton from the acid form water. For a weak base like NH3 since it doesn't have a OH- then the product of that and an acid like HCl will be NH4Cl (salt). But remember that these are all aqueous solutions so water is there and NH3 in water forms NH4OH, so depending on if you are writing NH3 then no water is formed, but if you are writing NH4OH, then water is formed. Also in equilibrium expressions you don't write solvents (water here). Maryam
For 11.13, how do we know the pkas for parts c and d?
pKas for polyprotic acids are listed in table 10.9 p391. Maryam
Regarding buffers, the books states that "the conjugate base of a weak acid is a weak base" pg 409. I thought that a weak acid produced a strong conj. base. So, if you add a strong base to acetic acid and sodium acetate, "the strong base has been replaced by a weak base", which is the acetate ion. But, I thought the acetate ion was a strong base, so will you explain how the conc. of OH- ions remains nearly unchanged.
What you are saying that a weak acid has a strong conj. base is relative. So if you are comparing two acids the weaker acid has a stronger conj. base compared to the other acid. What the book is saying is in general that a weak acid and its conj base (which is also a week base) form a buffer. Acetate ion is a weak base overall, NaOH, for example is a strong base. So when the book says a strong base has been replaced by a weak base, means that NaOH(which is OH-) has been replaced by acetate ion. Maryam
Will we ever do titrations with weak acids or bases as titrant?
What we covered is titration of a strong acid and a strong base, a weak acid and a strong base, a strong acid and a weak base. We didn't do any weak acid and weak base titrations. Maryam
If we know the pKa for a titration, is it true to say that we can only use the Henderson-H equation for a titration to find pH at all times except the beginning, the middle (stiochiometry point), and the end (these are the places where [A-]/[HA] is undefined or 0)? At all other times we must use an icebox, correct? Also, it is true that we can always use an icebox to find pH?
You can always use an icebox to find pH (or pOH), so you don't even need the Henderson-Hasselbalch equation. But since Henderson-Hasselbalch equation is an approximation it will make it easier to do the calculations, but keep in mind that you can only use it when you have a buffer. That is why when dealing with titration problems you can use Henderson-Hasselbalch equation only when you have a buffer at that point of titration. So at the begining there is no buffer, but once you add some titrant then it will be a buffer solution, and at stoichimetric point it is not a buffer, and after the stoichiomteric point is not a buffer anymore. Maryam
Do we use the Henderson-Hasselbalch equation with A- and HA concentrations at equilibrium for titration questions, or just their concentrations after we calculate the effect of the titrant?
If you could either use an ice box and calculate the equlibrium concentrations according to that or use the Henderson-Hasselbalch equation and just use the concentrations that you get after the effect of the titrant. Henderson-Hasselbalch equation is an approximation, Maryam
Question 10.61: Part C asks to calculate the pH of 0.055 M AlCl3 (aq). The olutions manual goes about this problem the usual way, but I don't understand the reaction they used to solve the problem. Where did the (H20)6 3+ come from? Similarly, I have the same problem understanding the (H20)6 3+ in 10.59 E + F.
AlCl3 in water will be Al3+ and 3Cl-, we know that Cl- is a neutral anion and wont do anything to water. But Al3+ (and aslo other small highly charged cations in water forms a octahedral complex, Al(H2O)6 3+ which will react with water to give hydronium ion and [Al(H2O)5OH]2+. That is why is acidic. This the case for other small highly charged cations take a look at table 10.7. Maryam
Question 10.55: Why do we multiply 6.7% times .10? Furthermore, in this problem we are supposed put 0.100 - 0.0067 in the equlibrium row of octylamine. Why don't we do the same thing in question number 10.53, and put .110 - .00264 in the equilibrium row of benzoic acid?
We multiply 6.7% by .10, because .10 was the concentration of octylamine initially, and it is saying that 6.7% of it will be protonated. So if you look at the ice box that they have in the solution manual, .067*.10= .0067 represents the change. The concept is the same for 10.53, except here 2.4% will be deprotonated and the initial concentration is .110, so the change in their ice box will be .024*0.110. Maryam
What is the correct way of writing this salt: NH3NH2+ or NH2NH3+?
If you write is as NH2NH3+ is better. But it is not a big deal, both are acceptable. Maryam
How do we know the sig figs for problems with -log's and pH's?
For a review on sig figs go to the chem14a voh website under everything you whant to know about SF. Anyhow, the mumber of digits following the decimal point in a pH value is what you count as the number of sigfigs. For example if pH is 14.1, this corresponds to one sig fig not 3. Maryam
Hello, what exactly is happening at the molecular level when a strong acid or a strong base is added to a buffer? For example in 11.19 part a, OH is added to a buffer solution. The OH is reacting with the hydroniums to form water. Therefore, the reaction would shift in the direction of creating more hydroniums. However, for each mol of hydronium that reacts with hydroxide, it would take more than one mol of weak acid to dissociate in order to compensate for the loss. How can we modify the initial concentrations of acid and conj. base before entering the values into the henderson-hassalbalch? Just because we add x mols of hydroxide, does it mean that the same number of mols of weak acid will dissociate as the answer suggests?
In 11.19 part a, when OH- is added to the buffer, it will react mostly with CH3COOH (acid) not hydronium (because CH3COOH is a weak acid and there are not much hydroniums in the buffer, compare to CH3COOH). So OH- reacts with CH3COOH and forms CH3COO- And in part b, upon adding a strong acid the resultiong hydronium ions will react with CH3COO- and will form CH3COOH. I hope this helps to show that what they do in the solution manual makes sence. Maryam
Dr. Lavelle's lecture notes explains the stoichiometry point as the point where the "#moles acid (base) added = #moles base (acid) sample," but the book defines it as the point where "the amount of OH- (or H3O+) added as titrant is equal to the amount of H3O+ (or OH-) initially present in the analyte." I believe that these two are different. If you have x moles of a weak acid, you will initially have less than x moles [H3O+], but to reach the stoichiometry point, you will need x moles of strong base. This contradicts the books definition that amount of OH- or H3O+ ion equals the amount of initial hydronium or hydroxide ion because in solution there are less initial moles OH- or H3O+ than the moles of original substance, x. Is the book not being clear in its definition, or is the definition somehow correct?
Book is not so carefull about the definition at the beginging, when is mentioning amount of H3O+ and OH-. This is true for the case of strong acid and strong base, but as you see later on when they are doing calutions regarding for example a weak acid and strong base then they consider that stoichiometric point is the point where the "#moles acid (base) added = #moles base (acid) sample. Maryam
Will the chart of weak acid/base, like the one in Ch 10, pg 377 be given on the final exam? and is there still going to be a problem from the homework on the final? thanks!
The whole table might not be given, but for sure the needed values Ka or Kb values to solve the problems will be given. You don't need to memorize and constants. Yes there will be one homework problem on the final. Maryam
Question 10.35: In part A, we are told that HCL is a stronger acid than HF because the HCL bond strength is much weaker than the bond in HF. However, when doing the Lewis Structure, the bonds are the same; next I looked at electronegativity, and F is more electronegative than Cl, so shouldn't HF be the stronger acid? If it isn't, why not?
Lewis structures don't give you bond lengths. The bond between H and F is short and strong, that is because there is good orbital overlap between H and F. This good overlap has to do with the size of H and F. They are small so good overlap and a short bond. In case of HCl, Cl is large, so there is no good orbital overlap and the resulting bond is longer and weaker. The weaker bond results in the stronger the acid, becuase they dissociate easier in water to hydronium and their correspoinding anion. Maryam
How is it that F- has an impact on pH, but salts of strong acids (Br-, I-, Cl-) don't? What are the molecular forces/interactions that cause this?
F- is the conjugate base of weak acid, so it has basic property. That is because it reacts with water to form HF and OH-. So since upon reaction with water it formed OH- it is a basic. As of Cl-,Br-, and I- they wont react with water, so they don't change the pH and they are neutral. Because they are anions of strong acids that are 100% dissociated in water, so these anions don't have any basic properties, they are neutral. Maryam
Is NaOH considered a salt? If so, why do we usually write the dissociation of salts in water (e.g. instead of NaCl, we would write reactants as Na+ and Cl-), but for NaOH, we generally leave NaOH without showing its dissociation. What are the generaly rules for knowing when to show dissociations for acidic/basic salts, and general salts? Thanks!
NaOH is a strong base not a salt. In water it 100% dissociates to Na+ and OH-. Maryam
For 10.67: In the first step of a, when calculating .020 moles/liter X .150 L divided by .500 L, i keep getting .006, but the solutions manual is saying .0400 moles/liter. Am I doing something wrong?
You are correct, it should be .006, the final answer correction is on the listed errors for the book. Maryam
For 10.73: How are we supposed to know that we have .15 M of H30 when we are setting up the icebox?
Since we didn't go over this in lecture you are not expected to know how to set up the icebox for this problem. But you can see that it is not that difficult. The reason that you have .15 M H30+ initially is that the first deprotonation is 100%. But anyhow, since we didn't cover this, if a question is asking about pH of a sulfuric acid solution, you could assume that it is 100% for both deprotonations. Maryam
for question 10.65, is there only two sigfigs for both a and b because the Ka value on the tables two sigfigs?
That is correct. Maryam
on the announcement section, it says that the final exam is on wednesday at 8:00 - 11:00... thats only for the 1st lecture right? for lecture two, is it at the same place but on thursday 11:30 - 2:30???
Yes, that is correct. This cofusion is because some of you are looking under the announcement for the wrong lecture. Maryam
Do we have to know how to problem 10.69 part B or problem 10.101? Finding the pH of an amphiprotic salt is covered in 10.15, and we were only required to read up to 10.14. Thank you.
Omit 10.69. Yes you are not required to know how to calculate the pH of an amphiprotic salt. As of 10.101 it is not a salt it is an acid. Maryam
I had a question regarding 10.69b. In determining the ph value of CH3CH(NH2)C02Na, when there are two Ka values given, Ka1 and Ka2, I was under the impression to always use the Ka1 value, but the book uses both. Which is correct?
Omit 10.69. These two Ka values, one refers to the carboxylic acid group and the other to ammonium group, so they are different than Ka1 and Ka2 that you see for polyprotic acids. Maryam
Hello, I have a question about # 10.69, in the solutions manual it said that ph=1/2(pKA1+pKA2) i dont recall learning that. What does that mean? And another question was #10.61b, how do you know where to look for the Ka for AlCl3, i know it is acidic, but how do you know its conjugat base? Thank you Verrrrrrry much!!!!!
This is an announcementm, Dr. Lavelle said to Omit 10.69. But anyhow, the answer PH=1/2(pKA1+pKA2) is nonsense. It is clearly giving you a wrong answer since you should be getting a basic pH not acidic (6.1!!!). For your second question I think you mean 10.61c, get the Ka from table 10.7. Also look at a previous question about 10.61.c Maryam
I'm a little confused. So on Monday the 12th there's two review sessions going on at the same time?
Yes, but the focus is on different material. Maryam
I know it wasn't assigned, but for 10.62c, how would you find the pH of 0.071M FeCl3? What would be the equation when you add water to it?
See the solution for 10.61c. It is the same concept, exept here you have Fe(H2O)6 3+, read charge 3 positive. Also read the answer to the previous question about 10.61c. Maryam
For 10.61c, how did the solutions manual come up with the chemical equation for the reaction of AlCl3
For AlCl3 in water you have Al3+, which will form an octahedral complex, Al(H2O)6 3+, this is supposed to read charge 3 positive. Then it will lose a proton to a water molecute and forms H3O+ and Al(H2O)5OH2+, this is supposed to read charge 2 pasitve. This is why it is acidic. A similar reaction happens for other small highly charged metal cations like Fe3+, Cr3+, ... (table 10.7). Maryam
this is question 10.62 and even though we dont have to do it, I have a question... it asks for the pH of .13M of Na2SO3. its going to be SO3^(2-) + H2O = HSO3^(-) + OH^(-)... how do you go from there b/c Table 10.1 only gives you the Ka of H2SO3
Your question is for 10.62 b. Table 10.9 gives that information. It is Ka2 of H2SO4. Maryam
I have a theoretical question about buffers. I don't understand what the problem is with just dissolving a weak acid or base and using that as a buffer. You would then have a little bit of H3O+ and A-. These could then react with any incoming acids or bases, and when used up, they would be regenerated to maintain equilibrium. Why then do we have to add the conjugate base? It doesn't affect pH does it? All it seems to do is put a ton of A- in solution, leaving you with lots of A- and AH, and still a small amount of H3O+. What am I missing?
Now that you have seen titration curves, it is easier to see why. That is correct that a weak acid or weak base will behave as buffers, as you have seen on the titration curves (buffering region). But optimum buffering capacity occures when [acid] = [conjugate base] or [base] = [conjugate acid] ie when you have a weak acid and it salf or a weak base and its salt. Buffering capacity of weak acids or bases without their salt is low especially when you are adding an acid to solution of a weak acid or a base to a solution of a weak base. Since they are weak acids or bases, the concentration of their conjugate base or acid is low. Maryam
are isomers goin to be on the final?
Transition metal complexes and their isomers will be discussed in chem14B. Maryam
are the sigfigs for number 10.21 a, d, e, and f right? I tried looking it up in textbook errors but it was acting odd...
For f it is correct, but for a, d, and e it is missing a digit. Maryam
how would you know if one is a strong acid/base or a weak acid/base if they dont give you the Ka or Kb value. if they do give it to you, what value of pka or pkb would be considered strong? for example...
Remember the strong acids and bases 100% dissociate in water, so Ka and Kb don’t have any meaning for them. You need to know which acid and bases are strong. For the rest, they are weak acids and bases, but in comparison some are stronger than the others. So you will be comparing acidity and basisity strengths. For this you need to be able to compare the acidity and basisity of the examples that we have covered (in the lecture and homework problems) by just looking at the structure of those acids and bases. For more difficult cases their corresponding pKa and pKbs will be given. Maryam
Hi Dr. Lavelle, I was just wondering what the average was for quiz #4. Thank you.
We don't have the class average yet. Maryam
hi! for the equilibrium equation: N2(g) + 3H2(g) <--> 2NH3(g) why does increasing the pressure decrease [H2] ? why does increasing the temberature increase [H2]?
Hi, On the right hand side of the equilibrium you have 4 moles of gas and on the left hand you have 2 moles of gas, so increase in pressure (or decrease in volume) will shift the equilibrium to the side with less number of moles of gas, to the right. By doing this the applied change will be relived (Le Chatelier’s Principle), since less number of moles of gasses occupy a smaller volume. So more NH3 will be formed and H2 and N2 will be used to form NH3. Remember change in pressure only affects gasses, so you need to only count the number of moles of gasses on each side of the equilibrium. As for increase in temperature more information about the reaction should be given, exo- or endothermicity of the reaction, or they can give K values for different temperatures so you can compare them and see the reaction is exo- or endothermic. Since this reaction is exothermic increase in temperature will shift it to the side that will relieve this change, to the left. [H2] and [N2] increase and [NH3] decrease. Maryam
for 10.51b, if you ignore the fact that the pH of NH2NH2 is 10.20, how do you know that it is a base and not an acid?
Usually when you see derivatives of N they are basic, for example ammonia, hydrazine and amines. That is because of the lone pair on N, which can react with water and produce OH- and the protonated form of that N derivative. Maryam
for question 37b, it says that formic acid is stronger than acetic acid b/c -ch3 has electron-donating properties... how would you know that it has electron-donating properties?
You didn't know that before but now you know it. You will be learning about electron-withdrawing and electron-donation groups when you study organic chemistry. Maryam
I was just wondering if you could tell us which one of the review sessions is going to be put on by Sanaz? Thanks.
Monday, December 12th from 7-9pm in CS 50 is by Sanaz and Wednesday (Dec. 7), 7-10pm is joint by Sanaz and Tom. Maryam
for 10.29, it asks for increasing base strength. Wouldn't that mean that pk(b) go towards a pH of 14 to be considered very strong? Why is it then that F- is considered the weakest and NH3 considered the strongest, when the pkb is 10.55 and 4.75 respectively? Shouldn't it be the other way around? I dont think i really get the pkb and pka...
The larger the Kb the stronger the base. Write the equilibrium equation for a base to convince yourself that why a larger Kb means a stronger base. That is that the reation of a base with water gives more product (OH- and the conjugate acid of the base) then Kb is bigger and the base is stronger. Since pKb is just -log of Kb, then the smaller the Kb, the stronger the base. Maryam
I not sure whether this was mentioned in class or not but I was just wondering what day the final is. On VOH announcements it says it will be Wednesday Dec. 14th at 8am. However on myUCLA it says it will be on Thursday Dec. 15th at 11:30am. Please clarify the final date. Thanks
Sections 1 and 2 have final exams scheduled at different times. 14A-1 final exam: 8-11am, Wednesday, December 14 14A-2 final exam: 11:30am-2:30pm, Thursday, December 15.
On the homework for chapter 10 number 53, I get the same answer as the one in the solutions manual. However, Dr. Lavelle has written that the solution manual is wrong but I cannot seem to get the answers he did on his "corrected answers" page. Could you tell me how to do it correctly?
The typo sheet is being updated. On 10.53 the Ka=6.5x10-5 and pH=2.58.
I believe that the correction made for 10.61d on Professor Lavelle's list of corrections may be a typo. I repeatedly get the answer 11.56 as pH instead of 11.056
The typo sheet is being updated. On question 10.61d the answer is pH=11.56
Hi! I have a question on pg.390 of Example 10.11. If there are 2 molecules of CH3 CO2- from Ca (CH3 CO2-)2, why isn't the reaction given there not balanced to reflect the 2 molcules of CH3 CO2-? And since you have to multiply the initial molar concentration by 2, shouldn't you do that with the CH3 COOH (making the change in molarity +2x) as well after the reaction is balanced? Please help!
The fact that each mole of Ca(CH3COO)2 will give us 2 moles of CH3COO- is only useful when calculating the initial concentration of CH3COO-. After that, we can use the balanced acid/base equation given to us in the example. I suppose you could use 2x throughout the problem but you would have to remember that your concetration of [-OH] is now going to be 2x.
Hello, I have a question regarding problem J.3 part B, with the ionic equations. Could you explain why HNO3 is broken up into H3O+ and NO3- instead of H+ and NO3-, and why there is water on the other end of the equation? Thanks.
These acid/base problems are all aqueous solutions. You can think of H+ and H3O+ as being the same thing, except that it is not reasonable to think of a H+ as being dissolved in water. The oxygens on the water are going to point toward the positive H+. The interaction is so strong that we consider a new molecule to have formed, H3O+. The water is included in the reaction because hydronium was used instead of H+.
How do significant figures apply to ph values? In other words, how do you determine the number of sig figs for a ph value= for example if you are given a concentration of 0.28. Thanks!
There is a special rule to remember with pH. The whole number portion does not count when figuring out sig figs. For example, if you calculate a pH=4.35467 and you needed two sig figs - you final answer would be 4.35
1) Is it possible for a conjugate base to have a positive charge? 2) How do I know that C6H5NH3 or NH2NH3 is an acid? I read section J of the Review Chapter on figuring out what is an acid based on looking at the molecular formula, but I don't recall this section explaining how I would be able to determine the aformentioned molecular formula is an acid. 3) Question 10.9d: Is't Bi2O3 Amphoteric because it is close to a metalloid? The answer doesn't say so. 4)Is Lewis Acid the same as conjugate acid?
1) It it is possible for a conjugate base to have a positive charge. However, with Bronsted acids, we usually expect the conjugate bases to be negatively charged. Check out the equilibrium equations in the solution to problem 10.59 and identify the conjugate bases. 2)In each of those molecules a nitrogen that has 4 bonds to it. That N will have a +1 formal charge. It would be very favorable for that nitrogen containing group to react with an H2O and give away an H+. After the reaction the N will only have 3 bonds and a lone pair of electrons and its formal charge will now be zero. All compounds with a positively charged N-H will act like an acid. 3)See Below 4)A Lewis acid is NOT the same as a conjugate acid. A Lewis acid is an electron pair acceptor. A conjugate acid is what is formed when a base reacts reacts with a proton.
I believe that an error has been made in the solutions manual that has not yet been added to Prof. Lavelle's list of corrections. In question 101b, the volume is incorrectly calculated. Volume is calculated by surface area x height. The surface area is 2.6 km^2, which converts to 2.6 km^2 (1000 m/km)^2= 2.6 x 10^6 m. The height is 2.0 cm which converts to 2.0 cm (1 m/100 cm)= .02 m. Surface area x height = 2.6 x 10^6 m^2 x .02 m= 5.2x 10^4 m^3. The solutions manual reports 5.2 x 10^7 m^3 which leads to an incorrect answer. The final answer, i believe, is a pH of 2.03.
The solutions manual reports a volume of 5.2x10^7 L. And when you convert 5.2x10^4 m^3 you will get a volume of 5.2x10^7 L.
Hi. Dr. Lavelle's "3rd Edition Corrections" lists 10.53 as incorrect as done in the solutions manual. I don't see what is wrong with the answers they give, and the only error I believe they made was that they improperly rounded. I get 2.58 as pH and 6.5 X 10-5 as Ka. Thank you!
Here are the numbers I came up with: Ka=6.5x10^-5 pKa=4.19 pH=2.58 I think you are correct.
Why is the answer for 10.9d listed as "basic" in the back of the book when figure 10.5 pg 369 lists Bi2O3 as an amphoteric oxide? Which answer is correct?
Bi2O3 is on the border between metals and non-metals and so you would expect it to be amphoteric. Why then is the answer for 10.9d basic? Figure 10.5 is correct and so I would consider Bi2O3 to be amphoteric.
With respect to equation (14)* pg 392, how is it possible for pH to be independent of molarity or concentration? Doesn't pH always depend on the amount of substrate available? How is a solution of an amphiprotic salt an exception to this trend? Thank you for satiating my curiosity :)
I think you will better understand once we talk about buffers. It's coming next.
Are we expected to be familiar with equation (14)* on page 392 although this concept was not covered in class?
I didn't see any HW problems that require you to use equation 14. In class, and in the notes, Dr. Lavelle said that when calculating the pH of a polyprotic acid, condiser only Ka1 because the other ioinization constants are much lower. Maybe this will help.
For 10.9 d, why is Bi2O3 basic and not amphoteric when it lies near the diagnoal frontier between metals and nonmetals?
It does lie close to the diagonal between metals and non-metals so I can see why it's not clear. The inside of the front cover colors the diagonal in a way that I would classify Bi as a metal.
for problem 10.65 why do they use the pH from H30 instead of OH-?
The definition of pH is -log10[H3O+] If you use [OH-] then you would be calculating pOH.
for question 10.59 part e and f and i dont understand why they changed the AlCl3 to Al(h2o)6^3+ or part f cor Cu
both parts (e) and (f) deal with Lewis acids. Cl- and NO3- are going to be completely ionized in water (they the are conjuate bases of strong acids). The metals can then react with water - they are Lewis acids and can accept a lone pair of electrons from a water molecule.
For question 10.37, what is the significance of an atom's electronegativity when determining an acid's strength? And for question 10.53, when calculating K for the reaction, the solutions manual has the concentration for benzoic acid as (0.110 - 0.24) x (0.110).....why is that? Thanks.
See below about 10.37 The solutions manual calculates the equilibrium concentration of benzoic acid as (1 - 0.024) x 0.110. Or think about it as 97.6% of 0.100M because (1 - 0.024) = 97.6%
For problem 10.37, as well as for other similar problems, what is the significance of electronegativity when determining the the strength of an acid? And for problem 10.31, why is it that acids are strong if their conjugate bases lie above water? Thanks.
For problems like 10.37: The conjugate bases of these compounds are negatively charged. Highly electonegative atoms will stabalize the negatively charged conjugate base by distributing the electron desity throughout the molecule. The more stable a conjugate base then the stronger the acid. Addressing your other question: A conjugate base that is "above water" is a weaker base than water. Or, in other words, the corresponding acid is stronger than H3O+. This is the definition of a strong acid.
I don't understand the explanation for 10.37b. What does it mean when the solution says, "The -CH3 group has electron-donating properties"? How could we have known that in advance based on what we have learned up to this point?
It would have been hard for you to know that a CH3 group is electron-donating. But now that you've done the homework you know that a CH3 group has electron-donating properties.
What are in general the best and fastest strategies when coping with the following problems (quiz #3 - CH2CH2 and CHCH)? - the shortest carbon carbon length - a higher energy carbon carbon length - a higher dissociation energy
The first thing I would do is draw the Lewis structures. After I have the correct Lewis structures I see that CH2CH2 has a carbon-carbon double bond while CHCH has a carbon-carbon triple bond. A carbon-carbon triple bond is going to be shorter than a carbon-carbon double bond. Shorter bonds are stronger and have a higer dissociation energy (the energy it takes to break the bond). So, one fast strategy would be to draw the Lewis structures.
Hi. I don't completely understand the definition of a Lewis Acid. Could you explain it more thoroughly to me? Also, what makes BF3 a Lewis Acid as opposed to a Bronsted Acid, and what should we generally look at to determine a substance is a Lewis Acid? Thank you!
Bronsted Acids contain hydrogens that will react with water to form H3O+. Lewis acids are electron pair acceptors. For example, the boron atom in the center of BF3 only has 6 electrons (remember that boron was an exception to the octet rule). The boron atom can bond to a pair of electrons from another compound (for ex. a compound with a lone pair on a nitrogen atom) to give boron an octet. BF3 is acting as a Lewis acid in the example.
I was just wondering when the review sessions are for the final and which ta will be leading each one
Keep an eye on the VOH announcement page. Information about review sessions will be posted as the sessions are scheduled.
When we are dealing with a strong acid or base, is it still possible for that acid or base to have a conjugate base or conjugate acid? For example, even though the reaction goes to completion, is Cl- considered a conjugate base in the reaction HCl + H2O -> H3O+ + Cl- ? Thank you!
Technically, Cl- would be the conjugate base of HCl, and if you were asked on an exam to identify the conjate base for HCl the answer would be Cl-. But do realize that the conjugate base of a strong acid is going to be completely ionized in water and will not act like a base.
How do you identify something as being either a strong or weak acid/base if you are given only the chemical formula? Is this something that we are expected to know?
You are expected to know which acids are strong acids and which bases are strong bases. All of the other acids and bases would be weak. Typically, alkili metal hydroxides, (such as NaOH, KOH, etc.) and the alkili earth metal hyrdoxides (Ca(OH)2, Ba(OH)2, Sr(OH)2) are strong bases. Some examples of strong acids are the the hydrogen halides (HCl, HBr, HI), sulfric acid (H2SO4), nitric acid (HNO3), and perchloric acid (HClO4).
do you prefer molarity or partial pressures for equilibrium, or is that contingent upon the molecules being in gas phase. Thanks.
It depends on how the information is given to you in the problem and in which form you feel most comfortable. Of course, partial pressures can only be used when the problem involves gases. Matt
What questions should we go up to for this week's homework assignments?
This week's quiz covers MO theory and Chapter 9. I was looking at the problems from Chapter 10 and it looks like, after Wednsday's lecture, you should be able to do everything up to problem 10.71 (the first problem involving polyprotic acids).
This one will be probably directed to Dr. Lavelle. I might have missed an announcement in the class, but what was the average score (at least approx.) on the midterm? It is good to know where I am compared to the class before the final exam. Thank You.
See Below
Sorry, and in the example NH3 + H20 <-> NH4+ + OH- , is OH- the conjugate base?
YES
In the reaction Hbr + H20 -> H30+ + Br-, HBr is the acid, H20 is the base, and Br- is the conjugate base. Is H30 the conjugate acid?
YES
I would really appreciate it if you could clarify what the valence e- configuration for C2 would be! Thanks!
sigma2s2 simga*2s2 pi2p4
i have a question about the midterm. Where/Who can we find the mean/standard deviation of the midterm? And is the midterm scores curved? or the curve will take place at the end after all the scores are collected? thanks!
The curve will take place at the end of the quarter. The average on the midterm was a 91/120.
why is it that an acid is a proton donor, while a Lewis acid as an electron pair acceptor and vice versa with the bases?
Try thinking about it this way. An acid, which is an electron acceptor, will give up a proton. In doing so, it keeps the electrons that were used in the bond to the hyrdogen. Think of it as the hydrogen donating the electron to the proton acceptor. For a base, it's the opposite. A hydrogen will bond to a pair of electrons on the base. The base has to donate the electrons to make the bond to the hydrogen.
Hello,I just want to double check...when writing out the molecular orbital configurations, and even drawing the MO, is it still acceptable for me iclude sigma 1s and sigma 1s*? Do we only HAVE to do the valence? Thanks :)
We usually only look at the valence shell because the lower energy levels are boring (because they are completely filled), but you can include the lower energy shells if your heart desires. Matt
Question # 9.11 Part a & b. For a, I got NO* NO2/ No2O3. Why is this not the right answer? Thank you.
I'm not quite sure what you are asking, but let me try to guess. You are talking about 9.11c. When dealing with gases, the partial pressures can be used instead of concentrations. So Q = (partial pressure of NO)*(partial pressure of NO2<\sub>) / (partial pressure of N2O3). It looks like you have the right idea from what you have written, but if I didn't answer your question then try to rephrase it. Matt
Hi...I am still completely confused about writing MO diagrams! When one writes the valence e- configuration, what do the superscripts suggest, what causes a change, ie. why is pi*2p^2 for O2, and pi*2p^4 for F2? Also, from reading the lecture notes, i realize the definitions of diamagnetic and paramagnetic, but what is the way to figure the difference when doing MO diagrams. I'd really appreciate it if you could guide me step by step and make this concept of MO diagrams clear to me! Thank u so much!!!
The supersripts in electron configurations indicate the number of electrons in that orbital. For a hydrogen atom - 1s1, the 1 tells you that there is 1 electron in the 1s orbital. It is the same for molecular orbitals. For a hydrogen molecule (H2) - simga1s2 - the 2 tells you that there are two electrons in the simga orbital.
After filling out a MO diagram, if there are any unpaired electrons then the molecule is paramagnetic. If all of the electrons are paired the molecule is diamagnetic.
For step-by-step guidence in creating MO diagrams you should come see a TA during office hours. It is just too much (and too hard to type) for VOH.
In the weekly outline it says to ommit section 9.3, yet the HW questions and the rest of the chapter makes references to Q, which is described in 9.3. Should we actually review this whole section or just the part containing Q?
Q was talked about in lecture and therefore you are responsible for knowing it. Part of section 9.3 talks about Q and you can read that part of section 9.3 if you'd like, but the information about Q was covered in the lecture notes.
w/ bronsteds acids and bases, how do you know when a molecule (ex: H2SO4)with water would turn to either a hydroxide or a hydronium ion?
Metal hydroxides are bases (-OH) and non-metal hydroxides are acids (H3O+). For example, since H2SO4 contains non-metal hydroxides (Sulfur is not a metal) it is going to produce H3O+ when reacted with water.
Hi, I had a question about 9.59, when filling out the ice box, how come the initial molar concentration of CH4 is 0 whereas the the change in molar concentration is .0478. And also how come the the initial molar concentrations for CO and 3H2 are 2.00 and 3.00 instead of being .2 and .3 since it is in 10L. Thank you verrrrry much :)
Initially, no CH4 was was placed into the reaction vessel and so the initial concentration of CH4 is zero.
The ICE box in the solutions manual uses moles instead of concentrations, but if you look under the ICE box you will see that they convert the equilibrium number of moles into equilibrium concentrations before plugging into the expression for Kc. Both ways will work.
(I hope my subscripts worked.) Matt
whats the difference between the complete ionic equation for a neutralization and its net ionic equation? more importantly how do u write them? for example j.5 part b
Section I.3 explains the differences between a complete ionic equation and a net ionic equation pretty well. For a complete ionic equation ALL of the ions are shown in the equation for the reaction. However, sometimes an ion will appear on both sides of the complete equation. When that happens, a net ionic equation can be used. A net ionic equation omits anything that appears on both sides of the complete equation. You can think of it like a math equation, anything that appears on both sides can be subtracted out. Again, I.3 has a good explaination and includes an examlpe.
In Problem j.5, the desired product is Zn(NO2)2. Remeber that metal hydroxides are bases and non-metal hydroxides are acids. Therefore, in the product, Zn will come from a base and nitrite will come from an acid.
Zn(OH)2 + 2HNO2 -> Zn(NO2)2 + 2H2O (I'm working on figuring out how to do subscripts, but the solution manual has that correct equation)
When we are using the quadratic equation in order to calculate the equilibrium composition, do we always reject the negative root because it is unphysical? (when we have the b + or -, do we reject the answer the negative asnwer?)
That is right, only one of the roots should make physical sense.
Hello, I have a question concerning 3.51 c. In the solutions manual, it appears that instead of filling the 2px and 2py, it instead fills the 2px* and then distributes one electron to the 2py* and 2pz*. Why are these distributed. Is this a rule that maybe I have overlooked, or is this a problem in the solutions manual? I think the pi orbitals should be filled before the sigma in the antibonding 2p set. Thanks.
You are correct, the pi*-Px and pi*-Py orbitals would fill before the sigma*-Pz orbital. The solutions manual is incorrect. However, the solution to 3.55 does includes the correct electron configuration for F2.
Hi. In the book, the examples require you to find the partial pressures of gases with PV=nRT before solving for K. However, we didn't have to do this in any of the homework problems. Will we be required to do this for the quiz/final? Thanks.
It is possible that a question may require you to calculate the partial pressures. However, many of the problems can be solved using concetrations. (if 1 mole of gas is placed in a 1 L flask then the concetration is 1M, right) If you have a question about a specific example I can go into more detail.
Matt
For question 9.27, would the correct answer have 2 or 3 signficant figures? The textbook and solution manual indicate 2 sig figs. All the values in the question have 3 sig figs, and the value in the table is written as 160. Aren'
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