Question

Hey guys! Stuck on some pH problems! Thanks so much in advanced:

A) Calculate the pH of a 0.200 M solution of potassium acetate, KCH3CO2, in water at 25 °C. At this temperature, Ka = 1.76 × 10-5 for acetic acid.

1. 2.73 2. 5.45 3. 9.02 4. 9.94 5. 10.16 6. none of these
Maybe Answer:
Kbacetate=10^-14/1.76*10^-5=5.68*10-16
[OH-]=sqroot{5.68*10^-10*.2M)=1.060*10^-5, so pOH=4.97
ph=14-4.97=9.02, maybe answer 3

B) The pH of which salt solution will be greater than 7.00 at 25 °C? (Assume that each solution has a molar concentration of 0.20 M.)
1.NH4Cl 2. NaI 3. KNO3 4. LiHSO4 5.KHCO3
Maybe Answer:
5)KHCO3 because it is basic due to its components

C)At 25 °C, the pH of a solution of NH4Cl is 5.18. The Ka value of NH4+ is
5.65 × 10-10 at 25 °C. What is the molar concentration of the NH4Cl solution?
1. 0.012 M 4. 0.077 M
2. 0.041 M 5. 0.28 M
3. 0.059 M
Maybe Answer: 10^-5.18=[H+]=6.607*10^-6 (square this to get the top value for the Ka function because H+ and OH- are the same in this case?):
5.65*10^-10=4.365*10^-11/x2
x=.00772M=4.0.077M

D) For the diprotic acid telluric acid, H2TeO4, at 18 °C the value of is
2.09 × 10-8 and is 6.46 × 10-12. What is the molar concentration of H+ in 0.150 M H2TeO4 at 18 °C?
1.7.48 × 10-3 4. 3.12 × 10-9
2. 4.99 × 10-2 5. 6.46 × 10-12
3. 5.60 × 10-5 6. none of the previous answers
Maybe Answer:
Step 1) 2.09*10^-8=x^2/.15-x
3.135*10^-9-2.09*10^-8x-x^2=0
x=5.598*10^-5M
Step 2) 6.46*10^-12=5.598*10^-5x+x2/5.598*10^-5-x
3.616*10^-16-5.598*10^-5-x2
x=6.46*10^-12, so my answer would be 5

E) Calculate the percent dissociation for a 1.00 × 10-6 M solution of HCN at 25 °C.
Ka = 4.90 × 10-10.
Maybe answer:
HA->H++A-
1.00*10^-6M-x x x
4.90*10^-10=x2/1.00*10^-6-x
4.9*10^-16-4.90*10^-10x-x2=0
x*100=percent dissociation (?)
2.19*10^-6%

Any help is appreciated! I worked really hard on this! Thanks so much!!!!

Answer 1

A. 4.97 and 9.02 are right but I obtained 1.066E-5 and not 1.060E-5.

B. ok for KHCO3 but I think your reasoning why is weak. The moon is made of green cheese, too, because of its components. The problem doesn't ask for a reason but the answer is because of the hydrolysis of the HCO3^- (and of course that's because of the components).

C. ok

D. I don't know where you're going on D. First, the second ionization never enters into the process since it is so much smaller than k1. Use k1 and stop there; i.e., H^+ = 5.598E-5 is the right answer (rounded of course to the correct number of s.f.).

Answer 2

Thanks for the help! Sorry about D; I've never experiences a problem like that before. I guess I have a question about D. I understand that k2 is a lot smaller than k1, but since you get a different answer when doing the second ionization doesn't that mean that it is still significant enough to impact the H+ concentration? I'm just confused on that. And if you get a chance, could you look over E? That would be a lot of help! Thanks again! I appreciate it!

Answer 3

On D. With k1 = about 10^-8 and k2 = about 10^-12, that is about 10,000 less for K2 so whatever (H^+) is from k1, the contribution from k2 is 10,000 less than that. So with 5.6E-5 for k1, trying to add a 10,000th of that is fruitless. I think part of your confusion stems from your statement that the H^+ from K2 is significant. It isn't and your calculation doesn't show that. Here is how you calculate H^+ from k2. In fact your answer for the k2 part is 6.46E-12 which is correct. I think you just added it in wrong. And you can shorten the way you calculate the contribution from k2.

HTeO4^- --> H^+ + TeO4^2-
The contribution of (H^+) by k2 = (TeO4^2-) so you need to find (TeO4^2-).
k2 = 6.46E-12 = (H^+)(TeO4^2-)/HTeO4^-
But your calculation from k1 (where you determined the (H^+) from k1) tells you that (H^+) = (HTeO4^-). So (H^+) in the numerator of k2 cancels with (HTeO4^-) in the denominator of k2 which makes (TeO4^2-) = k2 = 6.46E-12. So total (H^+) from the system is 5.6E-5 + 6.5E-12 = 5.6E-5 which shows very clearly that the H^+ from k2 really is quite negligible. Best of luck.