Question

A solution containing 0.0158 M maleic acid and 0.0226 M disodium maleate. The Ka values for maleic acid are 1.20 × 10-2 (Ka1) and 5.37 × 10-7 (Ka2). What is the pH of the solution ?

Answer 1

I'm sure you know to use the HH equation. So what's the trouble? Not know which k to use. Or something else.

Answer 2

I tried with both of them but the answer was still wrong.

Answer 3

So how did you do it? Show your work; probably I can find the error.
Did you try this?
k1 = (H^+)(HM^-)/(H2M)
k2 = (H^+)(M^2-)/(HM^-)

so k1*k2 = (H^+)^2(HM^-)(M^2-)/(HM^-)(H2M)
Cancel the HM^- in numerator and denominator, substitute the numbers for the others and solve for (H^+) and convert to pH.

Answer 4

K1*K2=(x^2*.0226)/.0158=.0000671

Which results in pH being 4.17

But it seems that my answer is still wrong. An i missing something ?

Answer 5

If you know the answer let me see it and I'll see if I can work around to the answer; otherwise, this last looks ok to me.

Answer 6

I really wish I knew the answer and I would've tried to work my way around but unfortunately I don't. My teacher didn't teach me how to do these calculations and I'm doing online Hw. Thanks for your help ! I'll post the answer if I get it right.

Answer 7

take 7- concentration B/ concentration A

Answer 8

Several of my students are having problems with this problem, and I do not like the way that Sapling worked this problem. The solution states that that the acid and 2nd conjugate base react to produce intermediate, ie
H2A + A^2-^ <----> 2 HA^1^
then
H2A + A^2-^ <----> 2 HA^1^
0.0158 0.0226 ~0
-0.0158 -0.0158 +2(0.0158)
= 0 = 0.0068 = 0.0316

Then calculate the pH as a buffer centered around K2.

pH = 6.27 + log(0.0068/0.0316) = 5.60

What I don't like about this problem: The solution treats H2A as a strong acid and/or A2- as a strong base. They are neither.

Answer 9

Figured it out.

H2A <--> HA- + H+ K1
HA- <----> A2- + H+ K2

H2A <--> HA- + H+ K1
A2- + H+ <-----> HA- 1/K2 (not a Kb because not forming OH-, merely flipping the K2 expression.)

H2A + A2- <---> 2 HA- K' = K1/K2
plug in the K values, K' = 2.2 x 10^4. VERY product favored equilibrium, so

" H2A + A^2-^ <----> 2 HA^1^
then
H2A + A^2-^ <----> 2 HA^1^
0.0158 0.0226 ~0
-0.0158 -0.0158 +2(0.0158)
= 0 = 0.0068 = 0.0316

Then calculate the pH as a buffer centered around K2.

pH = 6.27 + log(0.0068/0.0316) = 5.60 "

is correct!