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 ?
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.
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!
I tried with both of them but the answer was still wrong.
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 ?
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.
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.
take 7- concentration B/ concentration A
To find the pH of the solution, we need to consider the dissociation of maleic acid (H2C4H2O4) and the hydrolysis of disodium maleate (Na2C4H2O4).
1. Dissociation of Maleic Acid:
H2C4H2O4 ⇌ H+ + HC4H2O4–
From the given information, we know the concentration of maleic acid is 0.0158 M. Since maleic acid is a weak acid, we can assume that the concentration of H+ formed will be small compared to the initial concentration of maleic acid. Therefore, we can neglect the change in concentration of maleic acid (0.0158 - x) due to dissociation, where 'x' is the concentration of H+ formed.
This makes the concentration of HC4H2O4– the same as the initial concentration of maleic acid, i.e., 0.0158 M.
2. Hydrolysis of Disodium Maleate:
Na2C4H2O4 + H2O ⇌ NaOH + HC4H2O4–
Disodium maleate (Na2C4H2O4) reacts with water to form sodium hydroxide (NaOH) and HC4H2O4–. The concentration of disodium maleate is given as 0.0226 M. Similar to the previous case, the change in concentration of disodium maleate due to hydrolysis can be neglected. Hence, the concentration of NaOH formed is 0.0226 M.
Now, we have the concentrations of HC4H2O4– and NaOH, which will affect the pH of the solution.
To find the concentration of H+ ions and calculate the pH, we need to consider three equilibrium reactions:
1. Dissociation of H2C4H2O4: Ka1 = [H+][HC4H2O4–]/[H2C4H2O4]
2. Hydrolysis of Na2C4H2O4: Kw/[OH–] = [HC4H2O4–]/[NaOH]
3. Autodissociation of water: Kw = [H+][OH–]
By solving these equations together, we can find the concentration of H+ ions, which will give us the pH of the solution.
Alternatively, you can use a software application or online calculator specifically designed to calculate the pH of a solution with multiple equilibria to obtain a more accurate result.
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.