This is a question I am presently doing right now which I found in my homework. I am quite sure that in (a) I have to use the Hasselbach equation though I am not sure how to work it out. If anyone knows any of the answers, and how you got to them, please feel free to share. Any help is greatly appreciated :)

CH3COO- (aq) + H2O (l) <==> CH3COOH (aq) + OH- (aq)

A 0.01mol dm^-3 solution has a pH of 8.87

a. Calculate [H+] and [OH-] in the solution

b. Find [CH3COOH] in the solution.

c. Calculate the acid dissociation constant of CH3COOH.

Thanks! :)

a.

pH = 8.87
8.87 = -log(H^+)
Solve for (H^+), then
pH + pOH = pKw = 14
You know pH and pKw, solve for pOH and
pOH = -log(OH^-) which lets you solve for OH.

b.
This is a hydrolysis problem.
You know OH^- from part a. That = (CH3COOH).

c.
Kb for CH3COO^- = (Kw/Ka for CH3COOH)
(Kw/Ka) = (CH3COOH)(OH^-)/(CH3COO^-)
You know Kw, CH3COOH, OH, and CH3COO^-(hat's 0.01 in the problem), solve for Ka.

To solve this question, you can use the Henderson-Hasselbalch equation and the equation for the ionization constant of acetic acid.

a. Calculate [H+] and [OH-] in the solution:
The pH of the solution is given as 8.87. You can use the definition of pH to find the concentration of H+ ions.

pH = -log[H+]

Rearranging the equation, you get:

[H+] = 10^(-pH)

[H+] = 10^(-8.87)

[H+] ≈ 1.4 x 10^(-9) mol dm^(-3)

Since water dissociates into equal concentrations of H+ and OH- ions, the concentration of OH- can be calculated as:

[OH-] = (Kw / [H+])

Where Kw is the ion product of water, which is a constant value of 1 x 10^(-14) at 25°C.

[OH-] = (1 x 10^(-14)) / (1.4 x 10^(-9))

[OH-] ≈ 7.1 x 10^(-6) mol dm^(-3)

b. Find [CH3COOH] in the solution:
Since we have the concentration of the acetate ion (CH3COO-) and the concentration of the acetic acid (CH3COOH) is unknown, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([A-] / [HA])

Where pKa is the negative logarithm of the acid dissociation constant (Ka) of acetic acid. The value of pKa for acetic acid is 4.76.

Rearranging the equation, you get:

[HA] / [A-] = 10^(pH - pKa)

Substituting the values, you get:

[CH3COOH] / [CH3COO-] = 10^(8.87 - 4.76)

Simplifying, you find:

[CH3COOH] / [CH3COO-] ≈ 4.135

Since [CH3COO-] is given as 0.01 mol dm^(-3), you can calculate [CH3COOH] as:

[CH3COOH] = [CH3COO-] x ([CH3COOH] / [CH3COO-])

[CH3COOH] ≈ 0.01 mol dm^(-3) x 4.135

[CH3COOH] ≈ 0.04135 mol dm^(-3)

c. Calculate the acid dissociation constant of CH3COOH:
The acid dissociation constant (Ka) of acetic acid can be calculated using the equation:

Ka = [H+][A-] / [HA]

We have already calculated the concentration of [H+] and [CH3COO-] from part a and b. And the concentration of [CH3COOH] is given as 0.04135 mol dm^(-3).

Ka = (1.4 x 10^(-9)) x (0.04135) / (0.01)

Simplifying, you get:

Ka ≈ 5.779 x 10^(-6) mol dm^(-3)

These calculations should provide the answers you need for the question.