# Solutions

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1. Calculate the equilibrium concentrations and pH for a 0.20 M proprionic acid solution. Complete the equilibrium and express the equilibrium concentrations in terms of a single unknown 'x'.

 HC3H5O2 (aq) H+(aq) + C3H5O2- (aq) initial concentration 0.2 M 0 0 change - x + x + x equilibrium concentration 0.2 - x x x

Write the equilibrium expression (Ka) and express the Ka in terms of the equilibrium concentrations.

 [H+][C3H5O2-] (x)(x) Ka = _______________ = ___________ = 1.3 x 10-5 [HC3H5O2] (0.2 - x) (from table)

Solve for x; since the initial concentration is greater than 0.1 and the Ka less than 10-4, we can drop the '- x' term.

 x2 solving for 'x' gives _______ = 1.3 x 10-5 x = 1.61 x 10-3 0.2

Solve for the equilibrium concentrations by substituting the value of 'x' into the equilibrium concentrations: [H+] = [C3H5O2-] = x, [HC3H5O2] = 0.2 - x

[H+] = [C3H5O2-] = 1.61 x 10-3 M, [HC3H5O2] = 1.98 x 10-1 M;

pH = - log [H+] = - log(1.61 x 10-3)

pH = 2.79

2. Calculate the equilibrium concentrations and pH for a 0.20 M carbonic acid solution.
Compare the answer to problem 1. Note which has the higher Ka and lower pH.
This is done the same way as problem 1, using a Ka value of 4.2 x 10-7

 H2CO3 (aq) H+(aq) + HCO3- (aq) initial concentration 0.2 M 0 0 change - x + x + x equilibrium concentration 0.2 - x x x

 [H+][HCO3-] (x)(x) Ka = _______________ = ___________ = 4.2 x 10-7 [H2CO3] (0.2 - x) (from table)

 x2 solving for 'x' gives _______ = 4.2 x 10-7 x = 2.90 x 10-4 0.2

[H+] = [HCO3-] = 2.90 x 10-4 M, [H2CO3] = 2.0 x 10-1 M; pH = 3.54

The higher the Ka, the lower the pH for the same concentration of weak acid. The higher Ka value tells us we have more dissociation of the weak acid, giving a greater [H+].

3. Calculate the Ka for a 0.3 M solution of HA (weak acid) if the pH = 3.65
First calculate the [H+] concentration from the pH
[H+] = 10-pH; [H+] = 10-3.65; [H+] = 2.24 x 10-4

since 1:1 mole ratio

[H+] = [A-]; [HA] = 0.3 - [H+]

Substitute into the Ka expression

 [H+][A-] (2.24 x 10-4)(2.24 x 10-4) Ka = ________ = ________________________ [HA] (0.3 - 2.24 x 10-4)

Ka= 1.67 x 10-7

4. Calculate the pH for a 0.2 M pyridine solution.

 C5H5N (aq) OH-(aq) + C5H5NH+ (aq) initial concentration 0.2 M 0 0 change - x + x + x equilibrium concentration 0.2 - x x x

 [OH-][C5H5NH+] (x)(x) Kb = ________________ = ___________ = 2.0 x 10-9 [C5H5N] (0.2 - x) (from table)

Solve for 'x'

 x2 solving for 'x' gives _______ = 2.0 x 10-9 x = 2.0 x 10-5 0.2

[OH-] = [C5H5NH+] = 2.0 x 10-5 M, [C5H5N] = 2.0 x 10-1 M

Using the following equations: pOH = - log [OH-]; pH + pOH = 14

pOH = 4.70, pH = 9.30

5. Calculate the pH of a 0.34 M HCl solution.

pH = 0.47, strong acid, 100 % ionization; 1:1 mole ratio, [HCl] = [H+]

This page titled Solutions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark Draganjac via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.