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 (K_a) and express the K_a in terms of the equilibrium concentrations.

	[H⁺][C₃H₅O₂^-]		(x)(x)
K_a =	^{_______________}	=	^___________	= 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 K_a 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 K_a and lower pH.
This is done the same way as problem 1, using a K_a 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)
K_a =	^{_______________}	=	^___________	= 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 K_a, the lower the pH for the same concentration of weak acid. The higher K_a value tells us we have more dissociation of the weak acid, giving a greater [H⁺].

3. Calculate the K_a 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 K_aexpression

	[H⁺][A^-]		(2.24 x 10^-4)(2.24 x 10^-4)
K_a =	^________	=	^{________________________}
	[HA]		(0.3 - 2.24 x 10^-4)

K_a= 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)
K_b =	^{________________}	=	^___________	= 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⁺]

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