Arkansas State University Department of Chemistry and Physics

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)
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 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)
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 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_a expression

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