Question

What is the Ka and a (degree of ionization) of HC2H3O2 for the following pH and concentrations:

	1.0M	0.1M	0.01M
pH	2.70	3.2	3.8

Answer 1

Part a

pH = - log[H+]

[H+] = 10^-pH = 10^-2.70 = 0.001995 M

The balanced disassociation reaction with ICE TABLE

CH3COOH ---> CH3COO- + H+
I 1

C - x +x +x

E 1-x x x

At equilibrium

[H+] = x = 0.001995 M

degree of ionization = x*100/1

= 0.001995*100/1

= 0.1995%

Ka = [CH3COO-][H+] / [CH3COOH]

[CH3COO-]=[H+]=x

Ka = 0.001995^2 / (0.530-0.001995)

= 7.537 x 10^-6

Part b

pH = - log[H+]

[H+] = 10^-pH = 10^-3.2 = 0.0006309 M

The balanced disassociation reaction with ICE TABLE

CH3COOH ---> CH3COO- + H+
I 0.1

C - x +x +x

E 0.1-x x x

At equilibrium

[H+] = x = 0.0006309 M

degree of ionization = x*100/0.1

= 0.0006309*100/0.1

= 0.63%

Ka = [CH3COO-][H+] / [CH3COOH]

[CH3COO-]=[H+]=x

Ka = 0.0006309^2 / (0.530-0.0006309)

= 7.519 x 10^-7

Part c

pH = - log[H+]

[H+] = 10^-pH = 10^-3.8 = 0.0001584 M

The balanced disassociation reaction with ICE TABLE

CH3COOH ---> CH3COO- + H+
I 0.01

C - x +x +x

E 0.01-x x x

At equilibrium

[H+] = x = 0.0001584 M

degree of ionization = x*100/0.01

= 0.0001584*100/0.01

= 1.584%

Ka = [CH3COO-][H+] / [CH3COOH]

[CH3COO-]=[H+]=x

Ka = 0.0001584^2 / (0.530-0.0001584)

= 4.74 x 10^-8