**Problem Statement:**

Calculate the pH, pOH and [H^{+}] of 0.01M CH_{3}COOH solution when the percentage of dissociation is 12.5%.

**Solution:
**

Since CH_{3}COOH is a weak acid, it dissociates partially in aqueous solution as follows

Given,

Percentage of dissociation = 12.5%

so, [H^{+}] = 0.01 × 12.5%

= 0.01 × 0.125M

= 1.25 × 10^{-3} M

Now, pH = – log[H^{+}]
= – log (1.25 × 10^{-3})

= 2.903

and pOH = 14 – pH

= 14 – 2.903

= 11.097

**Problem Statement:
**

Calculate the pH, pOH and [OH^{–}] of 0.05M NH_{4}OH solution when the percentage of dissociation is 7.5%.

**Solution:
**

Since NH_{4}OH is a weak base, it dissociates partially in aqueous solution as follows

Given,

Percentage of dissociation = 7.5%

so, [OH^{–}] = 0.05 × 7.5%

= 0.05 × 0.075M

= 3.75 × 10^{-3} M

Now, pOH = – log[OH^{–}]
= – log (3.75 × 10^{-3})

= 2.426

and pH = 14 – pOH = 14 – 2.426

= 11.574