Problem Statement:
Calculate the pH, pOH and [H+] of 0.01M CH3COOH solution when the percentage of dissociation is 12.5%.
Solution:
Since CH3COOH 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 NH4OH solution when the percentage of dissociation is 7.5%.
Solution:
Since NH4OH 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