Question

A buffer solution of pH = 5.15 is to be prepared from acetic acid (Ka = `1.80 x 10^-5`), sodium acetate and water. What must be the molar ratio of acetate ion to acetic acid in this solution to obtain this pH?

Answer 1

`2.54`

Explanation:

Ethanoic acid is a weak acid which dissociates:

`CH_3COOH_((aq))rightleftharpoonsCH_3COO_((aq))^(-)+H_((aq))^(+)`

The expression for `K_a` is:

`K_a=[[CH_3COO_((aq))^(-)][H_((aq))^(+)])/([CH_3COOH_((aq))])`

So the molar ratio of acetate ion to acetic acid is:

`([CH_3COO_((aq))^(-)])/([CH_3COOH_((aq))])=K_a/[[H_((aq))^+]]`

The ratio of conentrations which I have given also corresponds to the molar ratio since the volume is common to both.

We are told that the `pH=5.15`

`:.-log[H_((aq))^(+)]=5.15`

`:.[H_((aq))^+]=7.08xx10^(-6)" ""mol/l"`

`:.([CH_3COO_((aq))^(-)])/([CH_3COOH_((aq))])=(nCH_3COO^(-))/(nCH_3COOH)=(1.8xx10^(-5))/(7.08xx10^(-6))=2.54`