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?

1 Answer
Jun 27, 2016

Answer:

#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#