Untuk membuat larutan dengan pH 6 kedalam 100 mL larutan asam asetat 0,1 M (ka=10^-5) ditambahkan natrium asetat (Mr=82) sebanyak ?

pH =6 ,[H+]=10^-6

n CH3COOH = 0,1×100 =10
[H+] = Ka.mol as lemah/[garam]
10^-6 = 10^-5 .10/[garam]
10^-1 = 10/[garam]
[garam] = 10/10^-1 =100 mmol

n [garam] =100/1000 = 0,1

M =n× Mr
=0,1 × 82
= 8,2 gram