Untuk membuat larutan dengan pH=6 ke dalam 100 ml larutan asam asetat 0,1 M (Ka=10^-5) di tambahkan natrium asetat (Mr=82) sebanyak....

mol CH3COOH = 100×0,1 = 10 mmol = 0,01 mol
pH=6 —> [H+] = 10⁻⁶ M
[H+] = (Ka × mol CH3COOH)/mol CH3COONa
10⁻⁶ = (10⁻⁵ × 0,01)/mol CH3COONa
mol CH3COONa = 0,1 mol
Massa CH3COONa = 0,1 × 82 = 8,2 gram

[tex]\displaystyle \left[\text{H}^+\right]=K_a\frac{n\text{CH}_3\text{COOH}}{n\text{CH}_3\text{COONa}}\\10^{-6}=10^{-5}\frac{100\cdot10^{-1}}{n\text{CH}_3\text{COONa}}\\10^{-1}=\frac{10}{n\text{CH}_3\text{COONa}}\\n\text{CH}_3\text{COONa}=\frac{10}{0,1}\\n\text{CH}_3\text{COONa}=100\,\text{mmol}\\\\n\text{CH}_3\text{COONa}=0,1\,\text{mol}\\\frac{m}{\text{Mr}}=\frac1{10}\\\frac{m}{82}=\frac1{10}\\m=\frac{82}{10}\\\boxed{\boxed{m=8,2\,\text{g}}}[/tex]