Berapa ml larutan ch3cooh 0,1 M harus ditambahkan kedalam 200ml larutan ch3coona 0,1 M untuk membuat larutan penyangga dengan pH 5 (ka=1,7X10^-5

n CH₃COONa = M x V = 0,1 x 200mL = 20 mmol
pH = - log [H+]
5 = - log [H+]
[H+] = 10⁻⁵

[H+] = Ka x [tex] \frac{n CH3COOH}{n CH3COONa} [/tex]
10⁻⁵ = 1,7 x 10⁻⁵ x [tex] \frac{n CH3COOH}{20} [/tex]
2 x 10⁻⁴ = 1,7 x 10⁻⁵ n CH₃COOH
n CH₃COOH = [tex] \frac{2 x 10^-^4}{1,7 x 10^-^5} [/tex]
n CH₃COOH = 11,7 mmol

V CH₃COOH = [tex] \frac{n}{M} [/tex] = [tex] \frac{11,7}{0,1} [/tex] = 117 mL