maka fokus larutan asam itu yakni …
pH = 3 – log 2
[H⁺] = 2 . 10⁻³
[H⁺] = √Ka . M
2 . 10⁻³ = √2 . 10⁻⁵ . M
(2 . 10⁻³)² = 2 . 10⁻⁵ . M
4 . 10⁻⁶ = 2 . 10⁻⁵ . M
M = 4 . 10⁻⁶
2 . 10⁻⁵
= 2 . 10⁻¹ M
Jadi, fokus larutan asam = 0,2 M