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pH = -㏒ [H⁺] = -log 2 x 10⁻¹ = 1 - log 2
b. α = 10⁻³
α = [H⁺]÷Ma
10⁻³ = [H⁺] ÷ 10⁻¹
[H⁺] = 10⁻⁴
pH = -log [H⁺] = -log 10⁻⁴ = 4