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pH = - log [H+]
3 = - log[H+]
[H+] = 10⁻³
[H+] = √Ka .M
10⁻³ = √10⁻⁵ x M
10⁻⁶ = 10⁻⁵ x M
M = 10⁻¹ = 0,1 M
CH3COOH + KOH => CH3COOK + H2O
mol CH3COOH = mol KOH
50 x 0,1 = v x 0,1
v KOH = 50 ml