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mol Ca(OH)2 = 0,1 × 0,1 = 0,01
2 CH3COOH + Ca(OH)2 → Ca(CH3COO)2 + 2 H2O
m 0,1 0,01
b 0,02 0,01 0,01 0,02
s 0,08 - 0,01 0,02
[H^+] = Ka × (mol CH3COOH ÷ mol Ca(CH3COO)2)
= 10^-5 × (0,08 ÷ 0,01)
= 8 × 10^-5
pH = 5 - log 8