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= 4 × 10^-5 × ( 0,01 ÷ 0,01)
= 4 × 10^-5
pH = 5 - log 4 ÷ 5 - 2 log 2
b. CH3COOH + NaOH ⇒ CH3COONa + H2O
m 0,015 0,01
b 0,01 0,01 0,01 0,01
s 0,05 - 0,01 0,01
H^+ = Ka × (mol asam ÷ mol garam)
= 10^-5 × ( 0,05 ÷ 0,01)
= 5 × 10^-5
pH = 5 - log 5