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mol NaOH = 40 per 1000 * 0,2 = 0,008 mol
HCOOH + NaOH -> HCOONa + H₂O
mula² 0,01 0,008 -
reaksi 0,008 0,008 0,008
sisa 0,002 - 0,008
[H⁺] = Ka * HCOOH per HCOONa
= 10⁻⁴ * 0,002 per 0,008
= 10⁻⁴ * 0,25
= 2,5 * 10⁻⁵
pH = -log 2,5 * 10⁻⁵
= 5 - log 2,5