Berapa mL larutan H2SO4 0,1 M diperlukan untuk menetralkan 200 mL larutan NaOH yang PH-nya 12+log 5

NaOH pH = 12 + log 5
pOH = 14 – (12 + log 5 ) = 2 – log 5
pOH = – log [OH-]
2 – log 5 = log [OH-]
[OH-] = 5 x 10⁻² = 0,05 M

mmol NaOH = 200 x 0,05 = 10 mmol

2 NaOH + H2SO4 => Na2SO4 + 2 H2O

mmol H2SO4 = 1/2 x mmol NaOH = 1/2 x 10 = 5 mmol
volume H2SO4 = 5 mmol / 0,1 mmol/ml = 50 ml