Answer is: 53 milliliters of KOH are required.
Balanced chemical reaction:
2KOH(aq) + H₂SO₄(aq) → 2H₂O(l) + K₂SO₄(aq).
N(KOH) = 0.0050 N; normality of potassium hydroxide solution.
V(H₂SO₄) = 53 mL; volume of sulfuric acid.
N(H₂SO₄) = 0.0050 N; normality of sulfuric acid.
Na·Va = Nb·Vb.
0.0050 N · 53 mL = 0.0050 N · V(KOH).
V(KOH) = 53 mL; volume of potassium hydroxide solution.
