Question

A volume of 90.0 ml of aqueous potassium hydroxide (koh) was titrated against a standard solution of sulfuric acid (h2so4). what was the molarity of the koh solution if 13.7 ml of 1.50 m h2so4 was needed? the equation is

Answer 1

2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

v₁=90.0 mL
c₁-?
v₂=13.7 mL
c₂=1.50 mol/L=1.50 mmol/mL

n(KOH)=v₁c₁
n(H₂SO₄)=v₂c₂
n(KOH)=2n(H₂SO₄)

v₁c₁=2v₂c₂

c₁=2v₂c₂/v₁

c₁=2*13.7*1.50/90.0=0.457 mol/L

0.457 M KOH