A volume of 80.0mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 21.7mL of 1.50 M H2SO4 was needed? The equation is

Question

A volume of 80.0mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 21.7mL of 1.50 M H2SO4 was needed? The equation is
2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

DrBob222 · Answer

2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l) mols H2SO4 = M x L mols KOH = 2*mols H2SO4 M KOH = mols KOH/L KOH