A volume of 55.6 ml of aqueous potassium hydroxide is titrated against a standard solution of sulfuric acid. What was the molarity of the potassium hydroxide solution if 15.1 ml of 1.50 m sulfuric acid was needed

The molar concentration of the potassium hydroxide solution was 0.815 mol/L.

Balanced equation: 2KOH + H_2SO_4 → K_2SO_4 + 2H_2O

Moles of H_2SO_4:

15.1 mL H_2SO_4 * (1.50 mmol H_2SO_4 / 1 mL H_2SO_4)

= 22.65 mmol H_2SO_4

Moles of KOH: 22.65 mmol H_2SO_4 * (2 mmol KOH/2 mmol H_2SO_4)

= 45.30 mmol KOH

Concentration of KOH: c = "moles"/"litres" = 45.30 mmol/55.6 mL

= 0.815 mol/L