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 14.2mL of 1.50 M H2SO4 was needed? The equation is

2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Answer 1

M1*V1=M2*V2;
One mole of sulfuric acid neutralizes two moles of potassium hydroxide. Therefore:
M1=(14.2*1.5*2)/80=0.532M