posted by Ellen
A volume of 50.0 mL of aqueous potassium hydroxide was titrated against a standard solution of sulfuric acid. What was the molarity of the KOH solution if 19.2 mL of 1.50 M H2SO4 was needed? The equation is
2KOH(aq) +H2SO4(aq)-» K2SO4(aq)+2H2O(l)
moles H2SO4 = M x L.
From the equation, moles KOH must be twice moles H2SO4.
M KOH = moles/L.