A volume of 60.0 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 22.2 mL of 1.50 M H2SO4 was needed? The equation is
2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)
Express your answer with the appropriate units.
moles H2SO4 used = M x L = 1.50 M x 0.0222 L = 0.0333
Convert to moles kOH = 0.0333 x (2 moles KOH/1 mol H2SO4) = 0.0333 x 2 = 0.0666
Then M KOH = moles KOH/L KOH = 0.0666/0.0600 L = ? moles/L