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posted by jose .
A volume of 40.0mL 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)

21.7mL of 1.50M H2SO4 contains .0217*1.50 = .03255 moles of H2SO4
Each mole of H2SO4 reacts with 2 moles of KOH.
So, the 40mL of KOH contained 0.0651 moles.
.0651mole/.040L = 1.6275M
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