Part A A volume of 80.0 mL of aqueous potassium hydroxide (KOH) was titrated aga

Question

Part A A volume of 80.0 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 16.7 mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq) + H2SO4(aq) K2SO4 (aq) 2H20(1) Express your answer with the appropriate units. molarity ValueUnits Submit Hints My Answers Give Up Review Part Part B Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H2O2, can be titrated against a solution of potassium permanganate, KMnO4. The following equation represents the reaction: 2KMnO4 (aq) + H,02 (aq) + 3H2 SO 4 (aq)

Explanation / Answer

mmol of acid = MV = 16.7*1.5 = 25.05 mmol of acid

we need 2x of base since ratio is 2:1

25.05*2 = 50.1 mmol of base needed

M = mmol/mL = 50.1/80 = 0.62625 M

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