Part A: A volume of 40.0 mL of aqueous potassium hydroxide (KOH) was titrated ag

Question

Part A: A volume of 40.0 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 18.2 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.

molarity = ???

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)+H2O2(aq)+3H2SO4(aq)3O2(g)+2MnSO4(aq)+K2SO4(aq)+4H2O(l)

A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. What mass of H2O2 was dissolved if the titration required 14.3 mL of the KMnO4 solution?

Express your answer with the appropriate units.

mass of H2O2 = ???

Explanation / Answer

Part A: A volume of 40.0 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 18.2 mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)K2SO4(aq)+2H2O(l)

Apply

mol of acid = 2mol of base

mol of acid = MV = 18.2*1.5 = 27.3

mol of base = 2*27.3 = 54.6 mmol of base

then

[KOH] = mmol/mL = 54.6/40 = 1.365 M

B)

similar to A,

identify ratio

2 mol of KMNO2 per 1 mol of H2O2

then

mol of KMnO2 = MV = 1.68*14.3 = 24.024 mmol of KMnO2

then

H2O2 = 1/2*24.024 = 12.012 mmol

mass = mol*MW = 12.012 *10^-3)(34) = 0.408408 g of H2O2

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