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)

what is the molarity =? Express your answer with the appropriate units.

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?

what is the mass of H2O2=? Express your answer with the appropriate units.

Explanation / Answer

A)

V = 40 ml KOH

M = ?

M = 1.5

V = 18.2

mmol of acid = MV = 1.5/18.2 = 0.08333 mmol

then we have 2*0.08333 = 0.16666 mmol of base

M = mmol/mL = 0.16666/40 = 0.0041665 M of KOH

B)

V = 100 ml of H2O2

M = 1.68 KMnO4

V2= 14.3 ml of kMNO4

mmol of KMNO4 = M2V2 = 1.68*14.3 = 1.68 mmol of KMnO4

ratio is 2:1

1.68/2 = 0.84 mmol of H2O2

mass = mol*MW = (0.84*10^-3)(34)= 0.02856 g of H2O2

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