A volume of 60.0mL of aqueous potassium hydroxide (KOH) was titrated against a s

Question

A volume of 60.0mL 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)

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)?O2(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 22.3mL of the KMnO4solution?

Explanation / Answer

First, find the moles of H2SO4:

1.50 M * .0217 L = .03255moles of H2S04

To complete the reaction, you need twice as many moles of KOH: .03255 * 2 = .0651

Finally, divide you moles of KOH by the volume (in liters) to get molarity.

.0651 / .060L = 1.085 M KOH

B.. Stoichiometry ratio of H2O2 and KMnO4 is 1:2
Molarity = moles / volume (in Liters) then
moles KMnO4 = (molarity of KMnO4)(volume of KMnO4)
moles KMnO4 = .0213(1.68)
moles of KMnO4 = 0.035784
for every 1 mole of H2O2, 2 moles of KMnO4 are needed
therefore:
moles H2O2 = 0.035784/2
moles H2O2 = 0.017892
mass = moles (molar mass)
mass H2O2 = 0.017892(34.016)
mass H2O2 = 0.608614272 grams

Therefore 0.608614272 grams of H2O2 was dissolved in 100.0 mL of water

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