Question

# A volume of 60.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution...

A volume of 60.0 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 11.7 mL of 1.50 M H2SO4 was needed? The equation is

2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Hints

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 17.8 mL of the KMnO4 solution?

part A)

C1 = 1.50 M

V1 = 11.7 mL

C2 = ?

V2 = 60.0 mL

2KOH(aq) + H2SO4(aq)------------------------------> K2SO4(aq) + 2H2O(l)

C2 V2 / n2 = C1 V1 / n1

C2 x 60 / 2 = 1.50 x 11.7 / 1

C2 = 0.585 M

molarity of KOH = 0.585 M

part B)

moles of KMnO4 = 1.68 x 17.8 / 1000

= 0.0299

2 mole KMnO4 --------------------------------> 1 mole H2O2

0.0299 mole KMnO4 ---------------------> 0.0299 / 2 = 0.01495

moles of H2O2 = 0.01495

mass of H2O2 = 0.01495 x 34

= 0.508 g

mass of H2O2 = 0.508 g

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