A volume of 80.0 mL of aqueous potassium hydroxide (KOH) was titrated against a

Question

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 23.7 mL of 1.50 M H2SO4 was needed? The equation is

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

PartB: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 H2O2was dissolved if the titration required 17.3 mL of the KMnO4 solution?

Explanation / Answer

a) 2KOH(aq)+H2SO4(aq)K2SO4(aq)+2H2O(l)

KOH :

Molarity M1 = ?

Volume V1 = 80.0 mL

moles n1 = 2

H2SO4:

Molarity M2 = 1.50 M

Volume V2 = 23.7 mL

moles n2 = 1

At equivalence point,

    M1V1/n1 = M2V2/n2

   M1 = (M2V2/n2) (n1/V1)

        = (1.50 M X 23.7mL/ 1) ( 2/ 80.0 mL)

          = 0.888 M

Molarity M1 = 0.888 M

Therefore, molarity of the KOH solution = 0.888 M

b) 2KMnO4(aq)+H2O2(aq)+3H2SO4(aq)3O2(g)+2MnSO4(aq)+K2SO4(aq)+4H2O(l)

H2O2:

Molarity M1 = ? M

Volume V1 = 100.0 mL

moles n1 = 1

KMnO4 :

Molarity M2 = 1.68 M

Volume V2 = 17.3 mL

moles n2 = 2

At equivalence point,

    M1V1/n1 = M2V2/n2

   M1 = (M2V2/n2) (n1/V1)

        = (1.68 M X 17.3 mL/ 2) ( 1/ 100.0 mL)

       = 0.145 M

        Molarity M1 = 0.145 M

Therefore, molarity of the H2O2 solution = 0.145 M

Mass of H2O2 :

Molarity of H2O2 = 0.145 M

Molar mass of H2O2 = 34.0 g/mol

Volume of H2O2 solution = 100.0 mL = 0.10 L

Molarity of H2O2 = (Mass of H2O2 / Molar mass of H2O2) X (1 / Volume of H2O2 solution in Litres)

0.145 M = (Mass of H2O2 / 34 gmol-1) X (1 / 0.10 L)

Mass of H2O2 = 0.493 g

Therefore, 0.493 g of H2O2 was dissolved in 100.0 mL of water.

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