A volume of 90.0mL of aqueous potassium hydroxide (KOH) was titrated against a s

Question

A volume of 90.0mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 14.7mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)?K2SO4(aq)+2H2O(l)

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 12.8mL of the KMnO4solution?

Explanation / Answer

(a)
Moles H2SO4 = 0.0147 L x 1.50 M = 0.02205
the ratio between H2SO4 and KOH in the balanced equation
H2SO4 + 2 KOH = K2SO4 +2 H2O
is 1 : 2
moles KOH = 2 x 0.02205 =0.0441
molar concentration of KOH = 0.0441 / 0.090 L = 0.49 M

(b)
moles KMnO4 = 0.0128 L x 1.68 M = 0.0215

6 H+ + 2MnO4- + 5 H2O2 = 2 Mn2+ + 5O2 + 8H2O
the ratio between MnO4- and H2O2 is 2 : 5
moles H2O2 = 0.0215 x 5/2 = 0.05376
mass H2O2 = 0.05376 mol x 34.014 g/mol = 1.83 gm

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