Question

Part A

A volume of 70.0 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 19.7 mL of 1.50 M H2SO4 was needed? The equation is

2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

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 13.8 mL of the KMnO4 solution?

Express your answer with the appropriate units.

Answer 1

Answer :

Useful formulaes :

Molarity = no. of moles of solute / volume of solution (L)

No. of moles of solute = Molarity * volume (L)

So,

No. of Millimoles of solute = Molarity * volume (ml)

Also,

No. of moles of solute = amount of solute taken / molar mass of solute

Part -A :

Reaction is :

2KOH + H_{​​​​​​2} SO_{​​​​​4} -------> K_​​​2​ SO ₄ + 2H₂O

Millimoles of H_{​​​​​​2} SO_{​​​​​4} = Molarity * volume (ml)

Millimoles of H_{​​​​​​2} SO_{​​​​​4} = 1.5 * 19.7 = 29.55 millimoles

From the reaction , we know that 1 mol of H_{​​​​​​2}​​​​​SO ₄ require 2 mol of KOH

So, Millimoles of KOH = 2 * 29.55 = 59.1 Millimoles

So, Molarity of KOH = Millimoles of KOH / volume (ml)

Molarity of KOH = 59.1 / 70 = 0.844 M

So, Molarity of KOH solution is 0.844 M .

Part - B :

Reaction is :

2KMnO₄ + H_{​​​​​​2}​​​​​O_​2  + 3H₂SO₄ -------> 3O₂ + 2MnSO₄ + K_{​​​​​​2} SO_{​​​​​4} + 4H₂ ​​​​​​O

Millimoles of KMnO_​​​4 = Molarity * volume (ml) = 1.68 * 13.8 = 23.184 Millimoles

From the reaction , we know that 2 mol of KMnO_​​​4 require 1 mol of H_{​​​​​​2}​​​​​O_{​ 2} . So, moles of H2O2 will be half than moles of KMnO_​​​4 .

Millimoles of H_{​​​​​​2} O_{​​​​​​2} = (1/2) * 23.184 = 11.592 Millimoles

Molar mass of H_{​​​​​​2} O_{​​​​​​2} = 34 g/mol

As

Millimoles of H_{​​​​​​2} O_{​​​​​​2} = mass taken (mg) / molar mass

Mass of H_{​​​​​​2} O_{​​​​​​2} (mg) = millimoles of H_{​​​​​​2} O_{​​​​​​2} * molar mass = 11.592 * 34 = 394.13 mg = 0.394 g

So, mass of H_{​​​​​​2} O_{​​​​​​2} used = 0.394 g