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 :
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 + H2 SO4 -------> K2 SO 4 + 2H2O
Millimoles of H2 SO4 = Molarity * volume (ml)
Millimoles of H2 SO4 = 1.5 * 19.7 = 29.55 millimoles
From the reaction , we know that 1 mol of H2SO 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 + H2O2 + 3H2SO4 -------> 3O2 + 2MnSO4 + K2 SO4 + 4H2 O
Millimoles of KMnO4 = Molarity * volume (ml) = 1.68 * 13.8 = 23.184 Millimoles
From the reaction , we know that 2 mol of KMnO4 require 1 mol of H2O 2 . So, moles of H2O2 will be half than moles of KMnO4 .
Millimoles of H2 O2 = (1/2) * 23.184 = 11.592 Millimoles
Molar mass of H2 O2 = 34 g/mol
As
Millimoles of H2 O2 = mass taken (mg) / molar mass
Mass of H2 O2 (mg) = millimoles of H2 O2 * molar mass = 11.592 * 34 = 394.13 mg = 0.394 g
So, mass of H2 O2 used = 0.394 g
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