Part A: A volume of 60.0 mL of aqueous potassium hydroxide (KOH) was titrated ag
ID: 906707 • Letter: P
Question
Part A:
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 23.2 mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)K2SO4(aq)+2H2O(l)
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 20.8 mL of the KMnO4 solution?
Explanation / Answer
Part A:
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 23.2 mL of 1.50 M H2SO4 was needed?
The equation is 2KOH(aq)+H2SO4(aq)K2SO4(aq)+2H2O(l)
V1 = 60 ml of KOH
M1 = ?
V2 = 23.2 ml
M2 = 1.5 H2SO4
Find moles of acid
mol = M*V = 23.2*1.5 = 34.8 mmol of acid
2 mol of base : 1 mol of acid so:
34.8 mmol * 2 = 69.6 mmol of KOH
M = mmol /ml = 69.6/60 = 1.16 M of KOH
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 20.8 mL of the KMnO4 solution?
V = 100 ml
M1 = ?
M2 = 1.68 KMnO4
V2 = 20.8 ml
Find moles of KMnO4
mol = M2*V2 = 1.68*20.8 = 34.94 mmol of KMnO4
2 mol of KMnO4 : 1 mol of H2O2
therefore
34.94/2 = 17.47 mmol of H2O2 are needed
MW of H2O2 = 34
mass = (17.47 *10^-3) * 34= 0.5934 g of H2O2