Part A A volume of 90.0mL of aqueous potassium hydroxide (KOH) was titrated agai

Question

Part A

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 19.7mL 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.

molarity = ?

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)?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 22.3mL of the KMnO4 solution?

Express your answer with the appropriate units.

mass of H2O2 = ?

Explanation / Answer

V1 = 90 ml KOH

M1 =

V2 = 19.7 ml of H2SO4

M2 = 1.5M

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

1 mol of H2SO4 will neutralize 2 mol of base

[H+] = 1.5*2 = 3.0

Mol of Base = M1*V1

Mol of Acid = M2*V2

M1 * 90ml = 3.0M * 19.7

M1 = 3*19.7/90 = 0.656 M

Molarity of KOH = 0.656

This means there are 0.656 mol of OH-

PART2

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

V1 = 100 ml of Water + H2O2

M2 = 1.68M of KMnO4

V2 = 22.3 ml of KMnO4

moles of KMNO4 = M2*V2 = 1.68*22.3 = 37.46 mmol of KMnO4

NOTE:

2 mol of KMnO4 will react with 1 mol of H2O2

Then 37.46 will react with 37.46/2 = 18.73 mmol of H2O2

1 mol of H2O2 = 34 g/gmol

if there are 18.73*10^-3 mol of H2O"

then

mass = mol * MW = (18.73*10^-3)*(34) = 0.6368 grams of H2O2

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