Answer to Question #347902 in Chemistry for Harley

Question

1. 1.275 liter of saturated silver chloride AgCl contains 0.01892 g of

dissolved AgCl at 25℃. Calculate the Ksp for AgCl.

2/ Balance the redox reaction in a) acidic and b) basic solution.

Fe2+ + Cr2O7 2- -> Fe3+ + Cr3+

3 A sample containing iron is titrated with a 0.025 M potassium

dichromate K2Cr2O7 solution with H3PO4/H2SO4. The titration required

26.74 mL of K2Cr2O7 for 1.45 g sample. What is the percent iron in the

sample? (Use the balanced reaction in #2).

Answer 1

Solution (1):

The molar mass of AgCl is 143.32 g/mol

Therefore,

Moles of AgCl = (0.001892 g AgCl) × (1 mol AgCl / 143.32 g AgCl) = 0.0000132 mol AgCl

Molarity = Moles of solute / Liters of solution

Therefore,

Molarity of AgCl = (0.0000132 mol) / (1.275 L) = 1.035×10⁻⁵ mol/L

Molarity of AgCl = 1.035×10⁻⁵ M

AgCl(s) → Ag⁺(aq) + Cl⁻(aq)

According to stoichiometry:

[Ag⁺] = [Cl⁻] = Molarity of AgCl = 1.035×10⁻⁵ M

The K_sp expression for AgCl:

K_sp = [Ag⁺] × [Cl⁻]

Therefore,

K_sp for AgCl = (1.035×10⁻⁵) × (1.035×10⁻⁵) = 1.07×10⁻¹⁰

The K_sp for AgCl is 1.07×10⁻¹⁰

Solution (2):

Fe²⁺ + Cr₂O₇²⁻ → Fe³⁺ + Cr³⁺

a) in acidic solution:

Oxidation half-reaction:

Fe²⁺ − e⁻ → Fe³⁺

Reduction half-reaction:

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Balance the redox reaction in acidic solution:

6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

b) in basic solution Cr₂O₇²⁻ is converted to CrO₄²⁻:

Cr₂O₇²⁻ + 2OH⁻ → 2CrO₄²⁻ + H₂O

Oxidation half-reaction:

Fe²⁺ − e⁻ → Fe³⁺

Reduction half-reaction:

CrO₄²⁻ + 4H₂O + 3e⁻ → Cr(OH)₃ + 5OH⁻

Balance the redox reaction in basic solution:

3Fe²⁺ + CrO₄²⁻ + 4H₂O → 3Fe³⁺ + Cr(OH)₃ + 5OH⁻

Solution (3):

Balance the redox reaction in acidic solution:

6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

According to stoichiometry:

Moles of Fe²⁺ / 6 = Moles of Cr₂O₇²⁻

or:

Moles of Fe²⁺ = 6 × Molarity of Cr₂O₇²⁻ × Volume of Cr₂O₇²⁻

Moles of Fe²⁺ = 6 × 0.025 M × 0.02674 L = 0.0040 mol Fe²⁺

Fe → Fe²⁺

According to stoichiometry:

Moles of Fe = Moles of Fe²⁺ = 0.0040 mol

The molar mass of iron (Fe) is 55.845 g/mol

Therefore,

Mass of Fe = (0.0040 mol Fe) × (55.845 g Fe / 1 mol Fe) = 0.22338 g Fe

%Fe in the sample = (Mass of Fe / Mass of sample) × 100%

%Fe in the sample = (0.22338 g / 1.45 g) × 100% = 15.4%

The percent iron (Fe) in the sample is 15.4%