Determine the number of grams of PbI2 that will dissolve in 0.500 L of water: (2 pts) b. Determine the number of grams of PbI2 that will dissolve in 0.500 L of 0.0634 M NaI: (3 pts) 3. A solution contains the following anions at a concentration of 0.10 M: Cl-, Br-, and CrO4^2-. Determine in what order the anions will precipitate if Ag+ is used as the precipitating agent: (4 pts) 4. A solution contains 0.020 M Mg^2+ and 0.10 M Ca^2+. Can 99.90% of Mg^2+ be precipitated by the addition of NaOH without precipitation of Ca^2+? (4 pts
The Correct Answer and Explanation is:
Let’s solve the chemistry problems shown in the image step-by-step:
2a. Determine the number of grams of PbI₂ that will dissolve in 0.500 L of water.
We use the solubility product (Ksp) of PbI₂.
- Ksp of PbI₂ = 7.1 × 10⁻⁹
- PbI₂ ⇌ Pb²⁺ + 2 I⁻
Let the molar solubility of PbI₂ be s mol/L.
Then, [Pb²⁺] = s, [I⁻] = 2s
Ksp = [Pb²⁺][I⁻]² = s × (2s)² = 4s³
So, 4s³ = 7.1 × 10⁻⁹
s³ = 1.775 × 10⁻⁹
s ≈ 1.21 × 10⁻³ mol/L
Now multiply by volume:
mol PbI₂ = 1.21 × 10⁻³ mol/L × 0.500 L = 6.05 × 10⁻⁴ mol
Molar mass of PbI₂ = 207.2 + 2(126.9) ≈ 461.0 g/mol
Mass = 6.05 × 10⁻⁴ mol × 461.0 g/mol ≈ 0.279 g
2b. Determine the number of grams of PbI₂ that will dissolve in 0.500 L of 0.0634 M NaI.
Common ion effect: [I⁻] = 0.0634 M (initial), so:
Ksp = [Pb²⁺][I⁻]²
Let [Pb²⁺] = x
Then, 7.1 × 10⁻⁹ = x × (0.0634)²
x = 7.1 × 10⁻⁹ / (0.0634)² ≈ 1.77 × 10⁻⁶ mol/L
mol PbI₂ = 1.77 × 10⁻⁶ mol/L × 0.500 L = 8.85 × 10⁻⁷ mol
Mass = 8.85 × 10⁻⁷ mol × 461.0 g/mol ≈ 0.000408 g
3. Order of precipitation with Ag⁺:
Solubility rules and Ksp values:
- AgCl (Ksp ≈ 1.8 × 10⁻¹⁰)
- AgBr (Ksp ≈ 5.0 × 10⁻¹³)
- AgI (Ksp ≈ 8.3 × 10⁻¹⁷)
- Ag₂CrO₄ (Ksp ≈ 1.1 × 10⁻¹²)
The lower the Ksp, the earlier it precipitates:
Order: AgI → AgBr → Ag₂CrO₄ → AgCl
4. Precipitation of Mg(OH)₂ and Ca(OH)₂:
Ksp Mg(OH)₂ = 5.6 × 10⁻¹²
Ksp Ca(OH)₂ = 6.5 × 10⁻⁶
To precipitate 99.90% of Mg²⁺:
Remaining Mg²⁺ = 0.0001 × 0.020 M = 2.0 × 10⁻⁶ M
Ksp = [Mg²⁺][OH⁻]² → 5.6 × 10⁻¹² = (2.0 × 10⁻⁶)[OH⁻]²
[OH⁻]² = 2.8 × 10⁻⁶ → [OH⁻] ≈ 1.67 × 10⁻³ M
Check Ca(OH)₂:
[OH⁻] = 1.67 × 10⁻³, [Ca²⁺] = 0.10 M
Q = [Ca²⁺][OH⁻]² = 0.10 × (1.67 × 10⁻³)² = 2.8 × 10⁻⁷
Since Q < Ksp (6.5 × 10⁻⁶), Ca²⁺ does not precipitate
Yes, 99.90% of Mg²⁺ can be precipitated without Ca²⁺ precipitation.
