Given:

• Compound: AgCl
• K_sp = 1.8 × 10⁻¹⁰

Write dissolution equation:

$AgCl\text{(s)} \rightleftharpoons Ag^+\text{(aq)} + Cl^-\text{(aq)}$

Set up ICE table:

Let s = molar solubility (mol/L that dissolve)

At equilibrium:

• [Ag⁺] = s
• [Cl⁻] = s

Write K_sp expression:

$K_{sp} = [Ag^+][Cl^-]$

Substitute:

$1.8 \times 10^{-10} = (s)(s) = s^2$

Solve for s:

$s^2 = 1.8 \times 10^{-10}$

$s = \sqrt{1.8 \times 10^{-10}}$

$s = 1.3 \times 10^{-5} \text{ M}$

Answer: Molar solubility = 1.3 × 10⁻⁵ M

Interpretation:

Very small solubility:

• AgCl is "insoluble" (K_sp very small)
• Only 1.3 × 10⁻⁵ mol/L dissolves
• This is why AgCl precipitates easily

In grams per liter:

• Molar mass AgCl = 143.5 g/mol
• s = (1.3 × 10⁻⁵ mol/L)(143.5 g/mol) = 1.9 × 10⁻³ g/L

Very low solubility!

Given:

• Compound: PbI₂
• K_sp = 7.9 × 10⁻⁹

Dissolution: PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq)

(a) Molar solubility in pure water

Let s = solubility:

Note: For every 1 Pb²⁺, get 2 I⁻

K_sp expression:

$K_{sp} = [Pb^{2+}][I^-]^2$

$7.9 \times 10^{-9} = (s)(2s)^2$

$7.9 \times 10^{-9} = s \cdot 4s^2$

$7.9 \times 10^{-9} = 4s^3$

$s^3 = \frac{7.9 \times 10^{-9}}{4} = 1.975 \times 10^{-9}$

$s = \sqrt[3]{1.975 \times 10^{-9}}$

$s = 1.25 \times 10^{-3} \text{ M}$

Answer (a): s = 1.3 × 10⁻³ M in pure water

(b) Molar solubility in 0.10 M NaI

NaI provides common ion (I⁻):

• NaI → Na⁺ + I⁻
• [I⁻]₀ = 0.10 M (from NaI)

ICE table:

K_sp expression:

$7.9 \times 10^{-9} = (s)(0.10 + 2s)^2$

Small s approximation:

Since K_sp is very small, s << 0.10

Assume: 0.10 + 2s ≈ 0.10

$7.9 \times 10^{-9} = (s)(0.10)^2$

$7.9 \times 10^{-9} = s(0.010)$

$s = \frac{7.9 \times 10^{-9}}{0.010}$

$s = 7.9 \times 10^{-7} \text{ M}$

Check: 2s = 1.6 × 10⁻⁶ << 0.10 ✓ (approximation valid)

Answer (b): s = 7.9 × 10⁻⁷ M in 0.10 M NaI

Comparison:

Pure water: s = 1.3 × 10⁻³ M 0.10 M NaI: s = 7.9 × 10⁻⁷ M

Common ion effect:

• Solubility decreased by factor of ~1600!
• I⁻ from NaI shifts equilibrium left
• Much less PbI₂ dissolves

Given:

• [Ag⁺] = 0.010 M
• [Pb²⁺] = 0.010 M
• K_sp(AgCl) = 1.8 × 10⁻¹⁰
• K_sp(PbCl₂) = 1.7 × 10⁻⁵

Adding NaCl → source of Cl⁻

Find [Cl⁻] needed to precipitate each:

For AgCl: AgCl(s) ⇌ Ag⁺ + Cl⁻

Precipitation when Q_sp = K_sp:

$K_{sp} = [Ag^+][Cl^-]$

$1.8 \times 10^{-10} = (0.010)[Cl^-]$

$[Cl^-] = \frac{1.8 \times 10^{-10}}{0.010}$

$[Cl^-] = 1.8 \times 10^{-8} \text{ M}$

AgCl precipitates when [Cl⁻] ≥ 1.8 × 10⁻⁸ M

For PbCl₂: PbCl₂(s) ⇌ Pb²⁺ + 2Cl⁻

Precipitation when Q_sp = K_sp:

$K_{sp} = [Pb^{2+}][Cl^-]^2$

$1.7 \times 10^{-5} = (0.010)[Cl^-]^2$

$[Cl^-]^2 = \frac{1.7 \times 10^{-5}}{0.010}$

$[Cl^-]^2 = 1.7 \times 10^{-3}$

$[Cl^-] = \sqrt{1.7 \times 10^{-3}}$

$[Cl^-] = 0.041 \text{ M}$

PbCl₂ precipitates when [Cl⁻] ≥ 0.041 M

Compare:

AgCl precipitates at: [Cl⁻] = 1.8 × 10⁻⁸ M PbCl₂ precipitates at: [Cl⁻] = 0.041 M

AgCl requires much less Cl⁻!

Answer: AgCl precipitates first

Separation:

Can separate Ag⁺ and Pb²⁺:

Step 1: Add Cl⁻ slowly to [Cl⁻] = 1.8 × 10⁻⁸ M

• AgCl precipitates
• Pb²⁺ stays in solution

Step 2: Continue adding to [Cl⁻] = 0.041 M

• Now PbCl₂ precipitates
• Most Ag⁺ already removed

Window for separation:

1.8 × 10⁻⁸ M < [Cl⁻] < 0.041 M

In this range:

• AgCl precipitated
• PbCl₂ still dissolved
• Successful separation!

General rule:

Compound with smaller K_sp precipitates first (when cations at equal concentration)