Advanced Chemical Equilibrium Problem

Problem Statement 🔴

A complex equilibrium system involves the simultaneous dissolution of magnesium hydroxide and the establishment of acid-base equilibrium in solution. The system contains:

Solid Phase: Mg(OH)₂(s)

Aqueous Phase: Initially contains 0.1 M NH₄Cl and 0.1 M NH₃·H₂O

Given Equilibrium Constants:

• $K_{sp}[Mg(OH)_2] = 5.61 \times 10^{-12}$ at 25°C
• $K_b[NH_3] = 1.8 \times 10^{-5}$ at 25°C
• $K_w = 1.0 \times 10^{-14}$ at 25°C

When the system reaches equilibrium, find:

1. The pH of the solution
2. The solubility of Mg(OH)₂ in this medium
3. The concentrations of all species at equilibrium
4. The effect of adding 0.05 M NaOH to the system

Solution 1: Systematic Equilibrium Approach

Step 1: Understanding the System

The system involves multiple equilibria:

1. Solubility Equilibrium: $Mg(OH)2(s) \rightleftharpoons Mg^{2+} + 2OH^-$ $K{sp} = [Mg^{2+}][OH^-]^2 = 5.61 \times 10^{-12}$

2. Base Equilibrium: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$ $K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = 1.8 \times 10^{-5}$

3. Water Equilibrium: $H_2O \rightleftharpoons H^+ + OH^-$ $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$

Step 2: Initial Analysis

Initial concentrations:

• $[NH_4^+]_0 = 0.1$ M (from NH₄Cl)
• $[NH_3]_0 = 0.1$ M
• $[Mg^{2+}]_0 = 0$
• $[OH^-]$ from ammonia base reaction

First, let’s find pH without Mg(OH)₂ dissolution:

For the buffer system: $NH_4^+/NH_3$

Using Henderson-Hasselbalch: $pOH = pK_b + \log\left(\frac{[NH_4^+]}{[NH_3]}\right)$ $pOH = 4.74 + \log\left(\frac{0.1}{0.1}\right) = 4.74$ $pH = 14 - 4.74 = 9.26$

Therefore, $[OH^-] = 10^{-4.74} = 1.82 \times 10^{-5}$ M

Step 3: Incorporating Mg(OH)₂ Solubility

Let $s$ be the solubility of Mg(OH)₂ in mol/L

When Mg(OH)₂ dissolves:

• $[Mg^{2+}] = s$
• $[OH^-]$ increases by $2s$

The equilibrium becomes:

• $[OH^-]_{total} = 1.82 \times 10^{-5} + 2s$
• $[Mg^{2+}] = s$

Using $K_{sp}$: $s(1.82 \times 10^{-5} + 2s)^2 = 5.61 \times 10^{-12}$

This is a cubic equation. Let’s solve it:

Let $x = s$ for simplicity: $x(1.82 \times 10^{-5} + 2x)^2 = 5.61 \times 10^{-12}$

Assuming $2x \ll 1.82 \times 10^{-5}$ initially: $x(1.82 \times 10^{-5})^2 = 5.61 \times 10^{-12}$ $x = \frac{5.61 \times 10^{-12}}{(1.82 \times 10^{-5})^2} = \frac{5.61 \times 10^{-12}}{3.31 \times 10^{-10}} = 0.0169$ M

Now check our assumption: $2x = 0.0338$ M, which is NOT $\ll 1.82 \times 10^{-5}$ M

Our assumption is wrong. We need to solve the full equation.

Step 4: Solving the Complete System

The presence of significant Mg(OH)₂ dissolution will change the buffer equilibrium. We need to solve all equilibria simultaneously.

Mass Balance for Ammonia Species: $[NH_4^+] + [NH_3] = 0.2$ M (total ammonia nitrogen)

Equilibrium Relationships:

1. $\frac{[NH_4^+][OH^-]}{[NH_3]} = 1.8 \times 10^{-5}$
2. $[Mg^{2+}][OH^-]^2 = 5.61 \times 10^{-12}$

Let $[OH^-] = x$ and $[Mg^{2+}] = s$

From mass balance and equilibrium (1): $\frac{[NH_4^+]x}{[NH_3]} = 1.8 \times 10^{-5}$

$[NH_4^+] = 0.2 - [NH_3]$

$\frac{(0.2 - [NH_3])x}{[NH_3]} = 1.8 \times 10^{-5}$

$0.2x - [NH_3]x = 1.8 \times 10^{-5}[NH_3]$

$0.2x = [NH_3](x + 1.8 \times 10^{-5})$

$[NH_3] = \frac{0.2x}{x + 1.8 \times 10^{-5}}$

Charge Balance: $[NH_4^+] + [H^+] + 2[Mg^{2+}] = [Cl^-] + [OH^-]$

$[NH_4^+] + \frac{K_w}{x} + 2s = 0.1 + x$

This is complex. Let me use a different approach.

Solution 2: Successive Approximation Method

Step 1: First Approximation

Assume the buffer dominates initially, then adjust for Mg(OH)₂ dissolution.

From buffer alone: $[OH^-]_1 = 1.82 \times 10^{-5}$ M

This gives $s_1 = \frac{K_{sp}}{[OH^-]_1^2} = \frac{5.61 \times 10^{-12}}{(1.82 \times 10^{-5})^2} = 0.0169$ M

This large solubility suggests the buffer will be significantly affected.

Step 2: Second Approximation

The additional $OH^-$ from Mg(OH)₂ dissolution: $[OH^-]_{from\ Mg} = 2s_1 = 0.0338$ M

Total $[OH^-]_2 = 1.82 \times 10^{-5} + 0.0338 = 0.0338$ M (approximately)

Now, let’s find the new buffer equilibrium:

At $[OH^-] = 0.0338$ M: From base equilibrium: $\frac{[NH_4^+][OH^-]}{[NH_3]} = 1.8 \times 10^{-5}$

$\frac{[NH_4^+] \times 0.0338}{[NH_3]} = 1.8 \times 10^{-5}$

$\frac{[NH_4^+]}{[NH_3]} = \frac{1.8 \times 10^{-5}}{0.0338} = 5.33 \times 10^{-4}$

With mass balance: $[NH_4^+] + [NH_3] = 0.2$

$[NH_4^+] = 5.33 \times 10^{-4}[NH_3]$

$5.33 \times 10^{-4}[NH_3] + [NH_3] = 0.2$

$[NH_3](1 + 5.33 \times 10^{-4}) = 0.2$

$[NH_3] = \frac{0.2}{1.000533} = 0.1999$ M

$[NH_4^+] = 0.0001066$ M

Step 3: Third Approximation

With these concentrations, let’s check the $K_{sp}$ condition:

$[Mg^{2+}] = \frac{K_{sp}}{[OH^-]^2} = \frac{5.61 \times 10^{-12}}{(0.0338)^2} = 4.91 \times 10^{-9}$ M

This is much smaller than our initial estimate. The large $[OH^-]$ suppresses Mg(OH)₂ solubility.

Step 4: Final Iteration

Let’s check if this $[OH^-]$ is consistent:

The $OH^-$ from ammonia base reaction: $[OH^-]_{ammonia} = \frac{K_b[NH_3]}{[NH_4^+]} = \frac{1.8 \times 10^{-5} \times 0.1999}{0.0001066} = 0.0338$ M

This is consistent! Our solution is:

Final Answer:

1. pH: $pH = 14 - pOH = 14 - (-\log[0.0338]) = 14 - 1.47 = 12.53$

2. Mg(OH)₂ Solubility: $s = 4.91 \times 10^{-9}$ M

3. Equilibrium Concentrations:

   • $[OH^-] = 0.0338$ M
   • $[NH_3] = 0.1999$ M
   • $[NH_4^+] = 0.0001066$ M
   • $[Mg^{2+}] = 4.91 \times 10^{-9}$ M
   • $[H^+] = 2.96 \times 10^{-13}$ M
   • $[Cl^-] = 0.1$ M

Solution 3: Mathematical Approach with Logarithms

Using pOH Approach

Let $pOH = -\log[OH^-]$

From base equilibrium: $pOH = pK_b + \log\left(\frac{[NH_4^+]}{[NH_3]}\right)$

From mass balance: $[NH_4^+] + [NH_3] = 0.2$

This approach leads to the same system but in logarithmic form, which can be useful for numerical solutions.

Effect of Adding 0.05 M NaOH

New Initial Conditions

• Additional $[OH^-] = 0.05$ M
• Total $[OH^-]_{initial} = 0.0838$ M

New Equilibrium

Following similar analysis:

New buffer equilibrium: $\frac{[NH_4^+]}{[NH_3]} = \frac{1.8 \times 10^{-5}}{0.0838} = 2.15 \times 10^{-4}$

$[NH_4^+] + [NH_3] = 0.2$

$[NH_3] = \frac{0.2}{1 + 2.15 \times 10^{-4}} = 0.19996$ M $[NH_4^+] = 4.3 \times 10^{-5}$ M

New Mg(OH)₂ solubility: $s = \frac{K_{sp}}{[OH^-]^2} = \frac{5.61 \times 10^{-12}}{(0.0838)^2} = 7.99 \times 10^{-10}$ M

Final pH: $pH = 14 - (-\log[0.0838]) = 14 - 1.08 = 12.92$

Key Concepts and Insights

Important Observations:

1. Common Ion Effect: High $[OH^-]$ from ammonia suppresses Mg(OH)₂ solubility
2. Buffer Capacity: The buffer gets consumed but still provides significant $[OH^-]$
3. Coupled Equilibria: Multiple equilibria must be solved simultaneously
4. Approximation Validity: Initial approximations may not be valid for complex systems

Problem-Solving Strategy:

1. Identify all equilibria involved in the system
2. Write mass balance equations for all components
3. Use charge balance as additional constraint
4. Make reasonable approximations and verify validity
5. Iterate if necessary to reach consistent solution

Common Mistakes

1. Ignoring coupled equilibria: Not considering how one equilibrium affects others
2. Invalid approximations: Making assumptions without checking validity
3. Mass balance errors: Forgetting to account for all species
4. Calculation errors: Mistakes in logarithmic calculations or algebraic manipulations
5. Unit consistency: Not maintaining consistent units throughout calculations

Related Problems

1. Different buffer system: Using acetate buffer instead of ammonia
2. Competing equilibria: Adding metal ions that form complexes with ammonia
3. Temperature effects: How changing temperature affects all equilibria
4. Multi-component systems: Adding additional salts or acids/bases
5. Kinetic considerations: Rate at which equilibrium is established

Video Solution

[Link to detailed video explanation with step-by-step solution and common mistake analysis]

Time Estimate: 15-20 minutes for complete solution Difficulty Level: Very Challenging 🔴 Success Rate: ~12% in first attempt Key Insight: Complex equilibrium systems often require iterative approaches and careful validation of assumptions