Question

What is [NH₄⁺] in a solution that is 0.02 M NH₃ and 0.01 M KOH? $$ K_b (\text{NH}_3) = 1.8 \times 10^{-5} $$ 

• A. \( 3.6 \times 10^{-5} \, {M} \)
• B. \( 1.8 \times 10^{-5} \, {M} \)
• C. \( 0.9 \times 10^{-5} \, {M} \)
• D. \( 7.2 \times 10^{-5} \, {M} \)

Hint:

When calculating the concentration of ions in a weak base solution, consider the contribution of \({OH}^-\) ions from both the base dissociation and any added hydroxide salts.

Answer 1

The Correct Option is A

In this solution, ammonia (NH₃) reacts with water to form ammonium ions (NH₄⁺) and hydroxide ions (OH⁻). KOH provides additional hydroxide ions to the solution, affecting the equilibrium. The equilibrium reaction for ammonia is: \[ {NH}_3 + {H}_2{O} \rightleftharpoons {NH}_4^+ + {OH}^- \] The \( K_b \) expression for ammonia is: \[ K_b = \frac{[{NH}_4^+][{OH}^-]}{[{NH}_3]} \] However, KOH also dissociates in water: \[ {KOH} \rightarrow {K}^+ + {OH}^- \] This adds to the concentration of OH⁻ ions already present from the ammonia dissociation. 1. Total OH⁻ concentration: The hydroxide ions come from both KOH and the ammonia equilibrium. The concentration of OH⁻ from KOH is 0.01 M. 2. Set up the equilibrium expression: Let \( x \) be the concentration of NH₄⁺ produced from ammonia dissociation. 
The equilibrium concentrations will be: - \( [{NH}_3] = 0.02 - x \) - \( [{NH}_4^+] = x \) - \( [{OH}^-] = 0.01 + x \) 3. Substitute into the expression for \( K_b \): \[ 1.8 \times 10^{-5} = \frac{x(0.01 + x)}{0.02 - x} \] Assume \( x \) is small compared to 0.01, so \( 0.01 + x \approx 0.01 \), and solve for \( x \): \[ 1.8 \times 10^{-5} = \frac{x \times 0.01}{0.02} \] \[ x = \frac{1.8 \times 10^{-5} \times 0.02}{0.01} = 3.6 \times 10^{-5} \, {M} \] 
Thus, the concentration of \( {NH}_4^+ \) is \( 3.6 \times 10^{-5} \, {M} \).