Question

What is \([\text{NH}_4^+]\) in a solution that is \(0.02\text{M } \text{NH}_3\) and \(0.01\text{ M KOH}\)? \([K_b(\text{NH}_3) = 1.8 \times 10^{-5}]\)

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

Hint:

When a weak base is mixed with a strong base, you can find the concentration of the conjugate acid directly using a shortened shortcut formula: \( [\text{Conjugate Acid}] = K_b \times \frac{[\text{Weak Base}]}{[\text{Strong Base}]} \). Here, \( 1.8 \times 10^{-5} \times \frac{0.02}{0.01} = 3.6 \times 10^{-5}\text{ M}\).

Answer 1

The Correct Option is A

Concept: This is a problem based on the common ion effect.
• Potassium hydroxide (\(\text{KOH}\)) is a strong base and dissociates completely in water to yield a high concentration of hydroxyl ions (\(\text{OH}^-\)).
• Ammonia (\(\text{NH}_3\)) is a weak base that partially ionizes in water to form \(\text{NH}_4^+\) and \(\text{OH}^-\).
• The presence of the common ion \(\text{OH}^-\) from the strong base shifts the equilibrium of the weak base backward, suppressing its dissociation even further. Step 1: Determining total ion concentrations at equilibrium.
First, write down the complete dissociation of the strong base: \[ \text{KOH}(aq) \rightarrow \text{K}^+(aq) + \text{OH}^-(aq) \] Since \([\text{KOH}] = 0.01\text{ M}\), the concentration of \(\text{OH}^-\) contributed by \(\text{KOH}\) is \(0.01\text{ M}\). Next, set up the equilibrium reaction table for the weak base \(\text{NH}_3\): \[ \begin{array}{lcccc} \text{Reaction:} & \text{NH}_3(aq) + \text{H}_2\text{O}(l) & \rightleftharpoons & \text{NH}_4^+(aq) & + & \text{OH}^-(aq) \text{Initial (M):} & 0.02 & & 0 & & 0.01 \text{Change (M):} & -x & & +x & & +x \text{Equilibrium (M):} & 0.02 - x & & x & & 0.01 + x \end{array} \] Since \(K_b\) is very small (\(1.8 \times 10^{-5}\)) and dissociation is highly suppressed due to the common ion effect, we can approximate: \[ 0.02 - x \approx 0.02 \quad \text{and} \quad 0.01 + x \approx 0.01 \]

Step 2: Solving for the ammonium ion concentration \([\text{NH}_4^+]\).
Write the expression for the base dissociation constant (\(K_b\)): \[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \] Substitute the equilibrium concentrations and the given value of \(K_b\): \[ 1.8 \times 10^{-5} = \frac{(x)(0.01)}{0.02} \] Simplify the fraction: \[ 1.8 \times 10^{-5} = x \times \frac{1}{2} \] \[ x = 1.8 \times 10^{-5} \times 2 = 3.6 \times 10^{-5}\text{ M} \] Since \(x = [\text{NH}_4^+]\): \[ [\text{NH}_4^+ ] = 3.6 \times 10^{-5}\text{ M} \]