Question

question 22 of 32
a solution of ammonia nh₃(aq) is at equilibrium. how would the equilibrium change if oh⁻ were added to the solution?

a. nh₄⁺ and oh⁻ would increase.

b. nh₄⁺ and oh⁻ would decrease.

c. nh₄⁺ and h⁺ would decrease.

d. nh₄⁺ would increase and oh⁻ would decrease.

Explanation:

1. First, recall the equilibrium reaction of ammonia in water: $\ce{NH_{3}(aq) + H_{2}O(l) <=> NH_{4}^{+}(aq) + OH^{-}(aq)}$.
2. According to Le Chatelier's principle, if we add $\ce{OH^{-}}$ (a product), the equilibrium will shift to the left to counteract the increase in $\ce{OH^{-}}$ concentration.
3. Shifting left means the formation of $\ce{NH_{3}}$ and $\ce{H_{2}O}$, so the concentration of $\ce{NH_{4}^{+}}$ will decrease.
4. Also, in an aqueous solution, $K_w = [\ce{H^{+}}][\ce{OH^{-}}]$ is constant at a given temperature. Since $[\ce{OH^{-}}]$ increases (from addition) and then slightly decreases due to equilibrium shift but is still higher than original, to keep $K_w$ constant, $[\ce{H^{+}}]$ must decrease (because if $[\ce{OH^{-}}]$ is relatively high, $[\ce{H^{+}}]=\frac{K_w}{[\ce{OH^{-}}]}$ will be low).
5. Analyzing options:

• Option A: $[\ce{NH_{4}^{+}}]$ should decrease, not increase. So A is wrong.
• Option B: $[\ce{OH^{-}}]$ initially increases (from addition) and then decreases a bit but is still higher than original, so it doesn't decrease overall in the way described. Also, $[\ce{NH_{4}^{+}}]$ decreases, but the reasoning for $[\ce{OH^{-}}]$ in B is incorrect.
• Option C: As explained, $[\ce{NH_{4}^{+}}]$ decreases (equilibrium shifts left) and $[\ce{H^{+}}]$ decreases (because $[\ce{OH^{-}}]$ is relatively high). This matches.
• Option D: $[\ce{NH_{4}^{+}}]$ should decrease, and $[\ce{OH^{-}}]$ is not decreasing (it was added and equilibrium shift only partially counteracts it). So D is wrong.

Answer:

C. $[\ce{NH_{4}^{+}}]$ and $[\ce{H^{+}}]$ would decrease.