Question

question 1 (2 points)
for the following reaction at equilibrium
nh₃ + h₂o ↔ nh₄⁺¹ + oh⁻¹
1 2 use the order given in the reaction
k_b = --------------------
3
k_b is a constant. what happens to the nh₃ if some base is added and the oh⁻¹
increases ? 4
what happens to the nh₃ if some acid is added and the oh⁻¹ decreases ? 5
a. nh₃ b. h₂o c. nh₄⁺¹ d. oh⁻¹
e. increases f. decreases

Explanation:

Step1: Recall \( K_b \) formula

For a base dissociation reaction \( \text{Base} + \text{H}_2\text{O}
ightleftharpoons \text{Conjugate Acid} + \text{OH}^- \), the \( K_b \) expression is \( K_b=\frac{[\text{Conjugate Acid}][\text{OH}^-]}{[\text{Base}]} \). For \( \text{NH}_3 + \text{H}_2\text{O}
ightleftharpoons \text{NH}_4^+ + \text{OH}^- \), the products are \( \text{NH}_4^+ \) (1st product) and \( \text{OH}^- \) (2nd product), reactant (denominator) is \( \text{NH}_3 \). So:

• 1: \( \text{NH}_4^{+1} \) (option C)
• 2: \( \text{OH}^{-1} \) (option D)
• 3: \( \text{NH}_3 \) (option A)

Step2: Le Chatelier’s Principle (Base added, \( [\text{OH}^-] \) increases)

When base is added, \( [\text{OH}^-] \) increases. The system shifts left to counteract (decrease \( [\text{OH}^-] \)). Shifting left means more \( \text{NH}_3 \) is formed (since reaction goes reverse: \( \text{NH}_4^+ + \text{OH}^-
ightarrow \text{NH}_3 + \text{H}_2\text{O} \)). So 4: increases (option E).

Step3: Le Chatelier’s Principle (Acid added, \( [\text{OH}^-] \) decreases)

Acid reacts with \( \text{OH}^- \), so \( [\text{OH}^-] \) decreases. System shifts right to produce more \( \text{OH}^- \) (reaction: \( \text{NH}_3 + \text{H}_2\text{O}
ightarrow \text{NH}_4^+ + \text{OH}^- \)). Shifting right means \( \text{NH}_3 \) is consumed, so \( [\text{NH}_3] \) decreases (option F).

Answer:

1: C. \( \text{NH}_4^{+1} \)
2: D. \( \text{OH}^{-1} \)
3: A. \( \text{NH}_3 \)
4: E. increases
5: F. decreases