To find the equilibrium constant \( K_C \) for the reaction involving ammonia (\( \text{NH}_3 \)) and ammonium ions (\( \text{NH}_4^+ \)), we first need to establish the reaction:
\[ \text{NH}_3(aq) + \text{H}_2O(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq) \]
This expression involves equilibrium concentrations, where \( K_C \) can be calculated as:
\[ K_C = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]
Step 1: Find the moles of ammonium ions (\( \text{NH}_4^+ \)) at equilibrium
The student placed 7 moles of ammonia into a 100 mL flask. Since 100 mL is equal to 0.1 L, the initial concentration of ammonia (\( \text{NH}_3 \)) is:
\[ \text{Initial concentration of } \text{NH}_3 = \frac{7 \text{ moles}}{0.1 \text{ L}} = 70 \text{ M} \]
Next, we add 50 mL of deionized water to this solution. The final volume of the solution is:
\[ V_{\text{final}} = 100 \text{ mL} + 50 \text{ mL} = 150 \text{ mL} = 0.15 \text{ L} \]
Now, the new concentration of ammonia (\( [\text{NH}_3] \)) after dilution is:
\[ [\text{NH}_3] = \frac{7 \text{ moles}}{0.15 \text{ L}} \approx 46.67 \text{ M} \]
Step 2: Determine the equilibrium concentrations
Given that at equilibrium, the concentration of ammonium ions (\( \text{NH}_4^+ \)) is \( 0.2 , \text{M} \).
The concentration of hydroxide ions (\( [\text{OH}^-] \)) will usually be approximately equal to the concentration of \( \text{NH}_4^+ \) in this case, assuming a simple reaction where each molecule of \( \text{NH}_4^+ \) contributes one \( \text{OH}^- \).
Thus, we have:
\[ [\text{NH}_4^+] = 0.2 , \text{M} \]
For \( [\text{OH}^-] \):
\[ [\text{OH}^-] \approx 0.2 , \text{M} \]
Step 3: Determine the concentration of ammonia (\( [\text{NH}_3] \)) at equilibrium
Initial \( [\text{NH}_3] = 46.67 \text{ M} \)
If 0.2 M of \( \text{NH}_4^+ \) is formed at equilibrium, we assume that \( 0.2 , \text{M} \) of \( \text{NH}_3 \) has been converted to \( \text{NH}_4^+ \):
\[ [\text{NH}3]{\text{equilibrium}} = [\text{NH}3]{\text{initial}} - [\text{NH}_4^+] = 46.67 , \text{M} - 0.2 , \text{M} = 46.47 , \text{M} \]
Step 4: Calculate \( K_C \)
Now we can substitute the equilibrium concentrations into the \( K_C \) expression:
\[ K_C = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]
\[ K_C = \frac{(0.2)(0.2)}{46.47} \approx \frac{0.04}{46.47} \approx 0.000860 \]
Final Results
- Equilibrium Constant (\( K_C \)): \( K_C \approx 0.000860 \)
- Moles of Ammonium (\( \text{NH}_4^+ \)) at Equilibrium: \[ \text{Moles of } \text{NH}_4^+ = 0.2 , \text{M} \times 0.15 , \text{L} = 0.03 , \text{moles} \]
Conclusion:
- Equilibrium constant (\( K_C \)): approximately 0.000860
- Moles of ammonium (\( \text{NH}_4^+ \)) at equilibrium: 0.03 moles.