First, let's write the dissociation equilibrium reaction of ammonia (\(NH_3\)) in water:
\[ NH_3 (aq) + H_2O (l) \rightleftharpoons NH_4^+ (aq) + OH^- (aq) \]
Step 1: Calculate the equilibrium concentration of \(NH_4^+\)
The equilibrium concentration of \(NH_4^+\) is given as \(0.2 , m\) (molality). To convert this to molarity (\(M\)), we need to determine the volume of the solution after the ammonia and water are mixed.
- The total volume of the solution is the sum of the volumes of the ammonia and the water: \[ \text{Total Volume} = 100 , \text{mL} + 50 , \text{mL} = 150 , \text{mL} = 0.150 , \text{L} \]
The molality is given in terms of moles per kg of solvent, so we need a little more information:
- We have 50 mL of deionized water. The density of water is approximately \(1 , \text{g/mL}\), therefore: \[ \text{Mass of water} = 50 , \text{g} = 0.050 , \text{kg} \]
Using the molality: \[ \text{moles of } NH_4^+ = 0.2 , \text{m} \times 0.050 , \text{kg} = 0.01 , \text{moles} \]
Step 2: Calculate the equilibrium concentration of \(NH_4^+\)
Now, to find the molarity of \(NH_4^+\): \[ [\text{NH}_4^+] = \frac{0.01 , \text{moles}}{0.150 , \text{L}} = 0.0667 , M \]
Step 3: Set up the equilibrium expression
The equilibrium expression for the reaction can be written as:
\[ K_C = \frac{[NH_4^+][OH^-]}{[NH_3]} \]
Since \(NH_4^+\) and \(OH^-\) are produced in a 1:1 molar ratio, we assume that at equilibrium, the concentrations of \(NH_4^+\) and \(OH^-\) are equal, thus \( [OH^-] = [NH_4^+] = 0.067 , M\).
Step 4: Determine the concentration of \(NH_3\)
We started with 7 moles of \(NH_3\) in 0.150 L: \[ [NH_3]_{\text{initial}} = \frac{7 , \text{moles}}{0.150 , \text{L}} = 46.67 , M \]
At equilibrium, the change in concentration of \(NH_3\) is equal to the amount that reacted to form \(NH_4^+\) and \(OH^-\), which we calculated as \(0.067 , M\): \[ [NH_3]_{\text{equilibrium}} = 46.67 , M - 0.067 , M \approx 46.6 , M \]
Step 5: Calculate \(K_C\)
Substituting the values we have into the equilibrium expression:
\[ K_C = \frac{(0.0667)(0.067)}{46.6} \]
Calculating gives us:
\[ K_C = \frac{0.00449}{46.6} \approx 0.0000963 \text{ or } 9.63 \times 10^{-5} \]
Conclusion
- The equilibrium constant (\(K_C\)) for the reaction is \(9.63 \times 10^{-5}\).
- The moles of \(NH_4^+\) at equilibrium is \(0.01\) moles.