A student place 7 moles of ammonia (NH3) in a 100 ml flask with 50 ml deionized water. At equilibrium, 0.2 m (NH4+) was found in a flask.what is the equilibrium constant (KC) for this reaction? What is the mole of ammonium (NH4+) at the equilibrium?

Answer 1

To determine the equilibrium constant \( K_c \) for the reaction involving ammonia (NH\(_3\)) and ammonium ion (NH\(_4^+\)), we start by writing the appropriate equilibrium reaction:

\[ \text{NH}_3 (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{NH}_4^+ (aq) + \text{OH}^- (aq) \]

Given data:

• Moles of NH\(_3\) = 7 moles
• Volume of the flask = 100 mL = 0.1 L
• Concentration of \(\text{NH}_4^+\) at equilibrium = 0.2 M

First, we calculate the total volume of the solution. Since we have 100 mL of the flask and 50 mL of deionized water, the total volume is:

\[ V_{total} = 100 , \text{mL} + 50 , \text{mL} = 150 , \text{mL} = 0.15 , \text{L} \]

Step 1: Calculate equilibrium concentration of NH\(_4^+\)

At equilibrium, the concentration of \(\text{NH}_4^+\) is given as 0.2 M. This concentration corresponds to the moles of \(\text{NH}_4^+\) in 0.15 L:

\[ \text{moles of } \text{NH}_4^+ = 0.2 , \text{M} \times 0.15 , \text{L} = 0.03 , \text{moles} \]

Step 2: Calculate change in moles

The mole of \(\text{NH}_4^+\) produced at equilibrium is 0.03 moles. To find the change in moles for \(\text{NH}_3\) and \(\text{OH}^-\), we can denote the change in concentration based on the stoichiometry of the reaction.

Assuming \( x \) moles of NH\(_3\) react to form 0.03 moles of \(\text{NH}_4^+\), we write:

\[ x = 0.03 , \text{moles of } \text{NH}_3 \]

Step 3: Calculate remaining moles of NH\(_3\)

The initial moles of NH\(_3\) was 7 moles. Thus, the moles of NH\(_3\) at equilibrium would be:

\[ \text{Moles of NH}_3 = 7 - 0.03 = 6.97 , \text{moles} \]

Step 4: Calculate the concentration of NH\(_3\) at equilibrium

Now, we convert the moles of \(\text{NH}_3\) at equilibrium to concentration:

\[ \text{Concentration of NH}_3 = \frac{6.97 , \text{moles}}{0.15 , \text{L}} = 46.47 , \text{M} \]

Step 5: Calculate concentration of OH\(^-\)

By stoichiometry, the concentration of OH\(^-\) produced would equal the concentration of \(\text{NH}_4^+\):

\[ [\text{OH}^-] = 0.2 , \text{M} \]

Step 6: Calculate \( K_c \)

The equilibrium constant \( K_c \) can be expressed as:

\[ K_c = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]

Substituting in the concentrations we calculated:

\[ K_c = \frac{(0.2)(0.2)}{46.47} = \frac{0.04}{46.47} \approx 0.000859 \]

Final Answers

1. The equilibrium constant \( K_c \) is approximately 0.000859.
2. The moles of ammonium (\(\text{NH}_4^+\)) at equilibrium is 0.03 moles.