Given `K_(a)` values of `5.76xx10^(-10)` and `4.8xx10^(-10)` for `NH_(4)^(+)` and HCN respectively. What is the equilibrium constant for the following reaction?
`NH_(4)^(+)(aq.)+CN^(-)(aq.)hArrNH_(3)(aq.)+HCN(aq.)`

Options

A. `0.83`
B. `1.2`
C. `8.0xx10^(-11)`
D. `27.6xx10^(-10)`

Answer

To find the equilibrium constant for the reaction:

\[ \text{NH}_4^+(aq) + \text{CN}^-(aq) \rightleftharpoons \text{NH}_3(aq) + \text{HCN}(aq) \]

we will use the given \( K_a \) values for the dissociation of \( \text{NH}_4^+ \) and \( \text{HCN} \).

### Step 1: Write the dissociation reactions and their \( K_a \) values

1. The dissociation of ammonium ion (\( \text{NH}_4^+ \)):
\[ \text{NH}_4^+(aq) \rightleftharpoons \text{NH}_3(aq) + \text{H}^+(aq) \]
The equilibrium constant for this reaction is given as:
\[ K_{a1} = 5.76 \times 10^{-10} \]

2. The dissociation of hydrogen cyanide (\( \text{HCN} \)):
\[ \text{HCN}(aq) \rightleftharpoons \text{H}^+(aq) + \text{CN}^-(aq) \]
The equilibrium constant for this reaction is given as:
\[ K_{a2} = 4.8 \times 10^{-10} \]

### Step 2: Reverse the second reaction

To find the equilibrium constant for the desired reaction, we need to reverse the dissociation of \( \text{HCN} \):
\[ \text{H}^+(aq) + \text{CN}^-(aq) \rightleftharpoons \text{HCN}(aq) \]
When we reverse the reaction, the new equilibrium constant (\( K_{a2}' \)) is the inverse of \( K_{a2} \):
\[ K_{a2}' = \frac{1}{K_{a2}} = \frac{1}{4.8 \times 10^{-10}} \]

### Step 3: Combine the reactions

Now, we can add the two reactions together:
1. \( \text{NH}_4^+(aq) \rightleftharpoons \text{NH}_3(aq) + \text{H}^+(aq) \) (with \( K_{a1} \))
2. \( \text{H}^+(aq) + \text{CN}^-(aq) \rightleftharpoons \text{HCN}(aq) \) (with \( K_{a2}' \))

The overall reaction becomes:
\[ \text{NH}_4^+(aq) + \text{CN}^-(aq) \rightleftharpoons \text{NH}_3(aq) + \text{HCN}(aq) \]

### Step 4: Calculate the equilibrium constant for the overall reaction

The equilibrium constant for the overall reaction (\( K_{eq} \)) is the product of the equilibrium constants of the individual reactions:
\[ K_{eq} = K_{a1} \times K_{a2}' \]
Substituting the values:
\[ K_{eq} = K_{a1} \times \frac{1}{K_{a2}} = 5.76 \times 10^{-10} \times \frac{1}{4.8 \times 10^{-10}} \]

### Step 5: Simplify the expression

Calculating the above expression:
\[ K_{eq} = \frac{5.76}{4.8} = 1.2 \]

### Final Answer

Thus, the equilibrium constant for the reaction is:
\[ K_{eq} = 1.2 \]

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