Concentration of `CN^(-) " in " 0.1` M HCN is (Given :`K_(a) = 4 xx 10^(-10)`)

Options

A. `2.5 xx 10^(-6) M`
B. `4.5 xx 10^(-6) M`
C. `6.3 xx 10^(-6) M`
D. `9.2 xx 10^(-6) M`

Answer

To find the concentration of `CN^(-)` in a `0.1 M` solution of `HCN`, we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of `HCN` can be represented as:
\[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \]

### Step 2: Define the initial concentration and change in concentration
Let:
- The initial concentration of `HCN` be `C = 0.1 M`
- The degree of dissociation be `α` (alpha)

At the start (initially):
- \[ [\text{HCN}] = 0.1 \]
- \[ [\text{H}^+] = 0 \]
- \[ [\text{CN}^-] = 0 \]

At equilibrium:
- \[ [\text{HCN}] = 0.1 - Cα \]
- \[ [\text{H}^+] = Cα \]
- \[ [\text{CN}^-] = Cα \]

### Step 3: Write the expression for the acid dissociation constant \( K_a \)
The expression for the acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} \]

Substituting the equilibrium concentrations:
\[ K_a = \frac{(Cα)(Cα)}{(C(1 - α))} \]

### Step 4: Substitute known values into the \( K_a \) expression
Given \( K_a = 4 \times 10^{-10} \) and \( C = 0.1 \):
\[ 4 \times 10^{-10} = \frac{(0.1α)(0.1α)}{0.1(1 - α)} \]

This simplifies to:
\[ 4 \times 10^{-10} = \frac{0.01α^2}{0.1(1 - α)} \]
\[ 4 \times 10^{-10} = \frac{0.1α^2}{1 - α} \]

### Step 5: Assume \( α \) is small
Since \( K_a \) is very small, we can assume that \( α \) is small compared to 1, which allows us to approximate \( 1 - α \approx 1 \):
\[ 4 \times 10^{-10} \approx 0.1α^2 \]

### Step 6: Solve for \( α \)
Rearranging gives:
\[ α^2 = \frac{4 \times 10^{-10}}{0.1} \]
\[ α^2 = 4 \times 10^{-9} \]
\[ α = \sqrt{4 \times 10^{-9}} \]
\[ α = 2 \times 10^{-5} \]

### Step 7: Calculate the concentration of \( CN^- \)
Now, we can find the concentration of \( CN^- \):
\[ [CN^-] = Cα = 0.1 \times 2 \times 10^{-5} \]
\[ [CN^-] = 2 \times 10^{-6} \, M \]

### Final Answer
The concentration of `CN^(-)` in `0.1 M HCN` is:
\[ [CN^-] = 2 \times 10^{-6} \, M \]

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