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 the cyanide ion \( CN^- \) in a 0.1 M solution of HCN, we can follow these steps:

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

### Step 2: Set up the expression for \( K_a \)
The acid dissociation constant \( K_a \) is given by the formula:
\[ K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} \]
Given that \( K_a = 4 \times 10^{-10} \).

### Step 3: Define the initial concentrations
Initially, we have:
- \([\text{HCN}] = 0.1 \, \text{M}\)
- \([\text{H}^+] = 0\)
- \([\text{CN}^-] = 0\)

### Step 4: Set up the change in concentration
Let \( x \) be the concentration of \( \text{H}^+ \) and \( \text{CN}^- \) formed at equilibrium. Thus, at equilibrium:
- \([\text{H}^+] = x\)
- \([\text{CN}^-] = x\)
- \([\text{HCN}] = 0.1 - x\)

### Step 5: Substitute into the \( K_a \) expression
Substituting these values into the \( K_a \) expression gives:
\[ K_a = \frac{x \cdot x}{0.1 - x} = \frac{x^2}{0.1 - x} \]

### Step 6: Assume \( x \) is negligible
Since HCN is a weak acid, we can assume that \( x \) is very small compared to 0.1 M, so \( 0.1 - x \approx 0.1 \). Thus, the equation simplifies to:
\[ K_a \approx \frac{x^2}{0.1} \]

### Step 7: Solve for \( x \)
Now substituting the value of \( K_a \):
\[ 4 \times 10^{-10} = \frac{x^2}{0.1} \]
Multiplying both sides by 0.1:
\[ 4 \times 10^{-11} = x^2 \]
Taking the square root of both sides:
\[ x = \sqrt{4 \times 10^{-11}} = 2 \times 10^{-5} \, \text{M} \]

### Step 8: Conclusion
Thus, the concentration of \( CN^- \) in the solution is:
\[ [\text{CN}^-] = 2 \times 10^{-5} \, \text{M} \]