Calculate the concentration of `H^(+)` (aq) in 0.2 M solution of HCN. Given that the dissociation constant of HCN in water is `4.9 xx 10^(-10)`.

Answer

To solve the problem of calculating the concentration of \( H^+ \) in a 0.2 M solution of HCN, we can follow these steps:

### Step 1: Write the dissociation equation
HCN dissociates in water as follows:
\[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \]

### Step 2: Set up the expression for the dissociation constant (K)
The dissociation constant \( K_a \) for the reaction is given by:
\[ K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} \]

### Step 3: Define the concentrations at equilibrium
Let \( x \) be the concentration of \( H^+ \) ions that dissociate from HCN. At equilibrium:
- The concentration of \( H^+ \) will be \( x \).
- The concentration of \( CN^- \) will also be \( x \).
- The concentration of undissociated HCN will be \( 0.2 - x \).

### Step 4: Substitute into the expression for \( K_a \)
Given \( K_a = 4.9 \times 10^{-10} \):
\[ K_a = \frac{x \cdot x}{0.2 - x} = \frac{x^2}{0.2 - x} \]

### Step 5: Neglect \( x \) in comparison to 0.2
Since \( K_a \) is very small, we can assume \( x \) is negligible compared to 0.2 M:
\[ K_a \approx \frac{x^2}{0.2} \]

### Step 6: Rearrange to find \( x^2 \)
Substituting the value of \( K_a \):
\[ 4.9 \times 10^{-10} = \frac{x^2}{0.2} \]
\[ x^2 = 4.9 \times 10^{-10} \times 0.2 \]

### Step 7: Calculate \( x^2 \)
Calculating \( x^2 \):
\[ x^2 = 4.9 \times 10^{-10} \times 0.2 = 9.8 \times 10^{-11} \]

### Step 8: Solve for \( x \)
Taking the square root to find \( x \):
\[ x = \sqrt{9.8 \times 10^{-11}} \approx 9.899 \times 10^{-6} \]

### Step 9: Final result
Thus, the concentration of \( H^+ \) in the solution is approximately:
\[ [H^+] \approx 9.9 \times 10^{-6} \, \text{M} \]