The concentraion of `CH_(3)COOH` and HCN is equal. Their pH 3.0 and 2.0 respectively. If `K_(a)` of `CH_(3)COOH` is `1.8 xx 10^(-5)` then `K_(a)` value of HCN is :

Options

A. `1.8 xx 10^(-7)`
B. `1.8 xx 10^(-3)`
C. `1.8 xx 10^(-5)`
D. None of these

Answer

To solve the problem, we need to find the \( K_a \) value of HCN given the \( K_a \) value of acetic acid (CH₃COOH) and the pH values of both acids. Here’s a step-by-step solution:

### Step 1: Calculate the concentration of hydrogen ions \([H^+]\) for CH₃COOH
Given the pH of acetic acid (CH₃COOH) is 3.0, we can find the concentration of hydrogen ions using the formula:
\[ [H^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

### Step 2: Calculate the concentration of hydrogen ions \([H^+]\) for HCN
Given the pH of HCN is 2.0, we can find the concentration of hydrogen ions similarly:
\[ [H^+] = 10^{-\text{pH}} = 10^{-2} \, \text{M} \]

### Step 3: Use the expression for \( K_a \)
The dissociation of a weak acid can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
The expression for the acid dissociation constant \( K_a \) is:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
For both acids, since the concentrations are equal, we can denote the concentration of the acids as \( C \).

### Step 4: Set up the equation for \( K_a \) of CH₃COOH
For acetic acid:
\[ K_{a1} = \frac{[H^+]^2}{C - [H^+]} \approx \frac{(10^{-3})^2}{C} \quad \text{(since } [H^+] \text{ is much smaller than } C\text{)} \]
Given \( K_{a1} = 1.8 \times 10^{-5} \):
\[ 1.8 \times 10^{-5} = \frac{(10^{-3})^2}{C} \]
\[ C = \frac{(10^{-3})^2}{1.8 \times 10^{-5}} = \frac{10^{-6}}{1.8 \times 10^{-5}} \approx 0.0556 \, \text{M} \]

### Step 5: Set up the equation for \( K_a \) of HCN
For HCN, we can express \( K_{a2} \) as:
\[ K_{a2} = \frac{[H^+]^2}{C - [H^+]} \approx \frac{(10^{-2})^2}{C} \]
Substituting the value of \( C \):
\[ K_{a2} = \frac{(10^{-2})^2}{0.0556} = \frac{10^{-4}}{0.0556} \approx 1.8 \times 10^{-3} \]

### Conclusion
Thus, the \( K_a \) value of HCN is approximately:
\[ K_{a2} \approx 1.8 \times 10^{-3} \]