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 find the \( K_a \) value of HCN given the \( K_a \) of acetic acid (CH₃COOH) and their respective pH values, we can follow these steps:

### Step 1: Understand the relationship between pH and \([H^+]\)
The pH of a solution is related to the concentration of hydrogen ions \([H^+]\) by the formula:
\[ [H^+] = 10^{-\text{pH}} \]

### Step 2: Calculate \([H^+]\) for both acids
1. For acetic acid (CH₃COOH) with pH = 3.0:
\[ [H^+]_{CH_3COOH} = 10^{-3} \, \text{M} \]

2. For HCN with pH = 2.0:
\[ [H^+]_{HCN} = 10^{-2} \, \text{M} \]

### Step 3: Set up the equilibrium expressions
For weak acids, the concentration \( C \) can be expressed in terms of \( K_a \) and \([H^+]\):
\[ C = \frac{[H^+]^2}{K_a} \]
Since both acids have equal concentrations, we can denote them as \( C \).

### Step 4: Write the equations for both acids
1. For acetic acid (CH₃COOH):
\[ C = \frac{(10^{-3})^2}{K_{a1}} \quad \text{where } K_{a1} = 1.8 \times 10^{-5} \]
\[ C = \frac{10^{-6}}{1.8 \times 10^{-5}} \]

2. For HCN:
\[ C = \frac{(10^{-2})^2}{K_{a2}} \quad \text{where } K_{a2 \text{ is what we need to find}} \]
\[ C = \frac{10^{-4}}{K_{a2}} \]

### Step 5: Set the two expressions for \( C \) equal to each other
Since both expressions for \( C \) are equal:
\[ \frac{10^{-6}}{1.8 \times 10^{-5}} = \frac{10^{-4}}{K_{a2}} \]

### Step 6: Solve for \( K_{a2} \)
Cross-multiplying gives:
\[ 10^{-6} \cdot K_{a2} = 10^{-4} \cdot (1.8 \times 10^{-5}) \]
\[ K_{a2} = \frac{10^{-4} \cdot 1.8 \times 10^{-5}}{10^{-6}} \]
\[ K_{a2} = 1.8 \times 10^{-4} \cdot 10^{2} \]
\[ K_{a2} = 1.8 \times 10^{-2} \]

### Final Answer
The \( K_a \) value of HCN is:
\[ K_{a2} = 1.8 \times 10^{-3} \]