The pH of `0.01M HCN` solution for which `pK_(a)` is 4 is

Options

A. `0.47`
B. `1.2`
C. `3.0`
D. `4.0`

Explanation

To find the pH of a 0.01 M HCN solution with a given pKa of 4, we can follow these steps:

### Step 1: Calculate Ka from pKa
The relationship between pKa and Ka is given by the formula:
\[ pK_a = -\log K_a \]
Given that \( pK_a = 4 \), we can find Ka:
\[ K_a = 10^{-pK_a} = 10^{-4} \]

### Step 2: Set up the equilibrium expression
For the dissociation of HCN in water:
\[ HCN \rightleftharpoons H^+ + CN^- \]
Let \( x \) be the concentration of \( H^+ \) ions produced at equilibrium. The initial concentration of HCN is 0.01 M, and at equilibrium, the concentrations will be:
- \([HCN] = 0.01 - x\)
- \([H^+] = x\)
- \([CN^-] = x\)

### Step 3: Write the expression for Ka
The expression for Ka is:
\[ K_a = \frac{[H^+][CN^-]}{[HCN]} = \frac{x \cdot x}{0.01 - x} = \frac{x^2}{0.01 - x} \]
Since \( K_a = 10^{-4} \), we can set up the equation:
\[ 10^{-4} = \frac{x^2}{0.01 - x} \]

### Step 4: Make an assumption
Since \( K_a \) is small, we can assume that \( x \) is much smaller than 0.01 M, allowing us to simplify:
\[ 10^{-4} \approx \frac{x^2}{0.01} \]
This leads to:
\[ x^2 = 10^{-4} \cdot 0.01 = 10^{-6} \]

### Step 5: Solve for x
Taking the square root of both sides:
\[ x = \sqrt{10^{-6}} = 10^{-3} \text{ M} \]
This means the concentration of \( H^+ \) ions is \( 10^{-3} \) M.

### Step 6: Calculate pH
Using the formula for pH:
\[ pH = -\log[H^+] = -\log(10^{-3}) = 3 \]

### Final Answer
The pH of the 0.01 M HCN solution is **3.0**.

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