The pH of `0.001 M HCN` is

Options

A. `3`
B. `11`
C. Between `3` & `7`
D. `7`

Explanation

To find the pH of a 0.001 M solution of HCN (hydrocyanic acid), we need to follow these steps:

### Step 1: Identify the nature of HCN
HCN is a weak acid, which means it does not completely dissociate in solution. Therefore, we need to use the acid dissociation constant (Ka) to find the concentration of hydrogen ions (H⁺) in the solution.

### Step 2: Write the dissociation equation
The dissociation of HCN in water can be represented as:
\[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \]

### Step 3: Set up the equilibrium expression
Let \( x \) be the concentration of H⁺ ions that dissociate from HCN at equilibrium. The initial concentration of HCN is 0.001 M, and at equilibrium, we have:
- [HCN] = 0.001 - x
- [H⁺] = x
- [CN⁻] = x

The equilibrium expression for the dissociation of HCN is given by:
\[ K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} \]

### Step 4: Use the Ka value
The Ka value for HCN is approximately \( 6.2 \times 10^{-10} \). Substituting the equilibrium concentrations into the expression gives:
\[ K_a = \frac{x \cdot x}{0.001 - x} \approx \frac{x^2}{0.001} \]
(since \( x \) is very small compared to 0.001, we can approximate \( 0.001 - x \approx 0.001 \)).

### Step 5: Solve for x
Now, substituting the Ka value:
\[ 6.2 \times 10^{-10} = \frac{x^2}{0.001} \]

Rearranging gives:
\[ x^2 = 6.2 \times 10^{-10} \times 0.001 \]
\[ x^2 = 6.2 \times 10^{-13} \]
\[ x = \sqrt{6.2 \times 10^{-13}} \]
\[ x \approx 7.87 \times 10^{-7} \]

### Step 6: Calculate the pH
Now that we have the concentration of H⁺ ions, we can calculate the pH:
\[ \text{pH} = -\log[\text{H}^+] = -\log(7.87 \times 10^{-7}) \]

Calculating the logarithm:
\[ \text{pH} \approx 6.10 \]

### Final Answer
The pH of 0.001 M HCN is approximately 6.10.

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