If `pK_(a)` for `CN^(-)` at `25^(@)C` is 4.7, the pH of 0.5 M aqueous NaCN solution is

Options

A. 10
B. 11.5
C. 11
D. 12

Answer

To find the pH of a 0.5 M aqueous NaCN solution given that the pKa for CN⁻ (the conjugate base of HCN) is 4.7, we can follow these steps:

### Step 1: Determine pKb from pKa
We know that:
\[ pK_a + pK_b = pK_w \]
At 25°C, \( pK_w = 14 \).

Given \( pK_a = 4.7 \):
\[ pK_b = pK_w - pK_a = 14 - 4.7 = 9.3 \]

### Step 2: Use the pKb to find the concentration of OH⁻
Since CN⁻ is a weak base, we can use the following equilibrium expression for the reaction:
\[ CN^- + H_2O \rightleftharpoons HCN + OH^- \]

Let \( x \) be the concentration of OH⁻ produced at equilibrium. The initial concentration of CN⁻ is 0.5 M, and at equilibrium, we have:
- [CN⁻] = 0.5 - x
- [HCN] = x
- [OH⁻] = x

Using the base dissociation constant \( K_b \):
\[ K_b = \frac{[HCN][OH^-]}{[CN^-]} \]

Substituting the equilibrium concentrations:
\[ K_b = \frac{x \cdot x}{0.5 - x} = \frac{x^2}{0.5 - x} \]

Since CN⁻ is a weak base, we can assume \( x \) is small compared to 0.5, so:
\[ K_b \approx \frac{x^2}{0.5} \]

### Step 3: Calculate Kb using pKb
We can find \( K_b \) from \( pK_b \):
\[ K_b = 10^{-pK_b} = 10^{-9.3} \]

### Step 4: Set up the equation
Setting the expression for \( K_b \) equal to the calculated \( K_b \):
\[ 10^{-9.3} = \frac{x^2}{0.5} \]

### Step 5: Solve for x
Rearranging gives:
\[ x^2 = 0.5 \cdot 10^{-9.3} \]
\[ x = \sqrt{0.5 \cdot 10^{-9.3}} \]

### Step 6: Calculate the concentration of OH⁻
Now calculate \( x \) (which is the concentration of OH⁻):
\[ x \approx \sqrt{0.5 \cdot 5.012 \times 10^{-10}} \]
\[ x \approx \sqrt{2.506 \times 10^{-10}} \]
\[ x \approx 5.006 \times 10^{-5} \, \text{M} \]

### Step 7: Calculate pOH
Now we can find the pOH:
\[ pOH = -\log[OH^-] = -\log(5.006 \times 10^{-5}) \]
\[ pOH \approx 4.3 \]

### Step 8: Calculate pH
Finally, we can find the pH:
\[ pH = 14 - pOH = 14 - 4.3 = 9.7 \]

### Final Answer
The pH of the 0.5 M aqueous NaCN solution is approximately **9.7**.

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