pH of 0.5 M aqueous NaCN solution is `(pK_a of HCN = 9.3, log 5 = 0.7)`

Options

A. 10.3
B. 9.5
C. 10.6
D. 11.5

Answer

To find the pH of a 0.5 M aqueous NaCN solution, we can use the following steps:

### Step 1: Identify the components of NaCN
NaCN is a salt formed from the weak acid HCN (hydrocyanic acid) and the strong base NaOH (sodium hydroxide). In solution, NaCN dissociates into Na⁺ and CN⁻ ions. The CN⁻ ion acts as a weak base.

### Step 2: Use the formula for pH of a salt solution
The pH of a solution of a salt derived from a weak acid and a strong base can be calculated using the formula:
\[ \text{pH} = 7 + \frac{1}{2} pK_a + \frac{1}{2} \log C \]
where \( C \) is the concentration of the salt solution and \( pK_a \) is the dissociation constant of the weak acid.

### Step 3: Substitute the values into the formula
Given:
- \( pK_a \) of HCN = 9.3
- Concentration \( C \) = 0.5 M

Now, we can substitute these values into the formula:
\[ \text{pH} = 7 + \frac{1}{2} (9.3) + \frac{1}{2} \log (0.5) \]

### Step 4: Calculate \( \frac{1}{2} pK_a \)
Calculate \( \frac{1}{2} \times 9.3 \):
\[ \frac{1}{2} \times 9.3 = 4.65 \]

### Step 5: Calculate \( \log (0.5) \)
Using the given information that \( \log 5 = 0.7 \), we can find \( \log (0.5) \):
Since \( 0.5 = \frac{5}{10} \), we can use the logarithmic identity:
\[ \log (0.5) = \log (5) - \log (10) = 0.7 - 1 = -0.3 \]

### Step 6: Substitute \( \log (0.5) \) into the equation
Now substitute \( \log (0.5) \) into the pH equation:
\[ \text{pH} = 7 + 4.65 + \frac{1}{2} (-0.3) \]
\[ \text{pH} = 7 + 4.65 - 0.15 \]

### Step 7: Calculate the final pH
Now, calculate the final pH:
\[ \text{pH} = 7 + 4.65 - 0.15 = 11.5 \]

### Final Answer
The pH of the 0.5 M aqueous NaCN solution is **11.5**.

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