What is the pH of a `0.50M` aqueous NaCN solution ? `(pK_(b)of CN^(-)=4.70)`

Explanation

To find the pH of a 0.50 M aqueous NaCN solution, we can follow these steps:

### Step 1: Understand the Hydrolysis of NaCN
NaCN is a salt formed from a strong base (NaOH) and a weak acid (HCN). In aqueous solution, NaCN will hydrolyze to produce hydroxide ions (OH⁻) and HCN:
\[ \text{NaCN} \rightarrow \text{Na}^+ + \text{CN}^- \]
\[ \text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^- \]

### Step 2: Use the Given pKb to Find pKa
We are given the pKb of CN⁻, which is 4.70. We can find the pKa using the relationship:
\[ \text{pKa} + \text{pKb} = 14 \]
Thus,
\[ \text{pKa} = 14 - \text{pKb} = 14 - 4.70 = 9.30 \]

### Step 3: Calculate pOH Using the Hydrolysis Equation
For the hydrolysis of a salt of a strong base and weak acid, we can use the formula:
\[ \text{pOH} = \frac{1}{2} \left( \text{pK}_w - \text{pK}_a - \log C \right) \]
Where:
- \( \text{pK}_w = 14 \)
- \( \text{pK}_a = 9.30 \)
- \( C = 0.50 \, \text{M} \)

Substituting the values:
\[ \text{pOH} = \frac{1}{2} \left( 14 - 9.30 - \log(0.50) \right) \]

### Step 4: Calculate the Logarithm
Calculate \( \log(0.50) \):
\[ \log(0.50) \approx -0.301 \]

### Step 5: Substitute and Calculate pOH
Now substitute back into the pOH equation:
\[ \text{pOH} = \frac{1}{2} \left( 14 - 9.30 + 0.301 \right) \]
\[ \text{pOH} = \frac{1}{2} \left( 4.701 \right) \]
\[ \text{pOH} = 2.3505 \]

### Step 6: Calculate pH
Finally, we can find the pH using:
\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 2.3505 = 11.6495 \]

Rounding to two decimal places, we get:
\[ \text{pH} \approx 11.65 \]

### Final Answer
The pH of a 0.50 M aqueous NaCN solution is approximately **11.65**.

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