In a mixture of weak acid and its salt, the ratio of concentration of salt to acid is increased ten fold. The pH of the solution

Options

A. increases by one
B. decreases by one
C. decreases ten fold
D. increased ten fold

Explanation

To solve the problem, we need to analyze how the pH of a buffer solution changes when the ratio of the concentration of salt (conjugate base) to the weak acid is altered.

### Step-by-Step Solution:

1. **Understanding the Buffer Solution**:
We have a mixture of a weak acid (HA) and its salt (A⁻). This forms an acidic buffer solution. The pH of a buffer solution is given by the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]

2. **Initial Ratio of Salt to Acid**:
Let the initial ratio of the concentration of salt to acid be \( X \):
\[ \text{pH} = \text{pK}_a + \log(X) \]

3. **Increasing the Ratio**:
According to the problem, the ratio of the concentration of salt to acid is increased tenfold. Therefore, the new ratio becomes:
\[ \frac{[\text{A}^-]}{[\text{HA}]} = 10X \]

4. **Calculating New pH**:
Substitute the new ratio into the Henderson-Hasselbalch equation:
\[ \text{pH}_{\text{new}} = \text{pK}_a + \log(10X) \]
Using the property of logarithms, we can separate this:
\[ \text{pH}_{\text{new}} = \text{pK}_a + \log(10) + \log(X) \]
Since \(\log(10) = 1\), we can simplify this to:
\[ \text{pH}_{\text{new}} = \text{pK}_a + 1 + \log(X) \]

5. **Comparing Initial and New pH**:
The initial pH was:
\[ \text{pH}_{\text{initial}} = \text{pK}_a + \log(X) \]
The change in pH when the ratio is increased tenfold is:
\[ \text{pH}_{\text{new}} - \text{pH}_{\text{initial}} = (\text{pK}_a + 1 + \log(X)) - (\text{pK}_a + \log(X)) = 1 \]

6. **Conclusion**:
Therefore, the pH of the solution increases by 1 unit when the ratio of the concentration of salt to acid is increased tenfold.

### Final Answer:
The pH of the solution **increases by 1**.

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