In a buffer solution of a weak acid and its salt, if the ratio of the concentratior of the salt to that of the acid is increased ten times, the pH of the buffer

Options

A. increases by 1
B. increases 10 times
C. decreases 10 times
D. decreases by 1

Answer

To solve the problem, we will use the Henderson-Hasselbalch equation, which is given by:

\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Let the initial concentration of the weak acid be \([HA]\).
- Let the initial concentration of the salt (conjugate base) be \([A^-]\).
- The initial pH of the buffer solution can be expressed as:
\[ \text{pH}_1 = \text{pK}_a + \log \left( \frac{[A^-]}{[HA]} \right) \]

2. **Increase the Concentration Ratio**:
- According to the problem, the ratio of the concentration of the salt to that of the acid is increased ten times. This means:
\[ \frac{[A^-]}{[HA]} \text{ becomes } 10 \times \frac{[A^-]}{[HA]} \]

3. **Express the New pH**:
- The new pH of the buffer solution can now be expressed as:
\[ \text{pH}_2 = \text{pK}_a + \log \left( 10 \times \frac{[A^-]}{[HA]} \right) \]

4. **Use Logarithmic Properties**:
- By using the property of logarithms, we can simplify:
\[ \text{pH}_2 = \text{pK}_a + \log(10) + \log \left( \frac{[A^-]}{[HA]} \right) \]
- Since \(\log(10) = 1\), we can rewrite it as:
\[ \text{pH}_2 = \text{pK}_a + 1 + \log \left( \frac{[A^-]}{[HA]} \right) \]

5. **Relate New pH to Initial pH**:
- Now, substituting the expression for the initial pH:
\[ \text{pH}_2 = \text{pH}_1 + 1 \]

6. **Conclusion**:
- Therefore, if the ratio of the concentration of the salt to that of the acid is increased ten times, the pH of the buffer solution increases by 1 unit.

### Final Answer:
If the ratio of the concentration of the salt to that of the acid is increased ten times, the pH of the buffer solution increases by 1 unit.

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