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

Options

A. Decreases by one
B. Increases by one-tenth
C. Increases by one
D. Increases ten-fold

Explanation

To solve the problem of how the pH of a solution changes when the ratio of the concentration of a weak acid to its salt is increased ten-fold, we can follow these steps:

### Step 1: Understand the Initial Condition
Initially, we assume the concentration of the weak acid (HA) and its salt (A-) are equal. This means:
\[ \frac{[HA]}{[A^-]} = 1 \]

### Step 2: Apply the Henderson-Hasselbalch Equation
For a buffer solution, the pH can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[A^-]}{[HA]}\right) \]

### Step 3: Calculate Initial pH
Since the initial ratio is 1:
\[ \text{pH}_1 = \text{pKa} + \log(1) \]
Since \(\log(1) = 0\):
\[ \text{pH}_1 = \text{pKa} \]

### Step 4: Change the Ratio of Acid to Salt
The problem states that the ratio of the concentration of acid to salt is increased ten-fold. Therefore:
\[ \frac{[HA]}{[A^-]} = 10 \]
This implies:
\[ \frac{[A^-]}{[HA]} = \frac{1}{10} \]

### Step 5: Calculate New pH
Now, we can calculate the new pH using the Henderson-Hasselbalch equation:
\[ \text{pH}_2 = \text{pKa} + \log\left(\frac{[A^-]}{[HA]}\right) \]
Substituting the new ratio:
\[ \text{pH}_2 = \text{pKa} + \log\left(\frac{1}{10}\right) \]
Since \(\log\left(\frac{1}{10}\right) = -1\):
\[ \text{pH}_2 = \text{pKa} - 1 \]

### Step 6: Conclusion
The pH of the solution decreases by 1 unit when the concentration of the weak acid is increased ten-fold compared to its salt.

### Final Answer
The new pH of the solution is:
\[ \text{pH}_2 = \text{pKa} - 1 \]