In a mixing of acetic acid and sodium acetate the ratio of concentration of the salts to the acid is increased ten times . Then the pH of the solution

Options

A. Increases by one unit
B. Decreases by one unit
C. Decrease ten fold
D. Increases ten fold

Answer

To solve the problem of determining the new pH when the ratio of the concentration of sodium acetate (salt) to acetic acid is increased ten times, we can follow these steps:

### Step 1: Understand the Buffer System
The mixture of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa) forms an acidic buffer solution. The pH of an acidic buffer can be calculated using the Henderson-Hasselbalch equation.

### Step 2: Write the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation for a weak acid and its salt is given by:

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

Where:
- \(\text{pKa}\) is the negative logarithm of the acid dissociation constant (Ka) of acetic acid.
- \([\text{Salt}]\) is the concentration of sodium acetate.
- \([\text{Acid}]\) is the concentration of acetic acid.

### Step 3: Initial Ratio of Salt to Acid
Let’s denote the initial concentration of sodium acetate as \([\text{Salt}]_0\) and the concentration of acetic acid as \([\text{Acid}]_0\). The initial ratio is:

\[ \frac{[\text{Salt}]_0}{[\text{Acid}]_0} \]

### Step 4: New Ratio After Increase
According to the problem, the ratio of the concentration of salt to acid is increased ten times. Therefore, the new ratio becomes:

\[ \frac{[\text{Salt}]_1}{[\text{Acid}]_1} = 10 \cdot \frac{[\text{Salt}]_0}{[\text{Acid}]_0} \]

### Step 5: Substitute into the Henderson-Hasselbalch Equation
Now, substituting the new ratio into the Henderson-Hasselbalch equation, we have:

\[ \text{pH}_1 = \text{pKa} + \log\left(10 \cdot \frac{[\text{Salt}]_0}{[\text{Acid}]_0}\right) \]

Using the logarithmic property \(\log(ab) = \log(a) + \log(b)\):

\[ \text{pH}_1 = \text{pKa} + \log(10) + \log\left(\frac{[\text{Salt}]_0}{[\text{Acid}]_0}\right) \]

Since \(\log(10) = 1\), we can simplify this to:

\[ \text{pH}_1 = \text{pKa} + 1 + \log\left(\frac{[\text{Salt}]_0}{[\text{Acid}]_0}\right) \]

### Step 6: Relate New pH to Initial pH
Let’s denote the initial pH as:

\[ \text{pH}_0 = \text{pKa} + \log\left(\frac{[\text{Salt}]_0}{[\text{Acid}]_0}\right) \]

Now, we can express the new pH in terms of the initial pH:

\[ \text{pH}_1 = \text{pH}_0 + 1 \]

### Conclusion
Thus, the new pH of the solution after increasing the ratio of salt to acid by ten times is:

\[ \text{pH}_1 = \text{pH}_0 + 1 \]

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

---