The `pH` of an acidic buffer can be raised by `2`units by

Options

A. Increasing the concentration of both weak acid and salt by two moles
B. Increasing the concentration of both the acid and salt by `10` times.
C. Diluting the solution by `10` times.
D. Increasing the concentration of the salt by `10` times by decreasing concentration of the acid by `10` times.

Explanation

To solve the problem of how to raise the pH of an acidic buffer by 2 units, we can follow these steps:

### Step 1: Understand the Concept of Acidic Buffer
An acidic buffer is a solution that consists of a weak acid (HA) and its conjugate base (A⁻). It maintains a pH less than 7. The pH of an acidic buffer can be calculated using the Henderson-Hasselbalch equation:

\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]

Where:
- \([\text{A}^-]\) is the concentration of the conjugate base (salt).
- \([\text{HA}]\) is the concentration of the weak acid.

### Step 2: Set Up Initial and Final Conditions
Let:
- Initial concentrations: \([\text{A}^-] = X_1\) and \([\text{HA}] = Y_1\)
- Final concentrations after raising pH by 2 units: \([\text{A}^-] = X_2\) and \([\text{HA}] = Y_2\)

### Step 3: Write the Two pH Equations
Using the Henderson-Hasselbalch equation, we can write two equations for the initial and final states:

1. For initial pH:
\[ \text{pH}_1 = \text{pK}_a + \log \left( \frac{X_1}{Y_1} \right) \]

2. For final pH (after raising by 2 units):
\[ \text{pH}_2 = \text{pK}_a + \log \left( \frac{X_2}{Y_2} \right) \]

### Step 4: Subtract the Two Equations
Subtract the first equation from the second:

\[ \text{pH}_2 - \text{pH}_1 = \log \left( \frac{X_2}{Y_2} \right) - \log \left( \frac{X_1}{Y_1} \right) \]

This simplifies to:

\[ 2 = \log \left( \frac{X_2}{Y_2} \cdot \frac{Y_1}{X_1} \right) \]

### Step 5: Exponentiate Both Sides
Exponentiating both sides gives:

\[ 10^2 = \frac{X_2 \cdot Y_1}{Y_2 \cdot X_1} \]

This means:

\[ 100 = \frac{X_2}{X_1} \cdot \frac{Y_1}{Y_2} \]

### Step 6: Analyze the Options
To achieve a factor of 100, we can consider the following scenarios:
- If we increase the concentration of salt (A⁻) by 10 times, then \(X_2 = 10X_1\).
- If we decrease the concentration of the weak acid (HA) by 10 times, then \(Y_2 = \frac{Y_1}{10}\).

Substituting these into the equation gives:

\[ 100 = \frac{10X_1}{X_1} \cdot \frac{Y_1}{Y_1/10} = 10 \cdot 10 = 100 \]

Thus, the correct option is to **increase the concentration of salt by 10 times and decrease the concentration of acid by 10 times**.

### Final Answer
The pH of an acidic buffer can be raised by 2 units by **increasing the concentration of salt by 10 times and decreasing the concentration of acid by 10 times**.

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