Buffer capacity of a buffer solution is `x`, the volume of `1 M NaOH` added to `100 mL` of this solution if the change of `pH` by `1` is

Options

A. `0.1 x mL`
B. `10 x mL`
C. `100 x mL`
D. `x mL`

Explanation

To solve the problem, we need to find the volume of 1 M NaOH that can be added to a buffer solution of volume 100 mL, such that the pH changes by 1 unit. We are given the buffer capacity \( x \).

### Step-by-Step Solution:

1. **Understanding Buffer Capacity**:
The buffer capacity (\( \beta \)) is defined as the number of moles of H\(^+\) or OH\(^-\) ions that can be added per liter of buffer solution to cause a change in pH of 1 unit. Mathematically, it can be expressed as:
\[ \beta = \frac{\Delta n}{\Delta \text{pH}} \]
where \( \Delta n \) is the number of moles of acid or base added, and \( \Delta \text{pH} \) is the change in pH.

2. **Relating Buffer Capacity to Volume and Concentration**:
Since the buffer capacity is given as \( x \), we can express the number of moles of OH\(^-\) added (from NaOH) as:
\[ \Delta n = x \cdot \Delta \text{pH} \]
Given that \( \Delta \text{pH} = 1 \):
\[ \Delta n = x \cdot 1 = x \]

3. **Calculating Moles of NaOH**:
The volume of the buffer solution is 100 mL, which is 0.1 L. The number of moles of NaOH needed to achieve this change in pH in 0.1 L of the buffer solution can be calculated as:
\[ \text{Moles of NaOH} = \Delta n = x \]

4. **Finding the Volume of NaOH**:
Since we are using 1 M NaOH, the number of moles is equal to the volume in liters. Therefore, if \( V \) is the volume in liters of 1 M NaOH needed:
\[ V = \text{Moles of NaOH} = x \]

5. **Converting Volume to mL**:
To convert the volume from liters to milliliters:
\[ V_{\text{mL}} = V \times 1000 = x \times 1000 \text{ mL} \]

6. **Final Result**:
The volume of 1 M NaOH that can be added to the buffer solution to change the pH by 1 unit is:
\[ V_{\text{mL}} = 1000x \text{ mL} \]

### Summary:
The volume of 1 M NaOH added to 100 mL of the buffer solution to change the pH by 1 is \( 1000x \) mL.