`100mL` of a buffer solution contains `0.1M` each of weak acid `HA` and salt `NaA`. How many gram of `NaOH` should be added to the buffer so that it `pH` will be `6`? `(K_(a)` of `HA = 10^(-5))`.

Options

A. `0.328`
B. `0458`
C. `4.19`
D. None

Explanation

To solve the problem of how many grams of NaOH should be added to the buffer solution to achieve a pH of 6, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Volume of buffer solution = 100 mL = 0.1 L
- Concentration of weak acid (HA) = 0.1 M
- Concentration of salt (NaA) = 0.1 M
- Desired pH = 6
- \( K_a \) of HA = \( 10^{-5} \)

2. **Calculate \( pK_a \)**:
\[ pK_a = -\log(K_a) = -\log(10^{-5}) = 5 \]

3. **Use the Henderson-Hasselbalch Equation**:
The Henderson-Hasselbalch equation is given by:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
Substituting the known values:
\[ 6 = 5 + \log\left(\frac{[NaA]}{[HA]}\right) \]

4. **Rearranging the Equation**:
\[ 1 = \log\left(\frac{[NaA]}{[HA]}\right) \]
Taking the antilogarithm:
\[ 10 = \frac{[NaA]}{[HA]} \]

5. **Set Up the Concentrations**:
Let \( x \) be the amount of NaOH added (in moles). After adding NaOH:
- New concentration of HA = \( 0.1 - x \)
- New concentration of NaA = \( 0.1 + x \)

Substitute these into the equation:
\[ 10 = \frac{0.1 + x}{0.1 - x} \]

6. **Cross Multiply and Solve for \( x \)**:
\[ 10(0.1 - x) = 0.1 + x \]
\[ 1 - 10x = 0.1 + x \]
\[ 1 - 0.1 = 11x \]
\[ 0.9 = 11x \]
\[ x = \frac{0.9}{11} \approx 0.0818 \text{ moles} \]

7. **Convert Moles to Grams**:
The molar mass of NaOH is approximately 40 g/mol. Therefore, the mass of NaOH required is:
\[ \text{Mass} = x \times \text{Molar Mass} = 0.0818 \text{ moles} \times 40 \text{ g/mol} \approx 3.272 \text{ grams} \]

### Final Answer:
Approximately **3.272 grams** of NaOH should be added to the buffer solution to achieve a pH of 6.