How many gm of solid KOH must be added to 100 mL of a buffer solution to make the pH of solution 6.0, if it is 0.1 M each w.r.t. acid HA and salt K A.
`[Given : pK_(a)(HA)=5]`

Options

A. `0.458`
B. `0.327`
C. `5.19`
D. None of these

Answer

To solve the problem of how many grams of solid KOH must be added to 100 mL of a buffer solution to achieve a pH of 6.0, we can follow these steps:

### Step 1: Understand the buffer system
The buffer solution consists of a weak acid (HA) and its conjugate base (A⁻). The pKa of the weak acid is given as 5.

### Step 2: Use the Henderson-Hasselbalch equation
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
Given:
- pH = 6.0
- pKa = 5.0

### Step 3: Rearrange the equation to find the ratio of concentrations
Substituting the known values into the equation:
\[ 6.0 = 5.0 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
This simplifies to:
\[ 1 = \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
Taking the antilogarithm, we find:
\[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^1 = 10 \]
This means that the concentration of the conjugate base (A⁻) must be 10 times that of the weak acid (HA).

### Step 4: Calculate initial concentrations of HA and A⁻
The initial concentrations of HA and A⁻ in the buffer solution are both 0.1 M. Since the volume of the solution is 100 mL (0.1 L), the initial moles of HA and A⁻ are:
\[ \text{Moles of HA} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{moles} \]
\[ \text{Moles of A}^- = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{moles} \]

### Step 5: Set up the changes in moles after adding KOH
When KOH is added, it reacts with HA:
\[ \text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O} \]
Let \( x \) be the moles of KOH added. After the reaction:
- Moles of HA = \( 0.01 - x \)
- Moles of A⁻ = \( 0.01 + x \)

### Step 6: Substitute into the ratio and solve for x
Using the ratio we found earlier:
\[ \frac{0.01 + x}{0.01 - x} = 10 \]
Cross-multiplying gives:
\[ 0.01 + x = 10(0.01 - x) \]
Expanding and rearranging:
\[ 0.01 + x = 0.1 - 10x \]
\[ 11x = 0.1 - 0.01 \]
\[ 11x = 0.09 \]
\[ x = \frac{0.09}{11} \approx 0.00818 \, \text{moles} \]

### Step 7: Calculate the mass of KOH required
The molecular weight of KOH is approximately 56 g/mol. Therefore, the mass of KOH needed is:
\[ \text{Mass} = x \times \text{Molecular weight} = 0.00818 \, \text{moles} \times 56 \, \text{g/mol} \approx 0.458 \, \text{grams} \]

### Final Answer
To achieve a pH of 6.0, approximately **0.458 grams** of solid KOH must be added to the buffer solution.

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