A weak acid `(HA)` after treatment with `12 mL` of `0.1 M` strong base `(BOH)` has a `pH` of `5`. At the end point , the volume of same base required is `27 mL`. `K_(a)` of acid is `(log 2=0.3)`

Options

A. `1.8xx10^(-5)`
B. `8xx10^(-6)`
C. `1.8xx10^(-6)`
D. `8xx10^(-5)`

Explanation

To solve the problem step by step, we will follow the information given and apply the relevant concepts of weak acids, strong bases, and pH calculations.

### Step 1: Calculate the millimoles of the strong base (BOH) at the endpoint
At the endpoint, the volume of the strong base required is given as 27 mL of 0.1 M solution.

\[ \text{Millimoles of BOH} = \text{Volume (mL)} \times \text{Molarity (M)} \]
\[ = 27 \, \text{mL} \times 0.1 \, \text{mol/L} = 2.7 \, \text{mmol} \]

### Step 2: Calculate the millimoles of the strong base (BOH) used in the reaction
The volume of the strong base used before reaching a pH of 5 is 12 mL.

\[ \text{Millimoles of BOH used} = 12 \, \text{mL} \times 0.1 \, \text{mol/L} = 1.2 \, \text{mmol} \]

### Step 3: Calculate the initial millimoles of the weak acid (HA)
Since the strong base reacts with the weak acid, the initial millimoles of the weak acid can be calculated as follows:

\[ \text{Initial moles of HA} = \text{Millimoles of BOH at endpoint} = 2.7 \, \text{mmol} \]

### Step 4: Calculate the remaining millimoles of the weak acid (HA) after reaction
After the reaction with 1.2 mmol of BOH, the remaining amount of HA can be calculated:

\[ \text{Remaining HA} = \text{Initial HA} - \text{Millimoles of BOH used} \]
\[ = 2.7 \, \text{mmol} - 1.2 \, \text{mmol} = 1.5 \, \text{mmol} \]

### Step 5: Set up the pH equation
We know that the pH is given as 5. The relationship between pH, pKa, and the concentrations of the acid and its conjugate base is given by the Henderson-Hasselbalch equation:

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

Where:
- \([\text{A}^-]\) is the concentration of the conjugate base (which is equal to the millimoles of BOH used, 1.2 mmol).
- \([\text{HA}]\) is the concentration of the weak acid remaining (1.5 mmol).

### Step 6: Substitute values into the equation
Substituting the known values into the equation:

\[ 5 = \text{pKa} + \log\left(\frac{1.2}{1.5}\right) \]

### Step 7: Calculate the logarithm
Calculate the logarithm:

\[ \log\left(\frac{1.2}{1.5}\right) = \log(0.8) = -0.1 \quad (\text{using } \log 2 = 0.3) \]

### Step 8: Solve for pKa
Substituting back into the equation:

\[ 5 = \text{pKa} - 0.1 \]
\[ \text{pKa} = 5 + 0.1 = 5.1 \]

### Step 9: Calculate Ka
Now, convert pKa to Ka:

\[ \text{Ka} = 10^{-\text{pKa}} = 10^{-5.1} = 8 \times 10^{-6} \]

### Final Answer
Thus, the value of \( K_a \) of the weak acid \( HA \) is:

\[ \boxed{8 \times 10^{-6}} \]