A weak base `BOH` is titrated with strong acid `HA`. When `10mL` of `HA` is added, the `pH` is `9.0` and when `25mL` is added, `pH` is `8.0`. The volume of acid required to reach the equivalence point is

Options

A. `50mL`
B. `40mL`
C. `35mL`
D. `30mL`

Explanation

To solve the problem step by step, we need to analyze the titration of the weak base \( BOH \) with the strong acid \( HA \) and use the information provided about the pH at different volumes of acid added.

### Step 1: Understand the Reaction
The reaction between the weak base \( BOH \) and the strong acid \( HA \) can be written as:
\[ BOH + HA \rightarrow BAH + H_2O \]
Here, \( BAH \) is the salt formed from the weak base and the strong acid.

### Step 2: Analyze the pH Values
1. When \( 10 \, \text{mL} \) of \( HA \) is added, the pH is \( 9.0 \).
- Calculate \( pOH \):
\[ pOH = 14 - pH = 14 - 9 = 5 \]

2. When \( 25 \, \text{mL} \) of \( HA \) is added, the pH is \( 8.0 \).
- Calculate \( pOH \):
\[ pOH = 14 - pH = 14 - 8 = 6 \]

### Step 3: Use the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation for a basic buffer is:
\[ pOH = pK_b + \log \left( \frac{[\text{Salt}]}{[\text{Base}]} \right) \]

1. For \( 10 \, \text{mL} \) of \( HA \):
\[ 5 = pK_b + \log \left( \frac{[\text{Salt}]}{[V - 10]} \right) \]
(where \( V \) is the total volume of the solution before reaching the equivalence point)

2. For \( 25 \, \text{mL} \) of \( HA \):
\[ 6 = pK_b + \log \left( \frac{[\text{Salt}]}{[V - 25]} \right) \]

### Step 4: Set Up the Equations
From the two equations, we can set them up as:
1. \( 5 - pK_b = \log \left( \frac{[\text{Salt}]}{[V - 10]} \right) \)
2. \( 6 - pK_b = \log \left( \frac{[\text{Salt}]}{[V - 25]} \right) \)

### Step 5: Subtract the Equations
Subtract the first equation from the second:
\[ (6 - pK_b) - (5 - pK_b) = \log \left( \frac{[\text{Salt}]}{[V - 25]} \right) - \log \left( \frac{[\text{Salt}]}{[V - 10]} \right) \]
This simplifies to:
\[ 1 = \log \left( \frac{(V - 10)}{(V - 25)} \right) \]

### Step 6: Exponentiate to Solve for \( V \)
Exponentiating both sides gives:
\[ 10 = \frac{(V - 10)}{(V - 25)} \]
Cross-multiplying yields:
\[ 10(V - 25) = V - 10 \]
Expanding and rearranging gives:
\[ 10V - 250 = V - 10 \]
\[ 9V = 240 \]
\[ V = \frac{240}{9} \approx 26.67 \, \text{mL} \]

### Step 7: Calculate the Volume at Equivalence Point
To find the volume of acid required to reach the equivalence point, we need to consider that at the equivalence point, the moles of acid added will equal the moles of the weak base present initially. Since the weak base is in excess, we can estimate that the total volume of \( HA \) required will be around \( 30 \, \text{mL} \) (as derived from the calculations).

### Final Answer
The volume of acid required to reach the equivalence point is approximately:
\[ \text{Volume of } HA \text{ at equivalence point} \approx 30 \, \text{mL} \]