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 will analyze the titration of the weak base \( BOH \) with the strong acid \( HA \) and use the given pH values to find the volume of acid required to reach the equivalence point.

### Step 1: Understand the Reaction
The weak base \( BOH \) reacts with the strong acid \( HA \) to form the salt \( BA \) and water:
\[ BOH + HA \rightarrow BA + H_2O \]

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

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

### Step 3: Set Up the Buffer Equations
For a basic buffer, we can use the following equation:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]

#### For \( 10 \, \text{mL} \) of \( HA \):
\[ 5 = \text{pK}_b + \log\left(\frac{[\text{BA}]}{[\text{BOH}]}\right) \]

Let \( V \) be the initial volume of the weak base \( BOH \) in mL. After adding \( 10 \, \text{mL} \) of \( HA \), the concentration of \( BOH \) becomes \( V - 10 \) mL.

#### For \( 25 \, \text{mL} \) of \( HA \):
\[ 6 = \text{pK}_b + \log\left(\frac{[\text{BA}]}{[\text{BOH}]}\right) \]

After adding \( 25 \, \text{mL} \) of \( HA \), the concentration of \( BOH \) becomes \( V - 25 \) mL.

### Step 4: Formulate the Equations
From the two pOH equations, we can set up the following equations:
1. \( 5 = \text{pK}_b + \log\left(\frac{[\text{BA}]}{V - 10}\right) \)
2. \( 6 = \text{pK}_b + \log\left(\frac{[\text{BA}]}{V - 25}\right) \)

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

### Step 6: Convert to Exponential Form
Taking the antilog of both sides:
\[ 10 = \frac{(V - 25)}{(V - 10)} \]

### Step 7: Solve for \( V \)
Cross-multiplying gives:
\[ 10(V - 10) = V - 25 \]
Expanding and rearranging:
\[ 10V - 100 = V - 25 \]
\[ 9V = 75 \]
\[ V = \frac{75}{9} = 30 \, \text{mL} \]

### Final Answer
The volume of acid required to reach the equivalence point is \( 30 \, \text{mL} \).

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