`2.5 ` mL of `(2)/(5)` M weak monoacidic base `(K_(b)=1xx10^(-12) "at" 25^(@)C)` is titrated with `(2)/(15)` M HCl in water at `25^(@)` C. The concentration of `H^(+)` at equivalence point is `(K_(w) = 1 xx 10^(-14) "at" 25^(@)C)`

Options

A. `3.7xx10^(-13)M`
B. `3.2xx10^(-7) M`
C. `3.2xx10^(-2)M`
D. `2.7xx10^(-2)M`

Explanation

To solve the problem, we need to determine the concentration of \( H^+ \) ions at the equivalence point when titrating a weak monoacidic base with hydrochloric acid (HCl). We'll follow these steps:

### Step 1: Calculate the number of moles of the weak base.
The concentration of the weak base is given as \( \frac{2}{5} \) M, and the volume is \( 2.5 \) mL.

\[ \text{Moles of weak base} = \text{Concentration} \times \text{Volume} = \left(\frac{2}{5} \, \text{mol/L}\right) \times \left(\frac{2.5 \, \text{mL}}{1000 \, \text{mL/L}}\right) \]

\[ = \left(\frac{2}{5}\right) \times \left(0.0025\right) = 0.001 \, \text{mol} \]

### Step 2: Determine the number of moles of HCl required for neutralization.
Since the weak base is monoacidic, it will react with HCl in a 1:1 ratio. Therefore, the moles of HCl required will also be \( 0.001 \, \text{mol} \).

### Step 3: Calculate the volume of HCl needed to reach the equivalence point.
The concentration of HCl is given as \( \frac{2}{15} \) M. We can use the formula:

\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} \]

Setting the moles of HCl equal to the moles of the weak base:

\[ 0.001 \, \text{mol} = \left(\frac{2}{15} \, \text{mol/L}\right) \times V \]

Solving for \( V \):

\[ V = \frac{0.001}{\frac{2}{15}} = 0.001 \times \frac{15}{2} = 0.0075 \, \text{L} = 7.5 \, \text{mL} \]

### Step 4: Calculate the total volume at the equivalence point.
The total volume at the equivalence point will be the sum of the volumes of the weak base and HCl:

\[ \text{Total Volume} = 2.5 \, \text{mL} + 7.5 \, \text{mL} = 10 \, \text{mL} \]

### Step 5: Calculate the concentration of \( H^+ \) ions at the equivalence point.
At the equivalence point, all the weak base has been converted to its conjugate acid. The concentration of \( H^+ \) ions can be determined using the ion product of water \( K_w \) and the \( K_b \) of the weak base.

Using the relationship:

\[ K_w = K_a \cdot K_b \]

We can find \( K_a \):

\[ K_a = \frac{K_w}{K_b} = \frac{1 \times 10^{-14}}{1 \times 10^{-12}} = 1 \times 10^{-2} \]

Now, we can find the concentration of \( H^+ \) ions using the formula:

\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

At the equivalence point, \( [HA] \) is the concentration of the conjugate acid formed. Since we have \( 0.001 \, \text{mol} \) of the acid in \( 10 \, \text{mL} \):

\[ [HA] = \frac{0.001}{0.01} = 0.1 \, \text{M} \]

Assuming \( x \) is the concentration of \( H^+ \) ions produced:

\[ K_a = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1} \]

Setting \( K_a = 1 \times 10^{-2} \):

\[ 1 \times 10^{-2} = \frac{x^2}{0.1} \]

\[ x^2 = 1 \times 10^{-3} \]

\[ x = \sqrt{1 \times 10^{-3}} = 0.03162 \, \text{M} \]

Thus, the concentration of \( H^+ \) at the equivalence point is approximately \( 0.0316 \, \text{M} \).

### Summary of Steps:
1. Calculate moles of weak base.
2. Determine moles of HCl needed.
3. Calculate volume of HCl required.
4. Find total volume at equivalence point.
5. Calculate concentration of \( H^+ \) ions at equivalence point.