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

Options

A. `3.7 xx 10^(-13)M`
B. `3.2 xx 10^(-7)M`
C. `3.2 xx 10^(-2)M`
D. `2.7 xx 10^(-2)M`

Answer

To solve the problem step by step, we will follow the titration process of a weak monoacidic base with HCl and calculate the concentration of \( H^+ \) ions at the equivalence point.

### Step 1: Calculate the number of moles of the weak base
Given:
- Volume of weak base \( V_b = 2.5 \, \text{mL} = 0.0025 \, \text{L} \)
- Concentration of weak base \( C_b = \frac{2}{5} \, \text{M} = 0.4 \, \text{M} \)

The number of moles of the weak base \( n_b \) is calculated as:
\[ n_b = C_b \times V_b = 0.4 \, \text{mol/L} \times 0.0025 \, \text{L} = 0.001 \, \text{mol} \]

### Step 2: Calculate the volume of HCl required for titration
Given:
- Concentration of HCl \( C_{HCl} = \frac{2}{15} \, \text{M} \)

At the equivalence point, the moles of HCl will equal the moles of the weak base:
\[ n_{HCl} = n_b = 0.001 \, \text{mol} \]

Now, we can find the volume \( V_{HCl} \) required:
\[ V_{HCl} = \frac{n_{HCl}}{C_{HCl}} = \frac{0.001 \, \text{mol}}{\frac{2}{15} \, \text{mol/L}} = 0.001 \, \text{mol} \times \frac{15}{2} = 0.0075 \, \text{L} = 7.5 \, \text{mL} \]

### Step 3: Calculate the total volume at the equivalence point
The total volume \( V_{total} \) at the equivalence point is:
\[ V_{total} = V_b + V_{HCl} = 2.5 \, \text{mL} + 7.5 \, \text{mL} = 10 \, \text{mL} = 0.01 \, \text{L} \]

### Step 4: Calculate the concentration of the salt formed
At the equivalence point, the weak base forms a salt (metal chloride). The concentration of the salt \( C_{salt} \) is:
\[ C_{salt} = \frac{n_b}{V_{total}} = \frac{0.001 \, \text{mol}}{0.01 \, \text{L}} = 0.1 \, \text{M} \]

### Step 5: Calculate the hydrolysis constant \( K_h \)
Using the relationship:
\[ K_h = \frac{K_w}{K_b} \]
Where:
- \( K_w = 1 \times 10^{-14} \)
- \( K_b = 1 \times 10^{-12} \)

We find:
\[ K_h = \frac{1 \times 10^{-14}}{1 \times 10^{-12}} = 1 \times 10^{-2} \]

### Step 6: Set up the hydrolysis equilibrium expression
Let \( x \) be the concentration of \( H^+ \) produced by hydrolysis. The equilibrium expression is:
\[ K_h = \frac{[H^+][OH^-]}{[salt]} = \frac{x^2}{C_{salt} - x} \]
Substituting \( C_{salt} = 0.1 \, \text{M} \):
\[ 1 \times 10^{-2} = \frac{x^2}{0.1 - x} \]

### Step 7: Solve for \( x \)
Assuming \( x \) is small compared to 0.1, we can approximate:
\[ 1 \times 10^{-2} = \frac{x^2}{0.1} \]
\[ x^2 = 1 \times 10^{-2} \times 0.1 = 1 \times 10^{-3} \]
\[ x = \sqrt{1 \times 10^{-3}} = 0.03162 \, \text{M} \]

### Step 8: Calculate the concentration of \( H^+ \)
Since \( x \) represents the concentration of \( H^+ \):
\[ [H^+] = x = 0.03162 \, \text{M} \approx 3.16 \times 10^{-2} \, \text{M} \]

### Final Answer
The concentration of \( H^+ \) at the equivalence point is approximately \( 3.16 \times 10^{-2} \, \text{M} \).

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