Upon titrating `50mL` of `1.96%(w//v) H_(2)SO_(4)` with a `KOH` solution (containing `11.2g KOH` per litre of solution on adding `50mL KOH` solution.

Options

A. Equivalent point has been just reached.
B. Equivalent point has been crossed
C. `pH` of resulting solution `=1`
D. `pH` of resulting solution `=7`

Explanation

To solve the problem of titrating `50 mL` of `1.96% (w/v) H₂SO₄` with a `KOH` solution containing `11.2 g KOH` per liter, we will follow these steps:

### Step 1: Calculate the mass of H₂SO₄ in the solution
Given that the solution is `1.96% (w/v)`, this means that in `100 mL` of the solution, there are `1.96 g` of H₂SO₄.

To find the mass in `1000 mL` (or `1 L`):
\[
\text{Mass of H₂SO₄ in 1000 mL} = \frac{1.96 \, \text{g}}{100 \, \text{mL}} \times 1000 \, \text{mL} = 19.6 \, \text{g}
...