What volume of 0.10 M$H_{2}S{O}_4$ must be added to 50 ml of a 0.10 M NaOH solution to make a solution in which the molarity of the$H_{2}S{O}_4$ is 0.050 M?

The correct option is A

50 ml

Explanation of correct option.

2NaOH +$H_{2}S{O}_4$$⟶$ $N{a}_{2}S{O}_4$ + 2$H_{2}O$

0.0025 moles of$H_{2}S{O}_4$ will neutralise 0.005 moles NaOH.

So for 0.01 V moles H2SO4 there will be 0.01v-0.0025 moles H2SO4 left unused in the total volume(v+50ml) of solution .

The concentration here will be

$\frac{0.01V-0.0025}{v+50}$$\times$1000 = 0.05M

$\frac{100V-2.5}{V+50}$ =0.05

$\frac{100V-2.5}{V+50}$= 2.5+2.5

99.95V= 5

V= 0.05002L

V= 50.02ML

Hence, the correct option is B