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?

400 ml

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50 ml

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100 ml

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150 ml

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Solution

The correct option is B
50 ml

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

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

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

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

99.95V= 5

V= 0.05002L

V= 50.02ML

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What volume of 0.10 MH2SO4 must be added to 50 ml of a 0.10 M NaOH solution to make a solution in which the molarity of theH2SO4 is 0.050 M?

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?