What volume of $0.10 M$ $H_{2} S O_{4}$ must be added to $50 m L$ of a $0.10 M N a O H$ solution to make a solution in which the molarity of the $H_{2} S O_{4}$ is $0.050 M$ ?

400 m L

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50 m L

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100 m L

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150 m L

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Solution

The correct option is B $50 m L$
if you add V ml of 0.1 M $H_{2} S O_{4}$ then you are adding 0.01 moles of $H_{2} S O_{4}$ .

in 50mL of 0.1 M NaOH, there are 0.005 moles of NaOH.

The chemical equation for that is :

$2 N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O$ .

0.0025 moles of $H_{2} S O_{4}$ will neutralize 0.005 moles of NaOH.

So from 0.01V moles of $H_{2} S O_{4}$ added in VmL there will be
0.01V- 0.0025 moles of $H_{2} S O_{4}$ left unused in the total volume of V + 50mL of solutions.

The conc. here will be $\frac{0.01 V - 0.0025}{V + 50} \times 1000 = 0.05 M$ .

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

On solving for V, we get the value :

V = 0.05002 L

V = $50.02 m L$

Hence, the correct option is $B$

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What volume of 0.10 M H2SO4 must be added to 50 mL of a 0.10 M NaOH solution to make a solution in which the molarity of the H2SO4 is 0.050 M?

What volume of $0.10 M$ $H_{2} S O_{4}$ must be added to $50 m L$ of a $0.10 M N a O H$ solution to make a solution in which the molarity of the $H_{2} S O_{4}$ is $0.050 M$ ?