The volume of $0.1 M H_{2} S O_{4}$ to be added in $50$ mL of $0.10 M N a O H$ to give a solution of $0.05 M$ $H_{2} S O_{4}$ is :

75

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100

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25

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50

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Solution

The correct option is B $100$ mL
Let $V_{1}$ mL of $0.1 M$ $H_{2} S O_{4}$ is used so that, molarity of $H_{2} S O_{4}$ becomes $0.05$ M after adding $N a O H$ .

Meq. of $H_{2} S O_{4}$ taken $=$ Meq. of $N a O H$ used + Meq. of $H_{2} S O_{4}$ left

$0.1 \times 2 \times V_{1} = 50 \times 0.1 + 0.05 \times 2 \times (V_{1} + 50)$

$∴ 0.2 V_{1} = 5 + 0.1 V_{1} + 5$

$∴ 0.1 V_{1} = 10$

$∴ V_{1} = 100$ mL

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The volume of 0.1 MH2SO4 to be added in 50 mL of 0.10 MNaOH to give a solution of 0.05 M H2SO4 is :

The correct option is B 100 mLLet V1 mL of 0.1 M H2SO4 is used so that, molarity of H2SO4 becomes 0.05 M after adding NaOH.Meq. of H2SO4 taken = Meq. of NaOH used + Meq. of H2SO4 left0.1×2×V1=50×0.1+0.05×2×(V1+50)∴0.2V1=5+0.1V1+5∴0.1V1=10∴V1=100 mL

The volume of $0.1 M H_{2} S O_{4}$ to be added in $50$ mL of $0.10 M N a O H$ to give a solution of $0.05 M$ $H_{2} S O_{4}$ is :