What volume of 0-10 - -2-4 must be added to 50 - of a 0-10 - - solution to make a solution in which the molarity of -2-4 is 0-050 -

The reaction of 𝐻_2𝑆𝑂_4 reacting with 𝑁𝑎𝑂𝐻 is as follows: 𝐻_2𝑆𝑂_4 nbsp;+ 2𝑁𝑎𝑂𝐻 → 𝑁𝑎_2 𝑆𝑂_4 + 2𝐻_2𝑂 For a neutralization reaction, Milliequivalents of 𝑁𝑎𝑂𝐻 = ...

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Byju's Answer

Standard XII

Chemistry

Concentration Terms

What volume o...

Question

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 $H_{2} S O_{4}$ is 0.050 𝑀?
$50$ mL
$400$ mL
$150$ mL
$100$ mL

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Solution

The correct option is D $100$ mL
The reaction of $H_{2} S O_{4}$ reacting with 𝑁𝑎𝑂𝐻 is as follows:
$H_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + 2 H_{2} O$

For a neutralization reaction,
Milliequivalents of 𝑁𝑎𝑂𝐻 = Milliequivalents of
$H_{2} S O_{4}$
Volume of 𝑁𝑎𝑂𝐻 × Normality of 𝑁𝑎𝑂𝐻
= Volume of $H_{2} S O_{4}$ × Normality of $H_{2} S O_{4}$
We know that,
$N o r m a l i t y = M o l a r i t y \times n - f a c t o r$

Volume of 𝑁𝑎𝑂𝐻 × Molarity of 𝑁𝑎𝑂𝐻 ×
𝑛 − 𝑓𝑎𝑐𝑡𝑜𝑟 = Volume of $H_{2} S O_{4}$ × Molarity of
$H_{2} S O_{4} \times n - f a c t o r$
$50 m L \times 0.1 M \times 1 = y m L \times 0.1 M \times 2$
$∴ y = 25 m l$

So, for complete reaction with 50 𝑚𝐿 𝑜𝑓 0.10 𝑀 𝑁𝑎𝑂𝐻, 25 𝑚𝐿 of 0.1 𝑀 $H_{2} S O_{4}$ is required.

Thus, the total volume of the solution after neutralization will be 50 𝑚𝐿 +25𝑚𝐿 = 75 𝑚L
To obtain 0.05 𝑀 $H_{2} S O_{4}$ , more volume of 0.10 𝑀 $H_{2} S O_{4}$ must be added.
Suppose the volume of 0.1 𝑀 $H_{2} S O_{4}$ to be added to this solution to make the molarity of sulphuric acid equal to 0.05 𝑀, is 𝑥 𝑚𝐿,
Then,
𝑉1 = Extra volume of $H_{2} S O_{4}$ needed = 𝑥 𝑚𝐿
𝑀1 = Molarity of $H_{2} S O_{4}$ = 0.10 𝑀
𝑉2 = Total volume of the resultant solution
after adding extra $H_{2} S O_{4}$ = (75 + 𝑥) 𝑚𝐿
𝑀2 = Molarity of the resultant solution
= 0.05 𝑀

We know that $M_{1} V_{1} = M_{2} V_{2}$
$x \times 0.10 M = (75 + x) \times 0.05 M$
$0.10 x = 3.75 + 0.05 x$
$0.05 x = 3.75$
$x = \frac{3.75}{0.05} = 75 m L$

Thus, the total volume of $0.10 M H_{2} S O_{4}$
required = 25 𝑚𝐿 (for neutralizing
50 𝑚𝐿 𝑜𝑓 0.10 𝑀 𝑁𝑎𝑂𝐻) + 75 𝑚𝐿 (for
attaining the 0.05 𝑀 concentration)
= 100 𝑚𝐿
So, option (C) is the correct answer

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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 H2SO4 is 0.050 𝑀?50 mL400 mL150 mL100 mL

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 $H_{2} S O_{4}$ is 0.050 𝑀?
$50$ mL
$400$ mL
$150$ mL
$100$ mL