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Some Basic Concepts of Chemistry question

2019 · 11 Jan · Shift 2 · Q10

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Some Basic Concepts of Chemistry
2019 · 11 Jan · Shift 2 · Q10

Some Basic Concepts of Chemistry question

2019 · 11 Jan · Shift 2 · Q10

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1

25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2 M aqueous NaOH solutions ?

A
50 mL
B
12.5 mL
C
25 mL
D
75 mL

View written solutionFree

Correct answer: C

Find the concentration of the given HCl solution

The reaction between sodium carbonate and hydrochloric acid is:

$\mathrm{Na_2CO_3 + 2HCl \to 2NaCl + H_2O + CO_2}$

So,

$1 \text{ mol } \mathrm{Na_2CO_3} \text{ reacts with } 2 \text{ mol } \mathrm{HCl}$

Given:

Volume of $\mathrm{Na_2CO_3}$ solution $= 30\,\text{mL} = 0.030\,\text{L}$
Molarity of $\mathrm{Na_2CO_3}$ solution $= 0.1\,\text{M}$

Moles of $\mathrm{Na_2CO_3}$ used:

$n = MV = 0.1 \times 0.030 = 0.003\,\text{mol}$

Therefore moles of HCl required:

$n(\mathrm{HCl}) = 2 \times 0.003 = 0.006\,\text{mol}$

These $0.006$ mol HCl are present in $25\,\text{mL} = 0.025\,\text{L}$ of the given HCl solution.

Hence molarity of HCl:

$M(\mathrm{HCl}) = \frac{0.006}{0.025} = 0.24\,\text{M}$

Find moles of NaOH to be neutralized

Reaction:

$\mathrm{HCl + NaOH \to NaCl + H_2O}$

This is a $1:1$ reaction.

Given:

Volume of NaOH solution $= 30\,\text{mL} = 0.030\,\text{L}$
Molarity of NaOH $= 0.2\,\text{M}$

Moles of NaOH:

$n(\mathrm{NaOH}) = 0.2 \times 0.030 = 0.006\,\text{mol}$

So moles of HCl required are also:

$n(\mathrm{HCl}) = 0.006\,\text{mol}$

Calculate the volume of HCl needed

Using

$V = \frac{n}{M} = \frac{0.006}{0.24} = 0.025\,\text{L}$

$0.025\,\text{L} = 25\,\text{mL}$

Match with the options

The required volume of HCl solution is:

$\boxed{25\,\text{mL}}$

So the correct option is C.

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