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$25 \mathrm{~mL}$ of the given $\mathrm{HCl}$ solution requires $30 \mathrm{~mL}$ of $0.1 \mathrm{M}$ sodium carbonate solution. What is the volume of this $\mathrm{HCl}$ solution required to titrate $30 \mathrm{~mL}$ of $0.2 \mathrm{M}$ aqueous $\mathrm{NaOH}$ solution?
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The correct answer is:
$25 \mathrm{~mL}$
$25 \mathrm{~mL}$ of $\mathrm{HCl}$ solution requires $30 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{Na}_{2} \mathrm{CO}_{3}$ solution.
$$
\begin{array}{l}
\because \mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2} \\
\therefore 25 \times \mathrm{N}_{1}=30 \times 0.2\left(0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}=0.2 \mathrm{~N}\right. \\
\left.\mathrm{Na}_{2} \mathrm{CO}_{3}\right) \\
\mathrm{N}_{1}=\frac{6}{25}=0.24 \mathrm{~N}
\end{array}
$$
Now, HCl solution is titrated with $\mathrm{NaOH}$ solution.
$$
\begin{array}{l}
\mathrm{M}_{1} \mathrm{~V}_{1}=\mathrm{M}_{2} \mathrm{~V}_{2} ; 0.24 \mathrm{NHCl}=0.24 \mathrm{M} \mathrm{HCl} \\
\therefore \mathrm{V} \times 0.24 \times 1=30 \times 0.2 \times 1 \Rightarrow \mathrm{V}=25 \mathrm{~mL}
\end{array}
$$
$$
\begin{array}{l}
\because \mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2} \\
\therefore 25 \times \mathrm{N}_{1}=30 \times 0.2\left(0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}=0.2 \mathrm{~N}\right. \\
\left.\mathrm{Na}_{2} \mathrm{CO}_{3}\right) \\
\mathrm{N}_{1}=\frac{6}{25}=0.24 \mathrm{~N}
\end{array}
$$
Now, HCl solution is titrated with $\mathrm{NaOH}$ solution.
$$
\begin{array}{l}
\mathrm{M}_{1} \mathrm{~V}_{1}=\mathrm{M}_{2} \mathrm{~V}_{2} ; 0.24 \mathrm{NHCl}=0.24 \mathrm{M} \mathrm{HCl} \\
\therefore \mathrm{V} \times 0.24 \times 1=30 \times 0.2 \times 1 \Rightarrow \mathrm{V}=25 \mathrm{~mL}
\end{array}
$$
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