Question

$30.0\, mL$ of the given $HCl$ solution requires $20.0 \,mL$ of $0.1 M$ sodium carbonate solution for complete neutralisation. What is the volume of this $HCl$ solution required to neutralise $30.0\, mL$ of $0.2 \,M \,NaOH$ solution?

• A. 25 mL
• B. 50 mL
• C. 90 mL
• D. 45 mL

Answer 1

Answer: (D)

$( N \times V )_{ HCl }=( N \times V )_{\text {sodium carbonate }}$
$N _{ HCl } \times 30=$ Molarity $\times n _{ f } \times 20$
$n _{ f }$ for sodium carbonate $=2$
$N _{ HCl } \times 30=0.1 \times 2 \times 20$
$N _{ HCl }=\frac{4}{30}$
when $HCl$ neutralises with $NaOH$ :
$( N \times V )_{ HCl }=( N \times V )_{ NaOH }$
$ \begin{array}{l} \left(\frac{4}{30} \times V \right)_{ HCl }=(1 \times 0.2 \times 30)_{ NaOH } \\ V _{ HCl }=45 mL \end{array} $