Question

$30.0mL$ of the given $HCl$ solution requires $20.0mL$ of $0.1M$ sodium carbonate solution for complete neutralisation. What is the volume of this $HCl$ solution required to neutralise $30.0mL$ of $0.2MNaOH$ 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 $NaOH$ :
$(N\times V{)}_{HCl}=(N\times V{)}_{NaOH}$
$\begin{array}{l}{\left(\frac{4}{30}\times V\right)}_{HCl}=(1\times 0.2\times 30{)}_{NaOH}\\ V_{HCl}=45mL\end{array}$