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

The correct option is D 45 mL

$(N\times V{)}_{HCl}=(N\times V{)}_{sodiumcarbonate}$
$N_{HCl}\times 30=Molarity\times n_f\times 20$
$n_f$ for sodium carbonate $=$2
$N_{HCl}\times 30=0.1\times 2\times 20$
$N_{HCl}=\frac{4}{30}$
when $HCl$ neutralises with $NaOH$:
$(N\times V{)}_{HCl}=(N\times V{)}_{NaOH}$
$(\frac{4}{30}\times V{)}_{HCl}=(1\times 0.2\times 30{)}_{NaOH}$
$V_{HCl}=45mL$