Question

Stomach acid, a dilute solution of $HCl$ in water can be neutralised by reaction with sodium hydrogen carbonate.

$NaHC{O}_3\left(aq\right)+HCl\left(aq\right)⟶NaCl\left(aq\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)$

How many millilitres of $0.125M$ $NaH{CO}_3$ solution are needed to neutralize $18.0ml$ of $0.100M$ $HCl$?

• A. $14.4ml$
• B. $12.0ml$
• C. $14.0ml$
• D. $13.2ml$

Answer 1

The correct option is A $14.4ml$

Given $M_{HCl}=0.1M$; $V_{HCl}=18.0ml$
$M_{NaH{CO}_3}=0.125M$
On applying $M_{HCl}\times V_{HCl}=M_{NaH{CO}_3}\times V_{NaH{CO}_3}$
$\Rightarrow$ $0.1\times 18=0.125\times V_{NaH{CO}_3}$
Thus, $14.4ml$ of the $1.25M$ $NaH{CO}_3$ solution is needed to neutralise $18.0ml$ of $0.100M$ $HCl$ solution.