Q62 / 180 · AIIMS 2010 · Chemistry
Stomach acid, a dilute solution of $\mathrm{HCl}$ in water, can be neutralized by reaction with sodium hydrogen carbonate,
$$
\mathrm{NaHCO}_{3(a q)}+\mathrm{HCl}_{(a q)} \rightarrow \mathrm{NaCl}_{(a q)}+\mathrm{H}_2 \mathrm{O}_{(l)}+\mathrm{CO}_{2(g)}
$$
How many milliliters of $0.125 \mathrm{M} \mathrm{NaHCO}_3$ solution are needed to neutralize $18.0 \mathrm{~mL}$ of $0.100 \mathrm{M} \mathrm{HCl}$ ?
- Correct answerA$14.4 \mathrm{~mL}$
- B$12.0 \mathrm{~mL}$
- C$14.0 \mathrm{~mL}$
- D$13.2 \mathrm{~mL}$
Solution
$$
\begin{aligned}
& \text { Given, } M_{\mathrm{HCl}}=0.1 \mathrm{M}, V_{\mathrm{HCl}}=18.0 \mathrm{~mL} \\
& M_{\mathrm{NaHCO}_3}=0.125 \mathrm{M}, V_{\mathrm{NaHCO}_3}=\text { ? }
\end{aligned}
$$
On applying, $M_{\mathrm{HC}} \times V_{\mathrm{HCl}}=M_{\mathrm{NaHCO}_3} \times V_{\mathrm{NaHCO}_3}$ $\Rightarrow 0.1 \times 18=0.125 \times V_{\mathrm{NaHCO}_3}$ $\Rightarrow M_{\mathrm{NaHCO}_3}=14.4 \mathrm{~mL}$
Thus, $14.4 \mathrm{~mL}$ of the $1.25 \mathrm{M} \mathrm{NaHCO} \mathrm{M}_3$ solution is needed to neutralise $18.0 \mathrm{~mL}$ of the $0.100 \mathrm{M}$ $\mathrm{HCl}$ solution.
Hence, the correct answer is option: 'A'.