Question

Stomach acid, a dilute solution of $HCl$ in water, can be neutralized by reaction with sodium hydrogen carbonate, {NaHCO_3_{(aq)} +HCL_{(aq)} ->NaCL_{(aq)} +H_2O_{(l)} +CO_2_{(g)} } How many milliliters of 0.125 M NaHCO solution are needed to neutralize 18.0 mL of 0.100 M HCL?

• A. 14.4 mL
• B. 12.0 mL
• C. 14.0 mL
• D. 13.2 mL

Answer 1

Answer: (A)

14.4 mL

Answer 2

Given, $M_{HCl} = 0.1\, M,\, V_{HCl} = 18.0\, mL$ $M_{NaHCO_3}=0.125\, M,\, V_{NaHCO_3}= ?$ On applying, $M_{HCl} \times V_{HCl}=M_{NaHCO_3}\times V_{NaHCO_3}$ $\Rightarrow 0.1 \times 18 =0.125 \times V_{NaHCO_3}$ $\Rightarrow M_{NaHCO_3} =14.4\, mL$ Thus, $14.4\, mL$ of the $1.25\, M\, NaHCO_3$ solution is needed to neutralise $18.0\, mL$ of the $0.100\, M\, HCl$ solution.