1. Tardigrade
  2. Question
  3. Chemistry
  4. Stomach acid, a dilute solution of H C l in water, can be neutralized by reaction with sodium hydrogen carbonate, NaHCO 3( aq )+ HCL ( aq ) longrightarrow NaCL ( aq )+ H 2 O ( l )+ CO 2( g ) How many milliliters of 0.125 M NaHCO 3 solution are needed to neutralize 18.0 mL of 0.100 M HCL ?

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Stomach acid, a dilute solution of $H C l$ in water, can be neutralized by reaction with sodium hydrogen carbonate, $NaHCO _{3( aq )}+ HCL _{( aq )} \longrightarrow NaCL _{( aq )}+ H _2 O _{( l )}+ CO _{2( g )}$
How many milliliters of $0.125 M NaHCO _3$ solution are needed to neutralize $18.0 mL$ of $0.100\, M HCL$ ?

AIIMSAIIMS 2010Some Basic Concepts of Chemistry

Solution:

Given, $M_{H C l}=0.1 \,M, V_{H C l}=18.0 \, mL$
$M_{ NaHCO _3}=0.125 \, M , V_{ NaHCO _3}=?$
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.