Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with sodium hydrogen carbonate, NaHCO_ 3(a q) +HCl_ (a q) arrow NaCl_ (a q) +H_ 2 O_ (l) +CO_ 2(g) How many milliliters of 0.125 M NaHCO_3

Correct answer: A. 14.4 ~mL aligned & Given, M_ HCl =0.1 M,V_ HCl =18.0 ~mL & M_ NaHCO_3 =0.125 M,V_ NaHCO_3 = ? aligned On applying, M_ HC × V_ HCl =M_ NaHCO_3 × V_ NaHCO_3 0.1 × 18=0.125 × V_ NaHCO_3 M_ NaHCO_3 =14.4 ~mL Thus, 14.4 ~mL of the 1.25 M NaHCO M_3 solution is needed to neutralise 18.0 ~mL of the 0.100 MHCl solution.

AIIMS 2010

Exam: AIIMS
Year: 2010
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

AIIMS 2010 · Chemistry · Some Basic Concepts of Chemistry

Stomach acid, a dilute solution of HCl in water, can be neutralized...

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}

Options

A $14.4 \mathrm{~mL}$
B $12.0 \mathrm{~mL}$
C $14.0 \mathrm{~mL}$
D $13.2 \mathrm{~mL}$

Correct answer

A. $14.4 \mathrm{~mL}$

Step-by-step 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.

Practice Some Basic Concepts of Chemistry on Quantrex Academy →

More from Some Basic Concepts of Chemistry