Solution " $X$ " contains $\mathrm{Na}_2 \mathrm{CO}_3$ and $\mathrm{NaHCO}_3$. $20 \mathrm{~mL}$ of $X$…
Solution " $X$ " contains $\mathrm{Na}_2 \mathrm{CO}_3$ and $\mathrm{NaHCO}_3$. $20 \mathrm{~mL}$ of $X$ when titrated using methyl orange indicator consumed $60 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{HCl}$ solution. In another experiment, $20 \mathrm{~mL}$ of $X$ solution when titrated using phenolphthalein consumed $20 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{HCl}$ solution. The concentrations (in $\mathrm{mol} \mathrm{L}^{-1}$ ) of $\mathrm{Na}_2 \mathrm{CO}_3$ and $\mathrm{NaHCO}_3$ in $X$ are respectively
$0.01,0.02$
$0.1,0.1$
$0.01,0.01$
$0.1,0.01$
Solution
For titration of a basic solution of $\mathrm{Na}_2 \mathrm{CO}_3$ and $\mathrm{NaHCO}_3$ against $\mathrm{HCl}$, if phenolphthalein is used as indicator, the end point is indicated only for half neutralization of $\mathrm{Na}_2 \mathrm{CO}_3$, i.e., (upto $\mathrm{NaHCO}_3$ ).
$
\mathrm{Na}_2 \mathrm{CO}_3+\mathrm{HCl} \longrightarrow \mathrm{NaHCO}_3+\mathrm{NaCl}
$
The remaining solution then contains the unreacted $\mathrm{NaHCO}_3$ from this reaction plus the unreacted $\mathrm{NaHCO}_3$ originally in the solution. At the phenolphthalein end point, there is no reaction between $\mathrm{HCl}$ and $\mathrm{NaHCO}_3$.
From the equations
$
\begin{aligned}
\text { Mol of } \mathrm{HCl} \text { consumed } & =\mathrm{mol} \text { of } \mathrm{Na}_2 \mathrm{CO}_3 \\
20 \mathrm{~mL} \text { of } 0.1 \mathrm{M} & =20 \mathrm{~mL} \text { of } 0.1 \mathrm{M}
\end{aligned}
$
$\therefore$ The concentration of $\mathrm{Na}_2 \mathrm{CO}_3$ in solution $X=0.1 \mathrm{M}$.
Note that for a quantity of $\mathrm{Na}_2 \mathrm{CO}_3$, exactly half volume of the $\mathrm{HCl}$ is used at the phenolphthalein end point and the second half volume of the $\mathrm{HCl}$ is required for complete neutralization of $\mathrm{Na}_2 \mathrm{CO}_3$ at methyl orange end point.
$
\mathrm{NaHCO}_3+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}
$
$\therefore$ Volume of $\mathrm{HCl}$ required to neutralize
$
\begin{aligned}
\mathrm{Na}_2 \mathrm{CO}_3 \text { in original sample } & =2 \times 20 \mathrm{~mL} \\
& =40 \mathrm{~mL}
\end{aligned}
$
If methyl orange is used, the end point is indicated when all the alkali is neutralized.
$
\mathrm{NaHCO}_3+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}
$
As $40 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}$ is consumed in complete neutralization of $\mathrm{Na}_2 \mathrm{CO}_3$ at methyl orange end point, so the volume of $\mathrm{HCl}$ used to neutralized $\mathrm{NaHCO}_3$ from the original sample would be
Remaining $\mathrm{HCl}=60-40=20 \mathrm{~mL}$ of $0.1 \mathrm{M}$
As per equation $=1 \mathrm{~mol}$ of $\mathrm{NaHCO}_3=1 \mathrm{~mol}$ of $\mathrm{HCl}$
$\therefore 0.1 \mathrm{~mol}$ of $\mathrm{NaHCO}_3=0.1 \mathrm{~mol}$ of $\mathrm{HCl}$