Question

Solution X contains $N{a_2}C{O_3}$ ​and $NaHC{O_3}$. 20 ml of X when titrated using methyl orange indicator consumed $60{\text{ }}ml$ of $0.1{\text{ }}M$ $HCl\;$ solution. In another experiment, $20{\text{ }}ml$ of X solution when titrated using phenolphthalein consumed $20{\text{ }}ml {\text{ }} of {\text{ }}0.1{\text{ }}M\;$ solution. The concentrations ($in\;mol\;lit ^{- 1}$) of $N{a_2}C{O_3}$ and $NaHC{O_3}$in X are respectively:

$${A.{\text{ }}0.01,{\text{ }}0.02} \\\ {B.{\text{ }}0.1,{\text{ }}0.1} \\\ {C.{\text{ }}0.01,{\text{ }}0.01} \\\ D.{\text{ }}0.1,{\text{ }}0.01 \\\$$

Answer 1

We must remember about acid-base titration. In an acid-base titration, we can find the equivalence point with the help of an indicator which changes its color at the endpoint. In the case of poly acidic base or polybasic acid titration there is more than one endpoint. One indicator gives one color change at a specific endpoint. Therefore, we use more than one indicator.

Complete answer:
In the question, they given that
Solution $X{\text{ }} = N{a_2}C{O_3}\; + \;NaHC{O_3}$
In Experiment 1 ,
$20{\text{ }}ml$ X solution + methyl orange indicator = $60{\text{ }}ml{\text{ }}of{\text{ }}0.1M{\text{ }}HCl$
This shows that, when methyl orange indicator is used, sodium carbonate is titrated to sodium bicarbonate and sodium bicarbonate is titrated to carbonic acid.
In Experiment 2 ,
$20{\text{ }}ml$ X solution + phenolphthalein indicator = $20{\text{ }}ml {\text{ }}of{\text{ }}0.1M{\text{ }}HCl$
This shows that,when phenolphthalein indicator is used, sodium carbonate is titrated to sodium bicarbonate.
From this above data in experiment 2, we can conclude that
$20{\text{ }}ml{\text{ }}0.1{\text{ }}M{\text{ }}HCl$ correspond to titration of $N{a_2}C{O_3}$, Sodium carbonate
So, Number of millimoles of $N{a_2}C{O_3}$ in the solution will be equal to
$= 20 \times 0.1$
$= 2{\text{ }}mmol$
so, similarly in experiment 1,
$40{\text{ }}ml {\text{ }}0.1{\text{ }}M{\text{ }}HCl$ correspond to titration of $N{a_2}C{O_3}$, Sodium carbonate to $NaHC{O_3}$ and finally to carbonic acid, and remaining $20{\text{ }}ml{\text{ }}of\;0.1M{\text{ }}HCl$ for titration of actual $NaHC{O_3}$ in X solution.
So, number of millimoles of sodium bicarbonate present in the solution X is equal to
$= 20 \times 0.1$
$= 2{\text{ }}mmol$
Hence, the concentration of $N{a_2}C{O_3} = \;\dfrac{{{\text{2 mmol}}}}{{{\text{20 ml}}}} = 0.1M$
Similarly, the concentration of $NaHC{O_3}{\text{ }} = \;\dfrac{{{\text{2 mmol}}}}{{{\text{20 ml}}}} = 0.1M$
$\left[ {Na2CO3} \right] = \left[ {NaHCO3} \right] = 0.1M$
Hence, the correct option is option B.

Note:
With phenolphthalein indicator, $NaHC{O_3}$ does not react with $HCl$ whereas with $N{a_2}C{O_3}$ will carry out only 50% reaction whereas with methyl orange indicator, $NaHC{O_3}$ reacts completely with $HCl$ and with $N{a_{2}}C{O_3}$ ​ will carry out 100% reaction.