Question

Given,0.1 g of a solution containing
$N{a_2}C{O_3}$ and $NaHC{O_3}$ requires 10 mL of 0.01 N HCl for neutralization using phenolphthalein as an indicator. Wt. % of $N{a_2}C{O_3}$ in the mixture is:
(A) 25
(B) 32
(C) 50
(D) 10.6

Answer 1

Hint: Phenolphthalein indicator shows colour change when pH of the solution comes in the rage of 8 to 10. When the mixture of $N{a_2}C{O_3}$ and $NaHC{O_3}$ reacts with HCl, first of all, carbonate ions get converted to bicarbonate ions.

Complete answer:
It is given that the neutralization titration is done using phenolphthalein indicators. Phenolphthalein indicator gives colour change in pH 8 to 10.
- In the titration of the mixture of sodium carbonate and sodium bicarbonate, the first carbonate anion is reduced to bicarbonate ions. The reaction can be given as
     \[C{O_3}^{2 - } + {H^ + } \to HC{O_3}^ - \]
Then, bicarbonate ions react with acid to give water and carbon dioxide. Thus, first of all, carbonate ions get converted to bicarbonate ions. So, we can say that phenolphthalein will give colour change from pink to colourless when all carbonate ions will get converted to bicarbonate ions.
- We can say that number of moles of HCl = Normality $ \times $ Volume in L
Number of moles of HCl = $0.01 \times 0.01$ = ${10^{ - 4}}$ moles
The reaction between $N{a_2}C{O_3}$ and HCl can be given as
     \[N{a_2}C{O_3} + HCl \to NaCl + NaHC{O_3}\]
Now, we can say that if ${10^{ - 4}}$ moles of HCl is used, then ${10^{ - 4}}$ moles of $N{a_2}C{O_3}$ would have been consumed.
Molecular weight of $N{a_2}C{O_3}$ = 2(Atomic weight of Na) + Atomic weight of C + 3(Atomic weight of O)
Molecular weight of $N{a_2}C{O_3}$ = 2(23) + 12 + 3(16) = 46+12+48 = 106$gmo{l^{ - 1}}$
We know that weight = Number of moles $ \times $ molecular weight
So, weight = ${10^{ - 4}} \times 106 = 0.0106$ g
That means that the weight of $N{a_2}C{O_3}$ in 0.1 g of sample is 0.0106g.
So, weight % of $N{a_2}C{O_3}$ = $\dfrac{{{\text{Weight of N}}{{\text{a}}_2}C{O_3}}}{{{\text{Total weight of sample}}}} \times 100 = \dfrac{{0.0106}}{{0.1}} \times 100 = 10.6\% $

Thus, the correct answer of the question is (D).

Note:
This is a tricky question and we need to predict from the indicator used that only the moles of sodium carbonate need to be taken in consideration. If methyl orange was used as an indicator in this question, then we need to assume that $NaHC{O_3}$ also gets converted to water and carbon dioxide.