Question

0.4 g mixture of NaOH, Na$_2$CO$_3$ and some inert impurities was first titrated with N/10 HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrated. 1.5 mL of same HCl was required for the next end point. The weight percentage of Na$_2$CO$_3$ in the mixture is _________

Hint:

In double indicator titrations, the volume of acid used between the phenolphthalein and methyl orange end points ($V_2$) always represents exactly half of the carbonate present.

Answer 1

Correct Answer: 4

Step 1: Phenolphthalein end point ($V_1 = 17.5$ mL) accounts for: All NaOH + half of Na$_2$CO$_3$ ($Na_2CO_3 \rightarrow NaHCO_3$).
Step 2: Methyl orange end point ($V_2 = 1.5$ mL) accounts for: The remaining half of Na$_2$CO$_3$ ($NaHCO_3 \rightarrow CO_2$).
Step 3: Total volume of HCl for Na$_2$CO$_3 = 2 \times V_2 = 2 \times 1.5 = 3.0$ mL.
Step 4: Equivalents of Na$_2$CO$_3$ = Equivalents of HCl = $N \times V = 0.1 \times (3/1000) = 3 \times 10^{-4}$.
Step 5: Mass of Na$_2$CO$_3$ = $\text{Equivalents} \times \text{Eq. weight} = 3 \times 10^{-4} \times (106/2) = 0.0159$ g.
Step 6: Weight % = $(0.0159 / 0.4) \times 100 = 3.975% \approx 4%$.