0.4 g mixture of NaOH, $$Na_2CO_3$$ and some inert impurities was first titrated with $$\frac{N}{10}$$ 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 $$Na_2CO_3$$ in the mixture is ______ (Rounded-off to the nearest integer)

This is a double indicator acid-base titration of a mixture containing NaOH, $$Na_2CO_3$$, and inert impurities with $$\frac{N}{10}$$ HCl.

In the first step using phenolphthalein indicator, all of the NaOH is neutralised and half of the $$Na_2CO_3$$ is converted to $$NaHCO_3$$. The volume of HCl used is $$V_1 = 17.5$$ mL.

In the second step after adding methyl orange indicator, the remaining half of $$Na_2CO_3$$ (now as $$NaHCO_3$$) is neutralised. The volume of HCl used is $$V_2 = 1.5$$ mL.

The volume of HCl used for the second half of $$Na_2CO_3$$ equals $$V_2 = 1.5$$ mL. So the total volume of HCl for complete neutralisation of $$Na_2CO_3$$ is $$2 \times V_2 = 2 \times 1.5 = 3.0$$ mL.

The milliequivalents of $$Na_2CO_3$$ are: $$3.0 \times \frac{1}{10} = 0.3$$ meq. Since the equivalent weight of $$Na_2CO_3$$ is $$\frac{106}{2} = 53$$ g/eq, the weight of $$Na_2CO_3$$ is: $$\frac{0.3 \times 53}{1000} = 0.0159$$ g.

The weight percentage of $$Na_2CO_3$$ in the mixture is: $$\frac{0.0159}{0.4} \times 100 = 3.975\% \approx 4\%$$

The weight percentage of $$Na_2CO_3$$ in the mixture is $$\mathbf{4}$$.