Question

Solve this:
A solution contains Na₂CO₃ and NaHCO₃, 10 ml of the solution required 2.5 ml of 0.1 M H₂SO₄ for neutralization using phenolphthalein as indicator. Methyl orange is then added a further 2.5 ml of 0.2 M H₂SO₄ was mixed. The amount of NaHCO₃ in one litre of the solution is
Na₂CO₃ = 5.3 g

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{With}\mathrm{phenolpthalene},\mathrm{the}\mathrm{reaction}\mathrm{occurring}\mathrm{is} \\ {\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}^{+}\to {\mathrm{NaHCO}}_3+{\mathrm{Na}}^{+} \\ \mathrm{To}\mathrm{neutralise}\mathrm{whole}\mathrm{of}{\mathrm{Na}}_2{\mathrm{CO}}_3,\mathrm{volume}\mathrm{of}{\mathrm{H}}_2{\mathrm{SO}}_4\mathrm{will}\mathrm{be}\mathrm{twice}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{volume} \\ {\mathrm{N}}_1=\frac{(2\times 2.5)(2\times 0.1)}{10}=0.1\mathrm{N} \\ \mathrm{Mass}\mathrm{of}{\mathrm{Na}}_2{\mathrm{CO}}_3\mathrm{in}1\mathrm{L}\mathrm{solution}\mathrm{is}=0.1\times 53=5.3\mathrm{g} \\ \mathrm{Volume}\mathrm{of}\mathrm{sulphuric}\mathrm{acid}\mathrm{solution}\mathrm{used}\mathrm{in}\mathrm{neutralising}{\mathrm{NaHCO}}_3==2.5\mathrm{ml} \\ \mathrm{Hence},\mathrm{normality}\mathrm{of}{\mathrm{NaHCO}}_3\mathrm{solution},{\mathrm{N}}_2=\frac{2.5\times (2\times 0.2)}{10}=0.1\mathrm{N} \\ \mathrm{Mass}\mathrm{of}{\mathrm{NaHCO}}_3=0.1\times 84=8.4\mathrm{g} \end{gathered}$$