Question

# Describe in each case, one chemical test that would enable you to distinguish between the following pairs of chemicals. Describe what happens with each chemical or state 'no visible reaction'. (i) Sodium chloride solution and sodium nitrate solution. (ii) Sodium sulphate solution and sodium chloride solution. (iii) Calcium nitrate solution and zinc nitrate solution.

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Solution

## i) Sodium chloride solution and sodium nitrate solution On adding conc. sulphuric acid (H​2SO4) to sodium chloride solution, a colourless gas, HCl, evolves. Evolution of HCl gas can be confirmed by bringing a glass rod dipped in ammonium hydroxide solution near the mouth of the test tube. Dense white fumes of ammonium chloride will be formed. $2\mathrm{NaCl}+{\mathrm{H}}_{2}{\mathrm{SO}}_{4}\to {\mathrm{Na}}_{2}{\mathrm{SO}}_{4}+2\mathrm{HCl}\phantom{\rule{0ex}{0ex}}\mathrm{Colourless}\mathrm{gas}\phantom{\rule{0ex}{0ex}}\mathrm{HCl}+{\mathrm{NH}}_{4}\mathrm{OH}\to {\mathrm{NH}}_{4}\mathrm{Cl}+{\mathrm{H}}_{2}\mathrm{O}\phantom{\rule{0ex}{0ex}}\mathrm{Dense}\mathrm{white}\mathrm{fumes}$ On adding conc. sulphuric acid (H​2SO4) to sodium nitrate solution, reddish brown fumes are observed. $2{\mathrm{NaNO}}_{3}+{\mathrm{H}}_{2}{\mathrm{SO}}_{4}\to {\mathrm{Na}}_{2}{\mathrm{SO}}_{4}+{\mathrm{HNO}}_{3}\phantom{\rule{0ex}{0ex}}\mathrm{Reddish}\mathrm{brown}\mathrm{fumes}$ ii) Sodium sulphate solution and sodium chloride solution On adding barium chloride solution to sodium sulphate solution, a white precipitate is formed. On the other hand, barium chloride will have no visible reaction with sodium chloride solution. ${\mathrm{BaCl}}_{2}+{\mathrm{Na}}_{2}{\mathrm{SO}}_{4}\to {\mathrm{BaSO}}_{4}+2\mathrm{NaCl}\phantom{\rule{0ex}{0ex}}\mathrm{White}\mathrm{precipitate}\phantom{\rule{0ex}{0ex}}{\mathrm{BaCl}}_{2}+\mathrm{NaCl}\to \mathrm{No}\mathrm{reaction}$ iii) Calcium nitrate solution and zinc nitrate solution On adding sodium hydroxide to both calcium nitrate and zinc nitrate, white precipitates are formed. For calcium nitrate, the precipitate will be insoluble in excess sodium hydroxide. But for zinc nitrate, the precipitate will be soluble in excess sodium hydroxide. $\mathrm{Ca}{\left({\mathrm{NO}}_{3}\right)}_{2}+2\mathrm{NaOH}\to \mathrm{Ca}{\left(\mathrm{OH}\right)}_{2}+2{\mathrm{NaNO}}_{3}\phantom{\rule{0ex}{0ex}}\mathrm{White}\mathrm{precipitate}\phantom{\rule{0ex}{0ex}}\mathrm{Ca}{\left(\mathrm{OH}\right)}_{2}+2\mathrm{NaOH}\mathit{\left(}excess\mathit{\right)}\to \mathrm{Insoluble}\phantom{\rule{0ex}{0ex}}\mathrm{Zn}{\left({\mathrm{NO}}_{3}\right)}_{2}+2\mathrm{NaOH}\to \mathrm{Zn}{\left(\mathrm{OH}\right)}_{2}+2{\mathrm{NaNO}}_{3}\phantom{\rule{0ex}{0ex}}\mathrm{White}\mathrm{precipitate}\phantom{\rule{0ex}{0ex}}\mathrm{Zn}{\left(\mathrm{OH}\right)}_{2}+2\mathrm{NaOH}\mathit{\left(}excess\mathit{\right)}\to {\mathrm{Na}}_{2}\left[\mathrm{Zn}{\left(\mathrm{OH}\right)}_{4}\right]$

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