Question

A salt on reaction with dil. ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ produces a colorless gas which turns lime water milky and an acidified orange coloured potassium dichromate solution green. The salt may be:
A.A carbonate
B.A sulphite
C.A sulphide
D.An acetate

Answer 1

Hint: All metal carbonates undergo one common reaction which is thermal decomposition. When a metal carbonate is heated, it undergoes breakdown to form the corresponding metal oxide and carbon dioxide gas.
-Sulphur dioxide can turn an acidified orange coloured potassium dichromate solution green.

Complete step by step answer:
A carbonate salt on reaction with dilute sulphuric acid and after being heated produces a colorless and odourless gas with brisk effervescence. This gas is carbon dioxide. When the carbon dioxide gas is passed through lime water which is calcium hydroxide, it turns milky due to the formation of calcium carbonate. But it doesn’t have any effect when passed through a potassium dichromate solution.
 For example, calcium carbonate on heating with dilute sulphuric acid gives carbon dioxide which turns lime water milky.
${\text{CaC}}{{\text{O}}_{\text{3}}}\left( {\text{s}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\left( {{\text{aq}}} \right) \to {\text{CaS}}{{\text{O}}_{\text{4}}}\left( {{\text{aq}}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right){\text{ + C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
\[{\text{Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}}\left( {{\text{aq}}} \right) + {\text{C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{CaC}}{{\text{O}}_3}\left( {\text{s}} \right) + {{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right)\]
Thus, option A is wrong.
When a sulphite salt reacts with dilute sulphuric acid, sulphur dioxide gas is produced. This sulphur dioxide gas when passed through lime water which is calcium hydroxide, it turns milky due to the formation of calcium sulphate. It also turns an acidified orange coloured potassium dichromate solution green due to the formation of chromium ions.
${\text{S}}{{\text{O}}_{\text{3}}}^{{\text{2 - }}}{\text{ + 2}}{{\text{H}}^{\text{ + }}} \to {\text{S}}{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
\[{\text{Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}}\left( {{\text{aq}}} \right) + {\text{S}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{CaS}}{{\text{O}}_3}\left( {\text{s}} \right) + {{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right)\]
${\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{{\text{2 - }}}{\text{ + 2}}{{\text{H}}^{\text{ + }}}{\text{ + 3S}}{{\text{O}}_{\text{2}}} \to {\text{C}}{{\text{r}}^{{\text{3 + }}}}{\text{ + 3S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
So, option B is correct.
Sulphide ion when treated with an acid liberates hydrogen sulfide gas.
${{\text{S}}^{{\text{2 - }}}}{\text{ + 2}}{{\text{H}}^{\text{ + }}} \to {{\text{H}}_{\text{2}}}{\text{S}}$
But the hydrogen sulfide gas produced when passed through lime water which is calcium hydroxide does not turn milky. So, option C is not correct.
Acetate ion treatment with dilute sulphuric acid, acetic acid will be produced by double decomposition reaction which will give the solution a sharp and distinct smell of vinegar.
${\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^{\text{ - }}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} + {\text{HS}}{{\text{O}}_{\text{4}}}^ - $
Acetic acid is not a gas and so option D is also incorrect.

Hence option C is correct.

Note:
If an excess of carbon dioxide is passed through lime water, then the solution will not be milky. This time it will be clear and colorless because calcium bicarbonate is formed which is soluble in water. The reaction is:
${\text{Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}}\left( {{\text{aq}}} \right) + {\text{C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{Ca}}{\left( {{\text{HC}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}\left( {{\text{aq}}} \right)$