1.
Potassium chromate solution is added to an aqueous solution of a metal chloride. The
precipitate thus obtained is insoluble in acetic acid. When precipitate is subjected to flame test the colour of the flame is
$$BaC{l_2} + {K_2}Cr{O_4} \to BaCr{O_4} + 2KCl.BaCr{O_4}$$
is insoluble in acetic acid and $$Ba$$ gives apple green colour in flame test.
2.
A gas $$“X”$$ is passed through water to form a saturated solution. The aqueous solutions on treatment with the $$AgN{O_3}$$ gives a white preciptate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of a colourless gas $$“Y”.$$ Identify $$'X'$$ and $$'Y'.$$
3.
Sodium carbonate cannot be used in place of $${\left( {N{H_4}} \right)_2}C{O_3}$$ for the identification of $$C{a^{2 + }},B{a^{2 + }}$$ and $$S{r^{2 + }}ions$$ ( in group $$V$$ ) during mixture analysis because :
A.
$$M{g^{2 + }}\,ions$$ will also be precipitated.
B.
Concentration of $$CO_3^{2 - }\,ions$$ is very low.
C.
Sodium ions will react with acid radicals.
D.
$$N{a^ + }\,ions$$ will interfere with the detection of $$C{a^{2 + }},B{a^{2 + }},S{r^{2 + }}\,ions.$$
Answer :
$$M{g^{2 + }}\,ions$$ will also be precipitated.
If $$N{a_2}C{O_3}$$ is used in place of $${\left( {N{H_4}} \right)_2}C{O_3}.$$ It will precipitate group $$V$$ radicals as well as magnesium radicals. The reason for this is the high ionization of $$N{a_2}C{O_3}$$ in water into $$N{a^ + }$$ and $$CO_3^{2 - }.$$ Now the higher concentration of $$CO_3^{2 - }$$ is available which exceeds the solubility product of group $$V$$ radicals as well as that of magnesium radicals.
4.
The sodium extract prepared from sulphanilic acid, contains $$SC{N^ - }.$$ It gives blood red colouration with
5.
A solution when diluted with $${H_2}O$$ and boiled, gives a white precipitate. On addition of excess $$N{H_4}Cl/N{H_4}OH,$$ the volume of precipitate decreases leaving behind a white gelatinous precipitate. Identify the precipitate which disolves in $$N{H_4}OH/N{H_4}Cl$$
When the sodium fusion extract is added with $$FeC{l_3}$$ and then the resulting solution is acidified with dilute hydrochloric acid, the appearance of Prussian blue colouration confirms the presence of nitrogen in the organic compound.
$$Na + C + N \to NaCN$$
$$FeS{O_4} + 2NaCN \to $$ $$Fe{\left( {CN} \right)_2} + N{a_2}S{O_4}$$
$$Fe{\left( {CN} \right)_2} + 4NaCN \to $$ $$N{a_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$$
$$3N{a_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] + 4FeC{l_3} \to $$ $$\mathop {F{e_4}{{\left[ {Fe\left( {C{N_6}} \right)} \right]}_3}}\limits_{{\text{Prussian blue}}} + 12NaCl$$
7.
The organic compound that gives following qualitative analysis is:
8.
A substance on treatment with $$dil.{H_2}S{O_4}$$ liberates a colourless gas which produces
(I) turbidity with baryta water and
(II) turns acidified dichromate solution green.
The reaction indicates the presence of
NOTE:
Only group $$I$$ cations are precipitated by dil. $$HCl$$
$$\mathop {SO_3^{2 - }}\limits_{\left( X \right)} + {H_2}S{O_4} \to \mathop {S{O_2}}\limits_{\left( Y \right)} + {H_2}O + SO_4^{2 - }$$
$$3S{O_2} + {K_2}C{r_2}{O_7} + {H_2}S{O_4} \to {K_2}S{O_4} + \mathop {C{r_2}{{\left( {S{O_4}} \right)}_3}}\limits_{\left( {{\text{green}}\,\,{\text{colour}}\,{\text{solution}}} \right)} + {H_2}O$$
Only $$PbC{l_2}\,{\text{and}}\,H{g_2}C{l_2}$$ will precipitate as $$P{b^{2 + }}\,{\text{and}}\,Hg_2^{2 + }$$ as first group basic radicals and their solubility product is less than the other radicals.
NOTE: dil. $$HCl$$ is the first group reagent.