Salt Analysis
4.0 Identification of Basic Radicals
4.0 Identification of Basic Radicals
All confirmatory tests for basic radicals are performed with the salt solution.
1. Group I ($A{g^ + },P{b^{2 + }},Hg_2^{2 + }$)
- $PbC{l_2}$ gives a yellow precipitate with ${K_2}Cr{O_4}$ . The precipitate is insoluble in acetic acid but soluble in $NaOH$ $$\begin{equation} \begin{aligned} Pb{(N{O_3})_2} + {K_2}Cr{O_4} \to PbCr{O_4} \downarrow (Yellow\ precipitate) + 2KN{O_3} \\ PbCr{O_4} + 4NaOH \to N{a_2}[Pb{O_2}] + N{a_2}Cr{O_4} + 2{H_2}O \\ \\\end{aligned} \end{equation} $$
- $$\begin{equation} \begin{aligned} Pb{(N{O_3})_2} + 2KI \to Pb{I_2} \downarrow (Yellow) + 2KN{O_3} \\ Pb{I_2} + 2KI(excess) \to {K_2}[Pb{I_4}] \\ \\\end{aligned} \end{equation} $$
- $AgCl$ is soluble in $N{H_4}OH$ forming a complex while $H{g_2}C{l_2}$ forms a black precipitate with $N{H_4}OH$. $$\begin{equation} \begin{aligned} AgCl + 2N{H_4}OH \to Ag{(N{H_3})_2}Cl + 2{H_2}O \\ H{g_2}C{l_2} + 2N{H_4}OH \to HgClN{H_2} + Hg \downarrow(Amino\ Mercuric\ Chloride) + N{H_4}Cl + 2{H_2}O \\\end{aligned} \end{equation} $$
2. Group IIA ($H{g^{2 + }},\ C{u^{2 + }},\ B{i^{3 + }},\ C{d^{2 + }}$)
- $H{g^{2+}}$ ions in solution on addition of $SnC{l_2}$ gives white precipitate turning black. $$\begin{equation} \begin{aligned} 2H{g^{2 + }} + SnC{l_2} \to S{n^{4 + }} + H{g_2}C{l_2} \downarrow (White) \\ H{g_2}C{l_2} + SnC{l_2} \to SnC{l_4} + 2Hg \downarrow (Black) \\\end{aligned} \end{equation} $$
- $C{u^{2 + }}$ ions in solution gives a pale blue precipitate which gives deep blue colour with excess of $N{H_4}OH$ . $$C{u^{2 + }} + 4N{H_4}OH \to {[Cu{(N{H_3})_4}]^{2 + }}(Deep\ colour\ in\ colour) + 4{H_2}O$$ $C{u^{2 + }}$ ions give chocolate precipitate with ${K_4}Fe{(CN)_6}$. $$2C{u^{2 + }} + {K_4}Fe{(CN)_6} \to C{u_2}[Fe{(CN)_6}] + 4{K^ + }$$ $C{u^{2 + }}$ reacts with $KCN$ to form a yellow precipitate of Copper(II) Cyanide. $$C{u^{2 + }} + 2C{N^ - } \to Cu{(CN)_2} \downarrow $$ The precipitate quickly decomposes into white copper(I) cyanide and cyanogens. $$2Cu{(CN)_2} \to 2CuCN + {(CN)_2} \uparrow $$ Excess reagent dissolves the precipitate, and a colourless complex is formed. $$Cu(CN) \downarrow + 3C{N^ - } \to {[Cu{(CN)_4}]^{3 - }}$$ The complex is stable and will not react with ${H_2}S$.
- $B{i^{3 + }}$ ions in solution of $HCl$ on addition of water gives cloudy precipitate. $$\begin{equation} \begin{aligned} BiC{l_3} + {H_2}O \to BiOCl \downarrow (White\ precipitate) + 2HCl \\ \\\end{aligned} \end{equation} $$ When treated with sodium stannite a black precipitate is obtained. $$2BiC{l_3} + 3NaSnO \to 2Bi \downarrow (Black) + 6NaCl + 3N{a_2}Sn{O_3} + 3{H_2}O$$Potassium iodide when added drop wise, black precipitate of Bismuth(III) Iodide. $$B{i^{3 + }} + 3{I^ - } \to Bi{I_3} \downarrow (Black)$$ The precipitate dissolves readily in excess reagent, when orange- coloured tetraiodobismuthate ions are formed. $$Bi{I_3} + {I^ - } \rightleftharpoons {[Bi{I_4}]^ - }(Orange)$$ When diluted with water, the above reaction is reversed and black bismuth iodide is reprecipitated. On heating with water, it turns orange. $$Bi{I_3} \downarrow + {H_2}O \to BiOI \downarrow (Orange) + 2{H^ + } + 2{I^ - }$$
- $C{d^{2 + }}$ ions in solution, with Ammonium hydroxide gives a white precipitate which dissolves. $$\begin{equation} \begin{aligned} C{d^{2 + }} + 2N{H_4}OH \to Cd{(OH)_2} \downarrow (Yellow) + 2NH_4^ + \\ Cd{(OH)_2} + 2N{H_4}OH \to [Cd{(N{H_3})_4}]{(OH)_2} \\ \\\end{aligned} \end{equation} $$ $C{d^{2 + }}$ reacts with $KCN$ to form white precipitate of Cadmium Cyanide. $$C{d^{2 + }} + 2C{N^ - } \to Cd{(CN)_2}$$ An excess of the reagent dissolves the precipitate, with the formation of a complex. $$Cd{(CN)_2} + 2C{N^ - } \to {[Cd{(CN)_4}]^{2 - }}$$ The colourless complex is not very stable, when ${H_2}S$ is introduced $CdS$ is precipitated. $${[Cd{(CN)_4}]^{2 - }} + {H_2}S \to CdS\ (Yellow) + 2{H^ + } + 4C{N^ - }$$
3. Group II B ($A{s^{3 + }},A{s^{5 + }},S{b^{3 + }},S{b^{5 + }},S{n^{2 + }},S{n^{4 + }}$)
- $A{s^{3+}}$ ions in solution gives yellow precipitate with ammonium molybadate and $HN{O_3}$. $$\begin{equation} \begin{aligned} A{s^{3 + }} \to A{s^{5 + }}(as\ {H_3}As{O_4}) \\ {H_3}As{O_4} + 12{(N{H_4})_2}Mo{O_4} + 21HN{O_3} \to {(N{H_4})_3}AsM{o_{12}}{O_{40}} \downarrow + 21N{H_4}N{O_3} + 12{H_2}O \\\end{aligned} \end{equation} $$
- $S{n^{2+}}$ ions in solution as $SnC{l_2}$ gives white precipitate $HgC{l_2}$, which turns black on standing. $$\begin{equation} \begin{aligned} 2HgC{l_2} + SnC{l_2} \to SnC{l_4} + H{g_2}C{l_2} \downarrow (White) \\ H{g_2}C{l_2} + SnC{l_2} \to SnC{l_4} + 2Hg \downarrow (Black) \\\end{aligned} \end{equation} $$
- $S{b^{3 + }}$ ions in solution as $SbC{l_3}$, on addition of water gives white precipitate. $$SbC{l_3} + {H_2}O \to SbOCl \downarrow (White) + 2HCl$$
4. Group III ($F{e^{3 + }},A{l^{3 + }},C{r^{3 + }}$)
- White precipitate of $Al{(OH)_3}$ is soluble in $NaOH$. $$Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O$$
- Precipitate of $Cr{(OH)_3}$ is soluble in $NaOH + B{r_2}$ water and addition of $BaC{l_2}$ to this solution gives yellow precipitate. $$\begin{equation} \begin{aligned} B{r_2} + {H_2}O \to 2HBr + (O) \\ 2Cr{(OH)_3} + 4NaOH + 3(O) \to 2N{a_2}Cr{O_4} + 5{H_2}O \\ N{a_2}Cr{O_4} + BaC{l_2} \to BaCr{O_4} \downarrow (Yellow\ precipitate) + 2NaCl \\\end{aligned} \end{equation} $$ $Fe{(OH)_3}$ is insoluble in $NaOH$.
- Brown precipitate of $Fe{(OH)_3}$ is dissolved in $HCl$ and addition of $KCNS$ to this solution gives blood red colour. $$\begin{equation} \begin{aligned} Fe{(OH)_3} + 3HCl \to FeC{l_3} + 3{H_2}O \\ FeC{l_3} + 3KCNS \to Fe{(CNS)_3}(Blood\ red) + 3KCl \\\end{aligned} \end{equation} $$ Also on addition of ${K_4}Fe{(CN)_6}$ to this solution, a prissian blue colour is obtained. $$3{K_4}Fe{(CN)_6} + FeC{l_3} \to F{e_4}{[Fe{(CN)_6}]_3}(Prussian\ blue\ colour) + 12KCl$$
5. Group IV ($N{i^{2 + }},C{o^{2 + }},M{n^{2 + }},Z{n^{2 + }}$)
- $N{i^{2 + }}$ & $C{o^{2 + }}$ ions in solution, on addition of $KHC{O_3}$ & $B{r_2}$ water gives apple green colour if $C{o^{2 + }}$ is present & black precipitate if $N{i^{2 + }}$ is present. $$\begin{equation} \begin{aligned} CoC{l_2} + 6KHC{O_3} \to {K_4}[Co{(C{O_3})_3}] + 2KCl + 3C{O_2} \uparrow + 3{H_2}O \\ 2{K_4}[Co{(C{O_3})_3}] + 2KHC{O_3} + [O] \to 2{K_3}[Co{(C{O_3})_3}](Apple\ green\ colour) + 2{K_2}C{O_3} + {H_2}O \\ NiC{l_2} + 2KHC{O_3} \to NiC{O_3} + 2KCl + {H_2}O + C{O_2} \uparrow \\ 2NiC{O_3} + 4NaOH + [O] \to N{i_2}{O_3} \downarrow (Black\ precipitate) + 2N{a_2}C{O_3} + 2{H_2}O \\\end{aligned} \end{equation} $$
- $Z{n^{2 + }}$ ions in solution give a white precipitate with $NaOH$, which dissolves in excess of $NaOH$ . $$\begin{equation} \begin{aligned} Z{n^{2 + }} + 2NaOH \to Zn{(OH)_2} \downarrow (White) + 2N{a^ + } \\ Zn{(OH)_2} + 2NaOH \to N{a_2}Zn{O_2} + 2{H_2}O \\\end{aligned} \end{equation} $$
- $M{n^{2 + }}$ ions in solution give pink precipitate with $NaOH$ turning black or brown on heating. $$\begin{equation} \begin{aligned} M{n^{2 + }} + 2NaOH \to Mn{(OH)_2} + 2N{a^ + } \\ Mn{(OH)_2} + [O] \to Mn{O_2}(Brown\ or\ Black) + {H_2}O \\\end{aligned} \end{equation} $$
6. Group V ($B{a^{2 + }},S{r^{2 + }},C{a^{2 + }}$)
(i) $B{a^{2 + }}$ ions in solution give
- Yellow precipitate with ${K_2}Cr{O_4}$ $$B{a^{2 + }} + {K_2}Cr{O_4} \to BaCr{O_4} \downarrow (Yellow) + 2{K^ + }$$
- White precipitate with ${(N{H_4})_2}S{O_4}$ $$B{a^{2 + }} + {(N{H_4})_2}S{O_4} \to BaS{O_4} \downarrow (White) + 2NH_4^ + $$
- White precipitate with ${(N{H_4})_2}{C_2}{O_4}$ $$B{a^{2 + }} + {(N{H_4})_2}{C_2}{O_4} \to Ba{C_2}{O_4} \downarrow (White) + 2NH_4^ + $$
(ii) $S{r^{2 + }}$ ions give white precipitate with ${(N{H_4})_2}S{O_4}$ & ${(N{H_4})_2}{C_2}{O_4}$ $$\begin{equation} \begin{aligned} S{r^{2 + }} + {(N{H_4})_2}{C_2}{O_4} \to Sr{C_2}{O_4} \downarrow (White) + 2NH_4^ + \\ S{r^{2 + }} + {(N{H_4})_2}S{O_4} \to SrS{O_4} \downarrow (White) + 2NH_4^ + \\\end{aligned} \end{equation} $$
(iii) $C{a^{2 + }}$ ions give white precipitate with ${(N{H_4})_2}{C_2}{O_4}$ only. $$C{a^{2 + }} + {(N{H_4})_2}{C_2}{O_4} \to Ca{C_2}{O_4} \downarrow (White) + 2NH_4^ + $$
7. Group VI ($NH_4^ + ,N{a^ + },{K^ + },M{g^{2 + }}$)
(i) All ammonium salts on heating with an alkali, say $NaOH$ gives a colourless, pungent smeeling gas ammonia($N{H_4}Cl$). $$N{H_4}Cl + NaOH \to NaCl + N{H_3} \uparrow + {H_2}O$$
- Gas evolved gives white fumes with a rod dipped in conc. $HCl$. $$N{H_3} + HCl \to N{H_4}Cl \uparrow (White\ fumes)$$
- Paper soaked in $ CuS{O_4}$solution, becomes deep blue due to complex formation with $N{H_4}Cl$. $$4N{H_3} + CuS{O_4} \to [Cu{(N{H_3})_4}]S{O_4}(Deep\ blue)$$
- With $H{g_2}{(N{O_3})_2}$ a black colour is obtained. $$2N{H_3} + H{g_2}{(N{O_3})_2} \to Hg \downarrow (Black) + Hg(N{H_2})N{O_3} \downarrow + N{H_4}N{O_3}$$
- An aqueous solution of Ammonium Chloride gives a brown precipitate with Nessler's reagent (alkaline solution of Potassium tetraiodomercurate(II)). $$N{H_4}Cl + 2{K_2}Hg{I_4} + 3KOH \to OHgHgN{H_2}I + 7KI + 2{H_2}O$$
(ii) Potassium salts give a yellow precipitate with Sodium Cobalt Nitrite. $$N{a_3}[Co{(N{O_2})_6}] + 3KCl \to {K_3}[Co{(N{O_2})_6}](Yellow) + 3NaC$$
(iii) Sodium salts gives a heavy white precipitate with Poyassium Dihydrogen Antimonite.
$$NaCl + K{H_2}Sb{O_4} \to Na{H_2}Sb{O_4} \downarrow (White\ precipitate) + KCl$$
(iv) $M{g^{2 + }}$ gives white precipitate of Magnesium Hydroxide with Sodium Hydroxide.
$$M{g^{2 + }} + 2N{H_3} + 2{H_2}O \to Mg{(OH)_2} + 2NH_4^ + $$ The precipitate obtained is sparingly soluble in water but readily soluble in Ammonium salt. $M{g^{2 + }}$ ions gives white precipitate with $N{H_4}OH$ & ${(N{H_4})_2}HP{O_4}$.
$$M{g^{2 + }} + {(N{H_4})_2}HP{O_4} + N{H_4}OH \to Mg(N{H_4})P{O_4} \downarrow + 2NH_4^ + + {H_2}O$$
This reaction is carried out in presence of $N{H_4}Cl$ because it prevents precipitation of Magnesium Hydroxide.
