Salt Analysis
3.0 Classification of Cations
3.0 Classification of Cations
For the purpose of systematic qualitative analysis, cations are classified into five groups on the basis of their behaviour with some reagents and classification is based on whether a cation reacts with these reagents by the formation of precipitate or not (solubility difference).
| Group | Group Reagent | Ions | Colour & precipitate |
| I | Dil.$HCl$ | $P{b^{2 + }},\ Hg_2^{2 + },\ A{g^ + }$ | $PbC{l_2},\ H{g_2}C{l_2},\ AgCl$ - white |
| II A | ${H_2}S\ in\ dil.HCl$ | $H{g^{2 + }},\ C{u^{2 + }},\ B{i^{3 + }},\ C{d^{2 + }}$ | $HgS,CuS$- Black $B{i_2}{S_3}$ - Brown $CdS$ - Yellow |
| II B | ${H_2}S\ in\ dil.HCl$ | $A{s^{3 + }},A{s^{5 + }},S{b^{3 + }},S{b^{5 + }},S{n^{2 + }},S{n^{4 + }}$ | $A{s_2}{S_3},A{s_2}{S_5},Sn{S_2}$ - Yellow $S{b_2}{S_5},S{b_2}{S_3}$ - Orange $SnS$ - Brown |
| III | $N{H_4}OH$ in presence of $N{H_4}Cl$ | $F{e^{3 + }},A{l^{3 + }},C{r^{3 + }}$ | $Fe{(OH)_3}$- Brown $Al{(OH)_3}$ - White $Cr{(OH)_3}$ - Green |
| IV | ${H_2}S$ in presence of $N{H_4}OH$ & $N{H_4}Cl$ or $N{H_4}$ | $N{i^{2 + }},C{o^{2 + }},M{n^{2 + }},Z{n^{2 + }}$ | $ZnS$ - White or Grey $CoS,NiS$ - Black $MnS$ - Buff ( light pink) |
| V | ${(N{H_4})_2}C{O_3}$ in presence of $N{H_4}OH$ & $N{H_4}Cl$ | $B{a^{2 + }},S{r^{2 + }},C{a^{2 + }}$ | $BaC{O_3},SrC{O_3},CaC{O_3}$ - White |
| VI | No common group reagent | $M{g^{2 + }},N{a^ + },{K^ + },NH_4^ + $ | - |
Points to Remember
1. Group I radicals ($A{g^ + },P{b^{2 + }},Hg_2^{2 + }$) are precipitated as chlorides because the solubility product of these chlorides ($AgCl,PbC{l_2},H{g_2}C{l_2}$) is less than the solubility products of all other chlorides which remain in solution.
2. Group II radicals are precipitated as sulphides because sulphides of other metals remain in solution because of their high solubility products, $HCl$ acts as a source of ${H^ + }$ and thus decreases the concentration of ${S^{2 - }}$ due to common ion effect. Hence decreased concentration of ${S^{2 - }}$ is only sufficient to precipitate the Group II radicals only.
3. Group IIIA radicals are precipitated as hydroxides and the $N{H_4}Cl$ supresses the ionisation of $N{H_4}OH$ so that only the group IIIA are precipitated because of their low solubility product.
Note:
- Excess of $N{H_4}Cl$ should be added otherwise manganese will be precipitate as $Mn{O_2}.{H_2}O$ .
- ${(N{H_4})_2}S{O_4}$ cannot be used in place of $N{H_4}Cl$ because the $SO_4^{2 - }$ will precipitate Barium as $BaS{O_4}$ .
- $N{H_4}N{O_3}$ can't be used in place of $N{H_4}Cl$ because $NO_3^ - $ ions will oxidise $M{n^{2 + }}$ to $M{n^{3+ }}$ and thus $Mn{(OH)_3}$ will be precipitated in IIIA group.
- Only $Al{(OH)_3}$ is soluble in excess of $NaOH$ followed by boiling to form Sodium Metaluminate while $Fe{(OH)_3}$ & $Cr{(OH)_3}$ are insoluble.
4. Ammonium hydroxide increases the ionisation of ${H_2}S$ by removing ${H^ + }$ from ${H_2}S$ as unionised water.
$${H_2}S \rightleftharpoons 2{H^ + } + {S^{2 - }}$$ $${H^ + } + O{H^ - } \to {H_2}O$$
Now excess of ${S^{2 - }}$ ions are available and hence the ionic product of hydroxides of group IIIB exceed their solubility product and precipitate will be obtained. In case ${H_2}S$ is passed through a neutral solution, incomplete precipitation will take place due to the formation of $HCl$ which decreases the ionization of ${H_2}S$.
$$MnC{l_2} + {H_2}S \to MnS + 2HCl$$
