# Method and Tips for Inorganic Qualitative Analysis

In a few weeks time, I will be doing an inorganic qualitative analysis test which is very pressed for time (a majority of people don't finish). The form of the test is most likely going to be that there will be around 9 unknown solutions (each contains 1 cation and 1 anion) which I have to identify. I will be given a list of all the possible cations and anions (there will be around 20 cations and 20 anions) and also we will have pH paper and a heat gun.

I wondering if anyone could suggest any tips on how to go about determining the composition of each unknown solution.

Currently, my method is to do it systematically by making a table and reacting all the solutions together. However, this can very time-consuming and probably not the most efficient way to identify the solutions.

Also, is there any distinctive reactions or properties that we will help me identify certain ions. For example, antimony and sulfide will give a unique peachy pink precipitate, or adding water to bismuth may cause a precipitate (as acid is usually required to make bismuth go into solution so by adding water, the acid is diluted and the bismuth precipitates out.)

• This is so open ended it is impossible to give any guidance. There are numerous books on inorganic qual. Generally the scheme is to identify multiple anions and cations in the same solution. // Can you prep any test solutions beforehand? Can you take in notes, or reference books? What sort of equipment will you have access to? (dozens of test tubes, centrifuge, ect..) How much solution will you get for each unknown? How long will the test last? – MaxW Apr 18 '16 at 5:13
• If you can prep test solutions and take in notes, I'd be develop a scheme using a couple of drops of test solution into a plate with multiple wells. So to well 1 to you add 1.0 M HCl, well 2 1.0 M NaOH, well 3 1.0 M Na2CO3, and so on. // The immediate problem that I see is that unless you start out with several gross of test tubes you could spend the majority of your time washing glassware. – MaxW Apr 18 '16 at 5:33
• @MaxW we get around 30 test tubes so spending time washing glassware isn't really such a big problem. Like I said above, the only other equipment we get is a heat gun and pH paper. There is enough solution and if it runs out it will be refilled. We aren't allowed to prepare any test solutions and bring any notes into the exam. – Nanoputian Apr 18 '16 at 8:20

This is a rough draft of the complete answer. A lot more will be added to the answer very soon. At present, it includes the common tests for sulfite and bisulfite ions. Please do not downvote just because it is not a complete answer yet.

## Identification Of Sulfite Anion: ($\ce{SO3^{2-}}$)

All sulfites are insoluble except "group IA sulfites " and$\ce{(NH4)2SO4}$. Sulfite ppt. is always white in color.

Test With Acid:

$\ce{SO3^{2-} + H^{+} -> SO2 (g)}$

Sulfur dioxide gas is evolved which gives lime water test as well as baryta water test:

$\ce{SO2 + Ca(OH)2 -> Ba(HSO3)2}[\text{soluble}]$

$\ce{SO2 + Ba(OH)2 -> Ca(HSO3)2}[\text{soluble}]$

Test with $\ce{BaCl2/CaCl2}$:

$\ce{SO3^{2-} + CaCl2 -> CaSO3 + 2Cl^{-}}$

$\ce{SO3^{2-} + BaCl2 -> BaSO3 + 2Cl^{-}}$

Note that $\ce{BaSO3}$ is slowly oxidized by atmospheric oxygen to form heavy white ppt of $\ce{BaSO4}$. Similarly:

$\ce{BaSO3 + Br2 -> BaSO4 + Br-}$

$\ce{BaSO3 + H2O2 -> BaSO4 + H2O}$

$\ce{BaSO3 + HNO3 -> BaSO4 + NO}$

Test With $\ce{Pb(NO3)2}$:

$\ce{Pb(NO3)2 + SO3^{2-} -> PbSO3 [\text{white ppt.}]}$

On boiling $\ce{PbSO3}$ in presence of $\ce{O2}$, $\ce{PbSO4}$ is produced.

(On boiling $\ce{PbS2O3}$ we get $\ce{PbS}[\text{black ppt}]$ and $\ce{H2SO4}$. Hence, this fact can be used to distinguish sulfites from thiosulfates).

Test With $\ce{AgNO3}$:

Schiff's Test/Fuschin's Test/Magenta Reagent Test:

Both $\ce{SO3^{2-}}$ and $\ce{SO2}$ are decolorized in this test.

Distinguishing b/w $\ce{SO3^{2-}}$ and $\ce{CO3^{2-}}$ ion:

## Some Important Redox Reactions:

$\ce{MnO4- + SO2/H2S -> Mn^{2+} + SO4^{2-}/S}$

$\ce{Cr2O7- + SO2/H2S -> Cr^{3+} + SO4^{2-}/S}$

$\ce{Br2 + H2O + SO2/H2S -> Br- + SO4^{2-}/S}$

$\ce{Fe^{3+} + SO2/H2S -> Fe^{2+} + SO4^{2-}/S}$

$\ce{IO3- + SO2/H2S -> I2 [\text{gives deep blue color in starch}] + SO4^{2-}/S}$

$\ce{SO2}$ can either be oxidized to $\ce{SO4^{2-}}$ or reduced to $\ce{S^{2-}}$. But, $\ce{H2S}$ can only be oxidized to $\ce{S}$.

Notable Reaction (Comproportionation) : $\ce{\ce{SO4^{2-}} + \ce{H2S} -> S + H2O}$

Points to remember:

• $\ce{CO2/CO/CO3^{2-}/HCO3^{-}}$ do not show any change on treatment with with oxidizers like $\ce{KMnO4/K2Cr2O7/Br2\text{ water}/Fe^{3+}/IO3^{-}}$

• $\ce{S^{2-},SO3^{2-}/HSO3^{-}}$ do show change on treatment with with oxidizers like $\ce{KMnO4/K2Cr2O7/Br2\text{ water}/Fe^{3+}/IO3^{-}}$.

• $\ce{SO3^{2-} + H2O2 -> SO4^{2-} + H2O}$ $[\text{no visible change is observed}]$

## Dvarda Alloy ($\ce{Zn}/\ce{Al}/\ce{Cu}$):

$\ce{Zn + NaOH -> Na2ZnO2 + 2H}$

In presence of dvarda alloy the following reactions can take place:

$\ce{SO3^{2-} -> S^{2-}}$

$\ce{SO2 -> H2S}$

$\ce{NO2^{-} -> NH3}$

$\ce{NO^{3-} -> NH3}$

$\ce{Zn}$ reacts with dil. $\ce{H2SO4}$ to produce $\ce{H2}$ gas which can also reduce $\ce{SO4^{2-}}$ to $\ce{S^{2-}}$.

## Test with Sodium Nitroprusside:

$\ce{Na2[Fe(CN)5NO] + ZnSO4 -> Zn[Fe(CN)5NO] [\text{salmon pink ppt }]}$

$\ce{Zn[Fe(CN)5NO] + \text{moist} SO2 -> \text{red color compound}}$

$\ce{S2O3^{2-}}$ and $\ce{S^{2-}}$ ions also give the test. So they can be removed on treating with $\ce{HgCl2}$ as follows:

$\ce{Hg^{2+} + S^{2-} -> HgS}$

$\ce{Hg^{2+} + S2O3^{2-} -> HgS2O3}$

$\ce{HgS2O3 [\text{on heating}]-> HgS [\text{black ppt}] + H2SO4}$

## Identification Of Bisulfite Anion ($\ce{HSO3^{-}}$):

Test With Acid:

$\ce{HSO3^{-} + H^{+} -> SO2 (g)}$

Test With $\ce{H2O2}$:

$\ce{SO3^{2-} + H2O2 -> SO4^{2-} + H2O}$

$\ce{HSO3^{-} + H2O2 -> H^{+} + SO4^{2-} + H2O}$ (acidic)

## Identification Of Carbonate Anion ($\ce{CO3^{2-}}$)

All carbonates are insoluble except group $\mathrm{IA/IIA}$ carbonates ($\ce{Li2CO3}$ is sparingly soluble) & $\ce{(NH4)2CO3}$.

Test With Acid:

$\ce{CO3^{2-} + H^{+} -> CO2}$ (Carbon dioxide can be identified using lime water or baryta water test)

$\ce{CO2 + Ca(OH)2 -> CaCO3}$ (white turbidity)

$\ce{CO2 + Ba(OH)2 -> BaCO3}$ (white turbidity)

When $\ce{CO2(g)}$ is in excess, $\ce{CaCO3/BaCO3 + CO2 + H2O -> Ca(HCO3)2/Ba(HCO3)2}$. This bicarbonate is water soluble.

Test Based On ppt:

$\ce{CO3^{2-} + CaCl2/BaCl2 -> BaCO3/CaCO3 + 2Cl^{-}}$

The formed carbonates are white precipitates which are soluble in $\text{(dil)}\ce{HCl}$, $\text{(dil)}\ce{HNO3}$, $\text{(dil)}\ce{CH3COOH}$ and soda water.

Remember that $\ce{HCl}$, $\ce{HNO3}$, $\ce{H2SO4}$ and $\ce{CH3COOH}$ is stronger than $\ce{H2CO3}$. So they can decompose the carbonate salt. For example $\ce{BaCO3 + HNO3 -> Ba(NO3)2 + CO2(g) + H2O}$.

Test With $\ce{Pb(NO3)2}$:

$\ce{CO3^{2-} + Pb(NO3)_2 -> PbCO3 + 2NO3^{-}}$

$\ce{PbCO3}$ is soluble in $\text{(dil)}\ce{HCl}$, $\text{(dil)}\ce{CH3COOH}$ and excess $\ce{NaOH}$.

$\ce{PbCO3 + 2NaOH -> Pb(OH)2 + Na2CO3}$

$\ce{Pb(OH)2 + 2NaOH -> Na2[Pb(OH)4]}$

Test With AgNO3:

$\ce{CO3^{2-} + AgNO3 -> Ag2CO3} \text{[white ppt.]}$

Soluble in $\text{(dil)}\ce{HCl}$ and $\text{(dil)}\ce{NH3}$.

$\ce{Ag2CO3 + NH3 <-> [Ag(NH3)2]^{+} + CO3^{2-}}$

$\ce{[Ag(NH3)2]^{+} -> Ag3N (furminating silver) -> Ag (black) + N2}$

The ppt. becomes yellow or brown upon addition of excess reagent owing to formation of $\ce{Ag2O}$.

Note: $\ce{Ag2CO3}$ on heating at above $1000^{0} C$ produces $\ce{Ag}$, $\ce{O2}$ and $\ce{CO2}$. Below that temperature if heated it produces $\ce{Ag2O}$, $\ce{CO2}$.

Test With $\ce{Hg(NO3)2}$:

$\ce{CO3^{2-} + Hg2(NO3)2 -> Hg2CO3 [yellow ppt.] -> Hg [black] + HgO [yellow] + CO2}$ (on heating)

$\ce{Hg^{2+}_{2} -> Hg + Hg^{2+}}$

Test With $\ce{HgCl2}$:

$\ce{CO3^{2-} + 4Hg^{2+} + 3H2O -> 3HgO.HgCO3 [reddish brown ppt] + 6H^{+}}$

Note: Phenopthalein turns pink in carbonate solutions and colorless in bicarbonate solutions .