Advance Stoichiometry Based on Redox Reactions

Stoichiometry is related to the number of atoms or molecules of reacting species and product formed during the course of reaction which is governed by the law of chemical combination (discussed previously in Some Basic Concepts).

Here, we focus our attention about the stoichiometry related to redox reaction and acid and base neutralization reaction. As stoichiometric calculations are necessary for both analytical purpose (to find out chemical composition of given sample) and for assesing required amount of raw material to obtain desired mass of products.

For analytical purpose, most reactions held in liquid state, (i.e. aqueous solution) as solution provides better homogenity which helps in accurate measurement with minimum error. This type of analysis in aqueous solution is known as volumetric anlalysis.

VOLUMETRIC ANALYSIS

Important definitions:

(I) Molarity = $\frac{No.ofmolesofsolute}{Vol.ofsolutioninL}$

(II) Normality = $\frac{No.ofgramequivalentsofsolute}{Vol.ofsolutioninL}$

No. of gram equivalents of solute = $\frac{Weightofsolute}{Equivalentweight}$

Equivalent weight = $\frac{Molecularweight}{{}^{\prime }{n}^{\prime }}$

It can be inferred that number of gram equivalents of a substance = n x number of moles and Normality= 'n' x Molarity

It is extremely convenient to use the law of gram equivalents to solve problems based on chemical reactions. According to this law the 'number of gram equivalents of all reactants are equal to each other in a reaction assuming none of them are in excess and is also equal to number of gram equivalents of all products' assuming that all the reactants are undergoing reaction in the reaction.

We can use this law conveniently to solve problems without requiring to know much about the reactions. For this we need to have good understanding of the 'n' factor of a substance. 'n' factor is the valency factor or conversion factor.

$\bullet$ n – factor of substance in redox reaction is equal to number of moles of lost or gain electron per molecule.

$\bullet$ n – factor of substance in non – redox reaction is equal to the product of displaced mole and its charge.

(1) Acids: The number of moles of replaceable H+ ions per mole of the acid e.g. for

HCl n = 1; H2SO4 n = 2; H3PO4 n = 3

(2) Bases: The number of moles of replaceable OH- ions per mole of the base e.g.

NaOH n = 1; Ba(OH)2 n = 2; Al(OH)3 n = 3

Illustration 1. H3PO4 is a tribasic acid and one of its salts is NaH2PO4. What volume of 1 M NaOH solution should be added to 12 g of NaH2PO4 to convert it into Na3PO4?

(A) 100 ml (B) 200 ml

(C) 80 ml (D) 300 ml

Solution. (B). Eq. of NaH2PO4 = Eq. of NaOH

$\frac{12}{M}\times 2=1\times V\times {10}^{-3}$

$\therefore V=200ml$

Illustration 2. A sample of H2SO4 (density 1.787 g ml–1) is labelled as 86% by weight. What is molarity of acid? What volume of acid has to be used to make 1 litre of 0.2 M H2SO4?

Solution: Let us take 1 lit of the solution

$\therefore$ Weight of the solution = 1000 x 1.787 g = 1787 g

$\therefore$ Weight of H2SO4 $=\left(1787\times \frac{86}{100}\right)$gm

Number of moles of H2SO4 = $\frac{1787\times 86}{100\times 98}$ = 15.68

Let V ml of this H2SO4 are used to prepare 1 litre of 9.2 M H2SO4

$\therefore$ milli mole of conc. H2SO4 = milli mole of dil. H2SO4

V × 15.68 = 1000 x 0.2 $\Rightarrow$ V = 12.75 ml

(3) Salts

Salts which reacts in such way that no atom in the salt undergoes change in oxidation state (oxidation state of an element in a molecule is the charge the element would have if all the bonds associated with the elements are assumed to be completely ionic.)

'n' factor = number of moles of cation in one mole of the salt × oxidation state of the cation.

$\therefore$ 'n' factor of MgCl2 = 1 x 2 = 2

(Note: whenever two substances react such that their 'n' factors are in the ratio of x : y, the molar ratio of these substances in the balanced chemical reaction would be in the ratio of y : x)

Illustration 3. Calculate the mass of anhydrous Ca3(PO4)2 present in 250 ml of 0.25 M solution.

Solution: Meq. of Ca3(PO4)2 = 250 x 0.25 x 6 = 375

$\therefore$ $\frac{W}{51.66}$x 1000 = 375

$\therefore$ W = 19.3725 g

'n' factor for a redox reaction is the moles of electrons released or acquired by 1 mole of the reactant in the reaction.

Illustration 4. If the 1.58g of $KMn{O}_4$ in acidic medium completely reacts with ferrous oxalate $\text{(Fe}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{),}$ what weight (in g) of ferrous oxalate is required?

(A) 2.73 (B) 4.73

(C) 11.19 (D) 8.5

Solution: (A). n-factor of ferrous oxalate = 3

n-factor of $KMn{O}_4$ in acidic medium = 5

$\therefore$No. of moles of KmnO4= $\frac{1.58}{158}={10}^{-2}$$\therefore$No. of eq. of $KMn{O}_4=5\times {10}^{-2}$;;;;;;;;;;;;;;;;;;;;;;$\therefore$No. of eq. of ferrous oxalate = $5\times {10}^{-2}$

$\therefore$No. of moles of ferrous oxalate = $\frac{5\times {10}^{-2}}{3}\times 164g$ = 2.73 g

Illustration 5. In the reaction, $VO+F{e}_2{O}_3\to FeO+V_2{O}_5,$the equivalent weight of V2O5 is equal to its

(A) Mol. weight (B) $\frac{Mol.weight}{8}$

(C) $\frac{Mol.weight}{6}$ (D) None of these

Solution: (C). n–factor for V2O5 = 2 (5 – 2) = 6 $E=\frac{M}{6}$

Titration: The procedure for determining the concentration of a solution by adding its known volume to react completely with other solution of known strength whose volume justified by experiment.

Sodium of known strength is called standard solution are classified as

(i) Primary standards

(ii) Secondary standards

Substances preferred primary standards must have following characteristics:

(i) Easily available and easy to preserve (made stable) {resistant to moisture and air)

(ii) Readily soluble in given solvent.

(iii) The reaction with a standard solution should be of constant stoichiometry.

(iv) Titration error should be negligible even at moderate concentration.

Some common examples with then use

Potassium hydrogen phthalate$C_8{H}_5{O}_{4}K\left[204.23\right]$ - Acid base

Anhydrous sodium carbonate$N{a}_{2}C{O}_3\left[106\right]$ - Acid base

Potassium dichromate [294.19] - Redox

Arsenic oxide As2O­3 [197.85] - Redox

Potassium iodate KIO3 - Redox

Sodium oxalate Na2C2O4 - Redox

EDTA [Na] [372.3] - Complexo method

The aim of titration is the addition of a quantity of standard and solution chemically equivalent to the quantity of unknown.

The completion of reaction is often indicated by a change in colour of reaction mixture which may be either due to change in colour of reactant or a substance mixed externally known as indicator.

TYPE OF TITRATIONS

(i) Acid – base or neutralization titration: The reaction in which an acid reacts with a base to give salt and water known as neutralization reaction and titration involving such a reaction known as neutralization titration.

The type of indicator used to indicate end point depend on the pH of end – point.

(ii) Redox titration: Titration involving oxidation-reduction known as redox – titration.

(iii) Precipitation titration: When two such solution are mixed during the course of titration a precipitate is formed and the end point is indicated by completion of precipitation.

Titration involving such type is known as precipitation titration. Some common example

$KCl+AgN{O}_3\to \underset{Whiteppt.}{AgCl\downarrow }+KN{O}_3$

$KI+AgN{O}_3\to \underset{ppt.}{AgI\downarrow }+KN{O}_3$

Some time titration involving silver nitrate also known as argentometric titration as all other salt of silver are insoluble in water thus form ppt.

Complexo metric titration: It involves the replacement of one or more of the co-ordinate covalent molecules which are co-ordinated to a central metal ion, M by some other groups. The groups getting attached to the central metal ion are known as ligand [L].

$M{\left(H_{2}O\right)}_n+nL\to M{L}_n+n{H}_{2}O$

For example ethylene diamine tetra acetic acid [EDTA] is used along with eriochromic indicator.

To find out the concentration or compositon of given samples, titration is used in different mode named as simple titration, double titration, back titration as discussed below.

SIMPLE TITRATION

In this, we can find the concentration of a substance with the help of the conc. of another substance which can react with it.

For example: Let there be a solution of a substance A of unknown concentration. We are given another substance B whose concentration is known$\text{(}{\text{N}}_{\text{1}}\text{)}$. We take a certain known volume of A in a flask $\text{(}{\text{V}}_{\text{2}}\text{)}$and then we add B to A slowly till all the A is consumed by B. (This can be known with the help of indicators). Let us assume that the volume of B consumed is${\text{V}}_{\text{1}}$. According to the Law of equivalents, the number of gm equivalents of A is equal to the number of gm equivalents of B.

$\therefore N_1{V}_1=N_2{V}_2,where{N}_2$is the conc. of A.

From this we can calculate the value of N2.

BACK TITRATION

Back titration is used to calculate % purity of a sample. Let us assume that we are given an impure solid substance C weighing w gms and we are asked to calculate the percentage of pure C in the sample. We will assume that the impurities are inert. We are provided with two solutions A and B, where the concentration of B is known (N1) and that of A is not known. This type of titration will work only if the following conditions are satisfied (a) A, B and C should be such compounds that A and B can react with each other, A and C can react with each other but product of A and C should not react with B.

Now we take a certain volume of A in a flask (A taken should be such that gm equivalents of A $⩾$ gm equivalents of C in the sample. This can be done by taking A in excess). Now we perform a simple titration using B. Let us assume that the volume of B used is${\text{V}}_{\text{1}}$. In another beaker, we again take the solution of A in the same volume as taken earlier. Now, C is added to this and after the reaction is complete, the solution is being titrated with B. Let us assume that volume of B used up is${\text{V}}_{\text{2}}$.

Gram equivalents of B used in the first titration = ${\text{N}}_{\text{1}}{\text{V}}_{\text{1}}$

$\therefore$gm. equivalents of A initially = ${\text{N}}_{\text{1}}{\text{V}}_{\text{1}}$

gm. equivalents of B used in the second titration is N1V2

$\therefore$gm. equivalents of A left in excess after reacting with C = N1V2

gm. equivalents of A that reacted with C = ${\text{N}}_{\text{1}}{\text{V}}_{\text{1}}\text{ - }{\text{N}}_{\text{1}}{\text{V}}_{\text{2}}$

gm. equivalents of pure C =${\text{N}}_{\text{1}}{\text{V}}_{\text{1}}\text{ - }{\text{N}}_{\text{1}}{\text{V}}_{\text{2}}$.

If the 'n' factor of C is x, then the moles of pure C = $\frac{N_1{V}_1-N_1{V}_2}{x}$

$\therefore$ the weight of C = $\frac{N_1{V}_1-N_1{V}_2}{x}$´ Molecular weight of C.

$\therefore$ percentage of C = $\frac{N_1{V}_1-N_1{V}_2}{x}\times \frac{Molecularwt.ofC}{w}\times 100$

DOUBLE TITRATION

The method involves two indicator (Indicators are substances that change their colour when a reaction is complete) phenolphthalein and methyl orange. This is a titration of specific compounds. Let us consider a solid mixture of NaOH, Na2CO3 and inert impurities weighing w g. You are asked to find out the % composition of mixture. You are also given a reagent that can react with the sample, say, HCl along with its concentration $\text{(}{\text{M}}_{\text{1}}\text{)}\text{.}$

We first dissolve the mixture in water to make a solution and then we add two indicators in it, namely phenolphthalein and methyl orange. Now, we titrate this solution with HCl.

NaOH is a strong base while $\text{N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}$is a weak base. So it is safe to assume that NaOH reacts with HCl first, completely and only then does $\text{N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}$ react.

$\text{NaOH + HCl }\to \text{ NaCl + }{\text{H}}_{\text{2}}\text{O}$

Once NaOH has reacted, it is the turn of Na2CO3 to react. It reacts with HCl in two steps:

$\text{N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}\text{ + HCl }\to \text{ NaHC}{\text{O}}_{\text{3}}\text{ + NaCl and then,}$

$\text{NaHC}{\text{O}}_{\text{3}}\text{ + HCl }\to \text{ NaCl + C}{\text{O}}_{\text{2}}\text{ + }{\text{H}}_{\text{2}}\text{O}$

As can be seen, when we go on adding more and more of HCl, the pH of the solution keeps on falling. When $\text{N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}$ is converted to $\text{NaHC}{\text{O}}_{\text{3}}$ completely, the solution is weakly basic due to the presence of $\text{N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}$(which is a weaker base as compared to$\text{NaHC}{\text{O}}_{\text{3}}$). At this instant phenolphthalein changes colour since it requires this weakly basic solution to change its colour. Therefore, remember that phenolphthalein changes colour only when the weakly basic $\text{N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}$ is present. As we keep adding HCl, the pH again falls and when all the $\text{N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}$ reacts to form $\text{NaCl, C}{\text{O}}_{\text{2}}\text{ and }{\text{H}}_{\text{2}}\text{O}$the solution becomes weakly acidic due to the presence of the weak acid${\text{H}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}\text{(C}{\text{O}}_{\text{2}}\text{ + }{\text{H}}_{\text{2}}\text{O)}$. At this instance methyl orange changes colour since it requires this weakly acidic solution to do so. Therefore, remember methyl orange changes colour only when H2CO3 is present.

Now, let us assume that the volume of HCl used up for the first and the second reaction,

i.e. $\text{NaOH+HCl }\to \text{NaCl+}{\text{H}}_{\text{2}}\text{O and N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}\text{+HCl }\to \text{NaHC}{\text{O}}_{\text{3}}\text{+NaCl be }{\text{V}}_{\text{1}}$ (this is the volume of HCl from the beginning of titration up to the point when phenolphthalein changes colour). Let the volume of HCl required for the last reaction, i.e., $\text{NaHC}{\text{O}}_{\text{3}}\text{+ HCl }\to \text{NaCl+C}{\text{O}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O be }{\text{V}}_{\text{2}}$ (this is the volume of HCl from the point where phenolphthalein had changed colour upto the point when methyl orange changes colour). Then, moles of HCl used for reacting with $\text{NaHC}{\text{O}}_{\text{3}}\text{ = moles of NaHC}{\text{O}}_{\text{3}}\text{ reacted = }{\text{M}}_{\text{1}}{\text{V}}_{\text{2}}$

$\text{moles of NaHC}{\text{O}}_{\text{3}}\text{ produced by the N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}\text{ = }{\text{M}}_{\text{1}}{\text{V}}_{\text{2}}$

moles of $\text{N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}\text{ that gave }{\text{M}}_{\text{1}}{\text{V}}_{\text{2}}\text{ moles of NaHC}{\text{O}}_{\text{3}}\text{ = }{\text{M}}_{\text{1}}{\text{V}}_{\text{2}}$

$\text{Mass of N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}\text{ = }{\text{M}}_{\text{1}}{\text{V}}_{\text{2}}\times \text{106}$

$\%N{a}_{2}C{O}_3=\frac{M_1{V}_2\times 106}{w}\times 100$

moles of HCl used for the first two reactions = ${\text{M}}_{\text{1}}{\text{V}}_{\text{1}}$

moles of $\text{N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}$= ${\text{M}}_{\text{1}}{\text{V}}_{\text{2}}$

moles of HCl used for reacting with $\text{N}{\text{a}}_{\text{2}}\text{C}{\text{O}}_{\text{3}}$= ${\text{M}}_{\text{1}}{\text{V}}_{\text{2}}$

moles of HCl used for reacting with only $\text{NaOH = }{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}\text{ = }{\text{M}}_{\text{1}}{\text{V}}_{\text{2}}$

$\therefore$ moles of $\text{NaOH = }{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}\text{ = }{\text{M}}_{\text{1}}{\text{V}}_{\text{2}}$

Mass of NaOH = (M1V1 – M1V2) × 40

$\%NaOH=\frac{\left(M_1{V}_1-M_1{V}_2\right)\times 40}{w}\times 100$

IODOMETRIC AND IODIMETRIC TITRATION

The reduction of free iodine to iodide ions and oxidation of iodide ions to free iodine occurs in these titration

${\text{I}}_{\text{2}}\text{ + 2}{\text{e}}^{\text{-}}\to \text{2}{\text{I}}^{\text{-}}\text{ (reduction)}$

$\text{2}{\text{I}}^{\text{-}}\to {\text{I}}_{\text{2}}\text{ + 2}{\text{e}}^{\text{-}}\text{ (oxidation)}$

These are divided into two types

$\bullet$ Iodometric titration: In iodometric titrations, an oxidising agent is allowed to react in neutral medium or in acidic medium with excess of potassium iodide to librate free iodine.

$\text{KI + oxidising agent }\to {\text{I}}_{\text{2}}$

Free iodine is titrated against a standard reducing agent usually with sodium thiosulphate, Halogen, dichromates, cupric ion, peroxides, etc can be estimated by this method.

${\text{I}}_{\text{2}}\text{ + 2N}{\text{a}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3 }}\to \text{2NaI + N}{\text{a}}_{\text{2}}{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}$

$\text{2CuS}{\text{O}}_{\text{4}}\text{ + 4KI }\to \text{C}{\text{u}}_{\text{2}}{\text{I}}_{\text{2}}\text{ + 2}{\text{K}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}\text{ + }{\text{I}}_{\text{2}}$

${\text{K}}_{\text{2}}\text{C}{\text{r}}_{\text{2}}{\text{O}}_{\text{7}}\text{ + 6KI + 7}{\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}\to \text{C}{\text{r}}_{\text{2}}{\text{(S}{\text{O}}_{\text{4}}\text{)}}_{\text{3}}\text{ + 4}{\text{K}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}\text{ + 7}{\text{H}}_{\text{2}}\text{O + 3}{\text{I}}_{\text{2}}$$\bullet$ Iodimetric titration: These are the titrations in which free iodine is used as it is difficult to prepare the solution of iodine (volatile and less soluble in water) it is dissolved in KI solution.$\text{KI + }{\text{I}}_{\text{2}}\to \text{K}{\text{I}}_{\text{3}}\text{ (Potassium triiodide)}$This solution is first standardised before use. With the standard solution of I2, substance such as sulphite, thiosulphate, arsenite etc. are estimated.In the iodimetric and iodometric titrations, starch solution is used as indicator. Starch solution gives blue or violet colour with free iodine. At the end point the blue or violet colour disappears when iodine is completely changed to iodide.Potassium permanganate titration: In these titrations, reducing agents like $FeS{O}_4,Moh{r}^{\prime }ssalt,{(N{H}_4)}_{2}S{O}_4.FeS{O}_4.6H_{2}O,H_2{O}_2,A{s}_2{O}_3,oxalic$acid${(COOH)}_2$and oxalates ${(COONa)}_{2}etc.$ are directly titrated against $KMn{O}_4$ as oxidising agent in acidic medium. For Example $5F{e}^{2+}+Mn{O}_4^{-}+8H^{+}\to$ $5F{e}^{3+}+M{n}^{2+}+4H_{2}O$

Illustration 12. 280 ml of $H_2{O}_2$ reacts completely with 15.8 g of $KMn{O}_4$ in acidic medium, then the volume strength of $H_2{O}_2$ is

(A) 10

(B) 100

(C) 5

(D)20

Solution: (A). Let the volume strength of $H_2{O}_2$ be x

According to the question

$\Rightarrow \text{ meq}\text{. of }{\text{H}}_{\text{2}}{O}_2=meq.\text{ of KMn}{\text{O}}_{\text{4}}$

$\Rightarrow 280\times \frac{x}{5.6}=\frac{15.8}{158/5}\times 1000$

$\Rightarrow 50x=\frac{15.8\times 5}{158}\times 1000$

50 x = 500 $\Rightarrow x=10$

Illustraton 6. 0.2 g of a sample of H2O2 required 10 ml of 1 N KMnO4 in a titration in the presence of H2SO4. Purity of H2O2 is

(A) 25%

(B) 85%

(C) 65%

(D) 95%

Solution: (B). meq. of H2O2 = 10

Weight of H2O2 $=10\times {10}^{-3}\times 17$

% purity of H2O2 $=\frac{0.17}{0.20}\times 100$ =85%;;

Illustration 7. A polyvalent metal weighing 0.1 gm and having atomic weight of 51 reacted with dilute H2SO4 to give 43.90 ml of hydrogen at N.T.P. This solution containing the metal in the lower oxidation state was found to require 58.8 ml of 0.02 M KMnO4 for complete oxidation. What are the oxidation states of the metal in the two reactions?

Solution: Let lower oxidation state of the metal be n

Equivalents of metal = $\frac{\text{0}\text{.1}\times \text{n}}{\text{51}}$ =;equivalents of H2 evolved;=$\frac{\text{43}\text{.90}}{\text{22400}}\times 2$

$\therefore$n = 2Let final oxidation state = n'Then equivalents of metal = equivalents of oxidant;;;;;;;;;$\frac{0.1\left(n^{\prime }-2\right)}{51}=\frac{58.8\times 0.02\times 5}{1000}$;;$\therefore$ n' = 5;

HARDNESS OF WATER

(iv) Ion exchange resins process:

Acidic resin \,\left[ \begin{align}  2{{R}_{n}}H\,\,\,\,\,+\,\,\,\,\,\,C{{a}^{2+}}\xrightarrow[{}]{}\,\,\,\,\,{{({{R}_{n}})}_{2}}Ca+2{{H}^{+}} \\  cation\,\,exchanger \\ \end{align} \right]

Degree of hardness: Degree of hardness is defined as number of parts by weight of CaCO3 (or its equivalent quantities of other substance) present in million parts by weight of water.

Hardness of water $=\left(\frac{WeightofCaC{O}_3}{Weightofwater}\times {10}^6\right)ppm$

Illustration 8. 200 ml of hard water was boiled with 100 ml $\frac{N}{50}N{a}_{2}C{O}_3$ reagent. After boiling the volume was again made to 200 ml and the solution filtered. 25 ml of the filtrate required 8.2 ml of $\frac{N}{50}$ HCl for neutralization. Calculate the permanent hardness in ppm.

Solution: Total Na2CO3 added in sample $=\frac{100\times 1}{50}=2Meq.$

Number of Meq of HCl reacted with 25 ml filtrate $=\frac{8.2\times 1}{50}Meq$

Total Meq. of HCl required for 200 ml filtrate = Meq. of Na2CO3 in sample $=\frac{8\times 8.2\times 1}{50}=1.312$

Used Na2CO3 = 2 – 1.312 = 0.688 Meq. = Meq. Of CaCO3

$\therefore WeightofCaC{O}_3=0.688\times 50\times {10}^{-3}=3.4\times {10}^{-2}gm$

Hardness of water $=\frac{WeightofCaC{O}_3}{Weightofwater}\times {10}^6=\frac{3.4\times {10}^{-2}}{200}\times {10}^6=172ppm$

SOME IMPORTANT REACTIONS REGARDING STOICHIOMETRY

(i) Effect of Heat on Carbonate and Bicarbonate

(a) (i) $N{a}_{2}C{O}_3\stackrel{\Delta }{\to }noeffect$

(ii) $K_{2}C{O}_3\stackrel{\Delta }{\to }noeffect$(as these are thermally stable)

(b) ${\left(N{H}_4\right)}_{2}C{O}_3\stackrel{\Delta }{\to }2N{H}_3\uparrow +C{O}_2\uparrow +H_{2}O\uparrow$

(c) Other carbonate decompose to gives oxides and $C{O}_2$

$MC{O}_3\underset{\Delta }{\to }\mathrm{MO}+C{O}_2$

(d) All bicarbonates decomposes to carbonates, CO2 and H2O

(i) $\underset{Solid}{2NaHC{O}_3}\stackrel{\Delta }{\to }N{a}_{2}C{O}_3+C{O}_2\uparrow +H_{2}O\uparrow$

(ii) $\underset{Solution}{Mg{\left(HC{O}_3\right)}_2}\stackrel{\Delta }{\to }MgC{O}_3+C{O}_2\uparrow +H_{2}O\uparrow$

(iii) $A{g}_{2}C{O}_3\underset{Strong}{\overset{\Delta }{\to }}2Ag+C{O}_2+\frac{1}{2}{O}_2$

(ii) Effect of Heat on Chlorides, Bromides and Iodides

(a) Generally these are not effected by heat.

(b) Some halides of higher oxidation states changes into halides of lower oxidation state on heating.

$2\stackrel{+3}{\mathrm{Fe}}C{l}_3\stackrel{\Delta }{\to }2\stackrel{+2}{\mathrm{Fe}}C{l}_2+C{l}_2\uparrow$(ii) $2\stackrel{+2}{\mathrm{Cu}}C{l}_3\stackrel{\Delta }{\to }2\stackrel{+1}{\mathrm{Cu}}C{l}_2+C{l}_2\uparrow$

(c) Hydrated halides on heating convert to oxides, $H_{2}O$and haloacids (HF, HCl, HBr, HI) such as

(i) $MgC{l}_2.6H_{2}O\stackrel{\Delta }{\to }MgO\downarrow +2HCl+5H_{2}O\uparrow$

(ii) $2AlC{l}_3.6H_{2}O\stackrel{\Delta }{\to }\underset{White}{A{l}_2{O}_3\downarrow }+6HCl+5H_{2}O\uparrow$

(iii) $2CrC{l}_3.6H_{2}O\stackrel{\Delta }{\to }C{r}_2{O}_3\downarrow +6HCl\uparrow +9H_{2}O\uparrow$

(d) $N{H}_{4}Cl\left(s\right)\underset{sublination}{\overset{\Delta }{\to }}N{H}_3\uparrow +HCl\uparrow$

(iii) Effect of heat on Sulphates and Bisulphates

(a) $4N{a}_2.S{O}_3\stackrel{\Delta }{\to }3N{a}_{2}S{O}_4+N{a}_{2}S$

(b)$2NaHS{O}_3\stackrel{\Delta }{\to }N{a}_{2}S{O}_3\uparrow +S{O}_2\uparrow +H_{2}O\uparrow$

(iv) Effect of Heat on Sulphates, Pyrosulphates and Bisulphates

Sodium and potassium sulphates are thermally stable. However other sulphates, anhydrous or hydrous, decompose to give different products. Such as

(a) $2{\left(N{H}_4\right)}_2.S{O}_4\stackrel{\Delta }{\to }{N}_2\uparrow +6H_{2}O\uparrow +3S{O}_2\uparrow +2N{H}_3\uparrow$

(b) $CaS{O}_4\stackrel{\Delta }{\to }CaO+S{O}_2\uparrow +\frac{1}{2}{O}_2\uparrow$

(c) $\underset{Yellow}{F{e}_2{\left(S{O}_4\right)}_3}\stackrel{\Delta }{\to }\underset{Blackishbrown}{F{e}_2{O}_3}\downarrow +3S{O}_3\uparrow$

(d) $\underset{lightgreen}{2FeS{O}_4}\stackrel{\Delta }{\to }\underset{\mathrm{Re}d}{F{e}_2{O}_3}\downarrow +S{O}_3\uparrow +S{O}_2\uparrow$

(e) $\underset{\left(Bluevirtriol\right)}{CuS{O}_4\cdot H_{2}O}\stackrel{\Delta }{\to }CuO\downarrow +S{O}_3\uparrow +H_{2}O\uparrow$

(f) $\underset{\begin{array}{c}Sodium\\ pyrosulphate\end{array}}{N{a}_2{S}_2{O}_7}\stackrel{\Delta }{\to }N{a}_{2}S{O}_4+S{O}_3\uparrow$

(g) $2NaHS{O}_4\stackrel{\Delta }{\to }N{a}_{2}S{O}_4+H_{2}O+S{O}_3$

(v) Effect of heat on Nitrates

They undergo thermal decomposition in following ways

(a) $N{H}_{4}N{O}_3\stackrel{\Delta }{\to }{N}_{2}O\uparrow +2H_{2}O$

(b) (i) $NaN{O}_3\stackrel{\Delta }{\to }NaN{O}_2+\frac{1}{2}{O}_2\uparrow$ (ii) $KN{O}_3\stackrel{\Delta }{\to }KN{O}_2+\frac{1}{2}{O}_2\uparrow$

(c) (i) $AgN{O}_3\stackrel{{450}^{\circ }C}{\to }AgN{O}_2\uparrow +\frac{1}{2}{O}_2\uparrow$ (ii) $AgN{O}_3\stackrel{\Delta }{\to }Ag\downarrow +N{O}_2\uparrow +\frac{1}{2}{O}_2\uparrow$

(iii)$HgN{O}_3\stackrel{\Delta }{\to }Hg\downarrow +N{O}_2\uparrow +\frac{1}{2}{O}_2\uparrow$

(d) (i)$Mg{\left(N{O}_3\right)}_2\stackrel{\Delta }{\to }MgO\downarrow +2N{O}_2\uparrow +\frac{1}{2}{O}_2\uparrow$,

(ii) $\underset{white}{Pb{\left(N{O}_3\right)}_2}\stackrel{\Delta }{\to }\underset{yellow}{PbO\downarrow }+2N{O}_2\uparrow +\frac{1}{2}{O}_2\uparrow$

(iii)$Cu{\left(N{O}_3\right)}_2\stackrel{\Delta }{\to }CuO\downarrow +2N{O}_2\uparrow +\frac{1}{2}{O}_2\uparrow$

(vi) Effect of heat on Oxide and Hydroxide

Hydroxide, and higher oxides and some metals undergo following change on heating strongly.

(i)$2Al{\left(OH\right)}_3\stackrel{\Delta }{\to }\underset{White}{A{l}_2{O}_3\downarrow }+3H_{2}O$ (ii) $\underset{Brown}{\stackrel{+4}{Pb{O}_2}}\stackrel{\Delta }{\to }\underset{yellow}{\stackrel{+2}{\mathrm{Pb}}O\downarrow }+\frac{1}{2}{O}_2\uparrow$

SOME OTHER IMPORTANT REACTIONS

(i) $2CuS{O}_4+4KI\to C{u}_2{I}_2+I_2+2K_{2}S{O}_4$

(ii) $O_3+2KI+H_{2}O\to 2KOH+I_2+O_2$

(iii) $2Mn{O}_4^{-}+5C_2{O}_4^{2-}+16H_{}^{+}\to 2M{n}^{2-}+10C{O}_2+8H_{2}O$

(iv) $2KMn{O}_4^{}+10FeS{O}_4^{}+8H_{2}S{O}_4\to 2MnS{O}_4+5F{e}_2{\left(S{O}_4\right)}_3+8H_{2}O+K_{2}S{O}_4$

(v) $BaC{O}_3+2HCl\to BaC{l}_2+C{O}_2+H_{2}O$

(vi) $BaC{l}_2+H_{2}Cr{O}_4\to BaCr{O}_4+2HCl$

(vii) $2BaCr{O}_4+6KI+8H_{2}S{O}_4\to 3I_2+2BaS{O}_4+3K_{2}S{O}_4+C{r}_2{\left(S{O}_4\right)}_3+8H_{2}O$

(viii) $M{n}_3{O}_4+2FeS{O}_4+4H_{2}S{O}_4\to 3MnS{O}_4+F{e}_2{\left(S{O}_4\right)}_3+4H_{2}O$

(ix) $KCl{O}_3\stackrel{\Delta }{\to }KCl+\frac{3}{2}{O}_2$

SOME IMPORTANT POINTS TO REMEMBER

(i) If a salt of strong acid and strong base (e.g. NaCl) is present in the solution it will not react with any acid or base added to the solution.

(ii) If a salt of strong acid and weak base (e.g.$N{H}_{4}Cl$) is present in the solution it will only react with strong base but it will not react with strong acid.

$N{H}_{4}Cl+NaOH\to NaCl+N{H}_{4}OH$

(iii) If a salt of strong base and weak acid $\left(e.g.C{H}_{3}COONa\right)$ is present in the solution it will react only with strong acid but it will not react with strong base.

$C{H}_{3}COONa+HCl\to NaCl+C{H}_{3}COOH$