+2 UNIT 3 PAGE- 2
COMMON ION EFFECT
Weak acids and bases are ionised only to small extent in their aqueous solutions. In their solutions, unionised molecules are in dynamic equilibrium with ions. The degree of ionisation of a weak electrolyte (weak acid or base) is further suppressed if some strong electrolyte which can furnish some common ions furnished by weak electrolyte, which is added to its solution. The effect is called Common ion effect. For example, degree of ionisation of NH4OH (a weak base) is suppressed by the addition of NH4Cl(a strong electrolyte). The ionisation of NH4OH and NH4Cl in solution is represented as follows:
Due to the addition of NH4Cl which is strongly ionised in solution the concentration of NH4+ ions increases in the solution, therefore, according to Le-Chatelier's principle, equilibrium equation (1) shifts in the backward direction in favour of unionised NH4OH. In this way, addition of NH4Cl suppresses the degree of ionisation of NH4OH. Thus, concentration OH ions in the solution is considerably reduced and the weak base NH4OH becomes still weaker base.
The suppression of the degree of ionisation of a weak electrolyte (weak acid or base) by addition of some strong electrolyte having a common ion is called Common ion effect.
The application of common ion effect in the qualitative analysis is illustrated below. The cations of Group II(Hg2+, Pb2+, Bi3+, Cu2+, Cd2+, As3+, Sb3+, Sn2+) are precipitated as their sulphides. Solubility product of sulphides of Group II radicals are very low. Therefore, even with low concentration of S2 ions , the ionic products exceed the value of their solubility products and the radicals of Group II get precipitated. The low concentration of S2 ion is obtained by passing H2S gas through the solution of the salts in the presence of dil. HCl which suppresses the degree of ionisation of H2S by common ion effect.
It is necessary to suppress the concentration of S2 ions, otherwise radicals of Group IV will also get precipitated along Group II radicals.
Radicals of Group IV (Ni2+, Co2+, Mn2+, Zn2+ ) are also precipitated as their sulphides. But the solubility products of their sulphides are quite high. In order that ionic products exceed solubility products, concentration of S2 ions should be high in this case. High concentration of S2 ions is achieved by passing H2S gas through the solution of the salts in the presence of NH4OH. Hydroxyl ions from NH4OH combine with H+ ions from H2S. Due to the removal of H+ ions the equilibrium of H2S shifts in favour of ionised form.
Hence concentration of S2 ions increases. With this increased concentration of S2 ions, ionic products exceed solubility products and the radicals of Group IV get precipitated as sulphides.
Radicals of Group III (Fe3+. Al3+, Cr3+) are precipitated as their hydroxides by NH4OH in the presence of NH4Cl. The purpose of NH4Cl is to suppress the degree of ionisation of NH4OH by common ion effect in order to decrease the concentration of OH ions.
The solubility product of hydroxides of Group III radicals are quite low. Therefore, even with this suppressed concentration of OH ions their ionic products exceed solubility products and hence they get precipitated. If the concentration of OH ions is not suppressed, the radicals of Group IV, V and Mg2+ will also be precipitated along with radicals of Group III.
Radicals of Group V (Ba2+, Sr2+, Ca2+) are precipitated as their carbonates by the addition of (NH4)2CO3 in the presence of NH4Cl and NH4OH. NH4Cl suppresses the degree of ionisation of (NH4)2CO3 by the common ion effect and hence decreases the concentration of CO32 ions.
But solubility products of carbonates of Group V are quite low and hence even with the suppressed concentration of CO32 ions their ionic products, exceed the solubility products and they get precipitated whereas Mg2+ and other radicals of Group VI having relatively high solubility products are not precipitated.
So in this way , we find the concept of solubility product and common ion effect enables us to selectively precipitate the cations of a particular group.
SALT HYDROLYSIS
It is commonly observed that different salts , on dissolution in water do not always form neutral solutions. For example aqueous solution of sodium acetate is basic, whereas the aqueous solution of copper sulphate is acidic. However , salts of strong acids and bases such as sodium chloride form neutral solution. This is due to the fact that a salt on dissolution in water undergoes dissociation to form ions. These ions can interact with water molecules and thereby produce either an acidic or alkaline solution. This process is called hydrolysis.
Salt + water Acid + Base
or BA + H2O HA + BOH
All salts are strong electrolytes and thus ionises completely in aqueous solution. If the acid (HA) produced is a strong acid and the base (BOH) produced is weak, we can write the above equation as :
B+ + A + H2O H+ + A + BOH
or B+ + H2O H+ + BOH
Thus , in this case the cation reacts with water to give an acidic solution. This is called cationic hydrolysis.
Again if acid produced is weak and base produced is strong, we can write,
B+ + A + H2O HA + B+ + OH
Or A + H2O HA + OH
Here the anion reacts with water to give the basic solution. This is called anionic hydrolysis.
Hence salt hydrolysis may be defined as the reaction of the cation or anion of the salt with water to produce acidic or basic solution.
Thus depending upon the relative strengths of the acid or base produced, the resulting solution is acidic , basic or neutral.
There are four distinct types of hydrolytic behaviour of various salts. These are :
(a) Salts of strong acids and strong bases.
(b) Salts of strong acid and weak bases.
(c) Salts of weak acids and strong bases.
(d) Salts of weak acids and weak bases.
1. SALTS OF STRONG ACID AND STRONG BASES
Examples are NaCl, NaNO3, Na2SO4, KCl, KNO3, K2SO4
As an illustration, let us discuss the hydrolysis of NaCl. We may write
NaCl + H2O NaOH + HCl
or Na+ + Cl + H2O Na+ + OH + H+ + Cl
or H2O H+ + OH
Thus it involves only ionisation and no hydrolysis. Further in the resulting solution [ H+ ] =[ OH ] So the solution is neutral. Hence it may be generalized that the salts of strong acids and strong bases do not undergo hydrolysis and the resulting solution is neutral.
2. SALTS OF WEAK ACIDS AND STRONG BASES
Examples are : CH3COONa, Na2CO3, K2CO3, Na3PO4 etc.
As an illustration , the hydrolysis of sodium acetate (CH3COONa) may be represented as follows:
CH3COONa + H2O CH3COOH + NaOH
or CH3COO + Na++ H2O CH3COOH + Na+ + OH
or CH3COO + H2O CH3COOH + OH
As it produces OH ions , the solution of such a salt is alkaline in nature.
3. SALTS OF STRONG ACIDS AND WEAK BASES
Examples are :
NH4Cl, CuSO4, NH4NO3, AlCl3, CaCl2 etc.
As an illustration , the hydrolysis of NH4Cl may be represented as follows:
NH4Cl + H2O ⇌ NH4OH + HCl
or NH4+ + Cl + H2O ⇌ NH4OH + H+ + Cl
or NH4+ + H2O ⇌ NH4OH + H+
As it produces H+ ions , the solution of such a salt is acidic in character.
4. SALTS OF WEAK ACIDS AND WEAK BASES
Examples are CH3COONH4, (NH4)2CO3, AlPO4 etc.
As an illustration, the hydrolysis of ammonium acetate may be represented as follows:
CH3COONH4 + H2O ⇌ CH3COOH + NH4OH
Or CH3COO + NH4+ + H2O ⇌ CH3COOH + NH4OH
Thus it involves both anionic and cationic hydrolysis.
The resulting solution may be neutral or slightly acidic or basic depending upon the relative degrees of ionisation of weak acid and weak base produced.
If ,
Ka = Kb ; the resulting solution is neutral
Ka > Kb ; the resulkting solution is acidic
Ka < Kb ; the resulting solution is basic. In the present example, the acid (CH3COOH) and base (NH4OH) formed are almost equally weak. Hence the resulting solution is almost neutral. HYDROLYTIC CONSTANT (Kh) The general equation for the hydrolysis of a salt ( BA ) may be written as : Applying the law of chemical equilibrium, we get K = the equilibrium constant. Since water is present in large excess in aqueous solution, its concentration [H2O] may be regarded as constant so that we have: where Kh is called hydrolysis constant. DEGREE OF HYDROLYSIS (h) The degree of hydrolysis of a salt is defined as the fraction (or percentage) of the total salt hydrolysed. i.e., CALCULATION OF HYDROLYSIS CONSTANT, DEGREE OF HYDROLYSIS AND pH OF SALT SOLUTIONS 1. SALTS OF WEAK ACID AND STRONG BASE (a) Hydrolysis Constant Representing salt by BA as usual, the hydrolysis may be represented as follows: i.e., it is a case of anion hydrolysis. The hydrolytic constant Kh of the above reaction will be given by : For weak acid HA , the dissociation equilibrium is The dissociation constant Ka of acid HA will be given by Further the ionic product of water Kw is given by Kw = [H+] [OH] ………(iii) Multiplying (i) with (ii) and dividing by (iii) , we get or, (b) DEGREE OF HYDROLYSIS Suppose the original concentration of the salt in solution is c moles per litre and h its degree of hydrolysis at this concentration. Then we have, The hydrolysis constant (Kh) will therefore be given by : If h is very small as compared to 1 , we can take ( 1 h ) = 1 so that the above expression becomes C h2 = Kh or Substituting the value of Ka from equation (iv) , we get (c) Calculation of pH In the present case we have i.e., [OH] = c h Substituting the value of h in equation (v) , we get Thus knowing the molar concentration C of the solution and dissociation constant Ka of the weak acid involved , the pH of the solution can be calculated. 2. SALTS OF STRONG ACID AND WEAK BASE (a) Hydolysis constant For a salt BA, the hydrolytic constant may be represented as : i.e., it is a case of cation hydrolysis. The hydrolytic constant Kh will be given by For the weak base BOH , the dissociation equilibrium is : The dissociation constant Kb of the weak base BOH will be given by : The ionic product of water , Kw is given by : Kw = [H+] [OH] ……… (ix) Multiplying equation (vii) with (viii ) and dividing by (ix) , we get, ………..(x) Thus , hydrolysis constant is inversely proportional to the dissociation constant of the base. Therefore the weaker the base, the greater is the hydrolysis constant of the salt. (b) DEGREE OF HYDROLYSIS Let the original concentration of the salt in the solution be c moles per litre and h is the degree of hydrolysis at that concentration. Hydrolytic constant Kh, If h is very small in comparison to 1, we may assume, 1 h 1, so that, Substituting the value of Kh from Eq. (x) (c) pH of the Hydrolysed salt solution Now, [H+] = c h Substituting the value of h from Eq (xi) ……….. (xii) 3. SALT OF WEAK ACID AND WEAK BASE In this case both the cation and anion undergo hydrolysis to the same or different extents. The resulting solution may be neutral, acidic or basic depending upon the relative strengths of acids and bases. Some common examples of such salts are CH3COONH4, (NH4)2CO3, AlPO4, etc. The hydrolysis may be written as: For a general reaction : i.e., it involves both anion hydrolysis as well as cation hydrolysis. (a) HYDROLYSIS CONSTANT The hydrolytic constant Kh may be written as : The following equilibria also exists in solution for weak acid: ……(xiii) For weak base, …….(xiv) Kw = [H+] [OH] …… (xv) Multiplying and dividing Eq (xii) by [H+] [OH] Calculation of degree of hydrolysis If h is very small in comparison to 1, we may assume, 1 h 1, so that, It may be noted that in this case the degree of hydrolysis is independent of the concentration of the solution. Further weaker the acid and base , the greater is the degree of hydrolysis of the salt. (b) pH of the hydrolysed salt solution According to Equ (xiii) , the dissociation constant of the weak acid , HA, Now , It is clear from the above equation that pH of the solution will depend upon the pK values of the acid and base. According to the above equation : • If pKa < PKb, the pH of the solution will be less than 7 and consequently solution will be acidic. • If pKa > pKb, then pH of the solution will be more than 7 and hence the solution will be alkaline.
• If pKa = pKb , the pH of the solution will equal to 7 and hence solution will be neutral.
4. SALT OF STRONG ACID AND STRONG BASE
Sodium chloride, sodium sulphate, potassium nitrate etc., are the salts of this type. Such salts do not undergo hydrolysis when dissolved in water and their solutions are neutral. This can be understood by considering the following example of sodium chloride.
Sodium chloride (NaCl) is the salt of a strong acid (HCl) and a strong base (NaOH) . It dissociates almost completely in solution to give Na+ and Cl ions.
NaCl Na+ + Cl
The Na+ and Cl ions practically have no tendency to interact with water because they have species (Cl ions act as a weak base, whereas Na+ ions as weak acid) because of being the conjugates of strong species. Thus Na+ ions practically have no tendency to take OH ions from water and Cl ions to take H+ ions furnished by water. Consequently, the equilibrium of H+ and OH ions remains almost undisturbed and the solution is neutral.
The important characteristics of the hydrolysis of different types of salts have been summarised in TABLE.
ACID BASE INDICATORS
In order determine the strength of an acid solution, a standard base solution is usually added drop by drop from a burette to a known volume of the given acid and the volume of the base required to exactly neutralise the given acid solution is measured. This process of determining the strength of an acid with the help of a standard base is called titration and the stage at which the added base exactly neutralises the acid is termed as the end point or equivalence point.
The end point of an acid- base titration is usually determined with the help of an organic compound which changes its colour at the end point. Such an organic compound is called an acid-base indicator and may be defined as follows.
The organic compound which indicates the end point of an acid base titration by visual change in its colour is called an acid-base indicator.
For example , phenolphthalein is commonly used in the titration of a strong acid against strong alkali. It changes its colour from colourless (in acid) to pink (in base) at the end point of the titration. Another important indicator is methyl orange which changes its colour from red (in acid) to yellow (in base).
Every indicator has a definite pH range in which it changes its colour. As soon as this range is attained during the course of a titration, the indicator changes its colour. The pH ranges of some important indicators are given in the TABLE.
Indicator
pH range Colour
Acid soln. Basic soln.
Methyl orange 3.1 – 4.5 Red Yellow
Bromocresol green 3.8 - 4.6 Yellow Blue
Methyl red 4.2 – 6.2 Pink Yellow
Bromocresol purple 5.0 – 6.8 Yellow Purple
Bromothymol blue 6.0 – 7.5 Orange Blue
Phenolphthalein 8.3 – 10.5 Colourless Pink
ACID BASE TITRATIONS USING INDICATORS
An acid – base titration always involves a change in the pH of the solution (acid or base placed in the receiver flask) with every addition of base or the acid from the burette. A plot of pH of the solution against the volume of the base or acid added is called pH curve or titration curve.
The pH curves or the titration curves are very helpful in choosing suitable indicators for different types of acid-base titrations. In most of the acid base titrations , the pH of the solution records a sharp change at the end point of the titration. A suitable indicator for a particular titration is one whose pH range lies within the range of pH in which the sharp change takes place at the end point of titration. For example, in the titration of a strong acid against a strong base , the pH suddenly changes from 3 to 10. Hence, for such a titration , the suitable indicators are phenolphthalein, methyl orange etc.
The pH curves for different types of acid-base titrations are discussed below.
1. Titration of a strong acid against a strong
base.
The pH curve for the titration of a strong acid (say HCl) against a strong base (say NaOH) is shown in Fig.
The pH curve for the titration of a strong acid against
a strong base.
From the curve, it is clear that the pH of the acid solution (kept in the receiver flask) increases slowly in the beginning on addition of the base from the burette. Near the end point of the titration, the pH of the solution increases sharply from 3 to 10 and the curve becomes almost vertical. Beyond this, the pH increases slowly and the curve is almost flat. Thus, any indicator having pH range between 3 and 10 will change its colour at the end point of the titration. Phenolphthalein, methyl orange, bromothymol blue, methyl red etc. are such indicators and can be used for this type of titrations. The commonly used indicators for this type of titrations are phenolphthalein and methyl orange.
2. Titation of a weak acid against a strong
base
The pH curve for for such titration is shown in Fig.
The pH curve for the titration of a weak acid against
a strong base.
Near the end point of the titration, pH of the solution changes sharply from 6 to 11 and the curve becomes almost vertical. Beyond this, the pH changes slowly and the curve is almost flat. Thus , any indicator having pH range in between 6 and 11 can be used for this type of titration. Phenolphthalein, thymolphthalein etc. are such indicators. The most commonly used indicator for this type of titration is phenolphthalein. It is to be noted that methyl orange, methyl red and bromophenol blue can not be used for such titrations, because they will change their colours much before the end point.
3. Titration of a strong acid against a weak
base
The pH curve for such titration is shown in Fig.
The pH curve for the titration of a strong acid against
a weak base.
Near the end point of the titration, the pH of the solution increases sharply roughly from 3 to 8 and the titration curve becomes almost vertical. Beyond this, the pH rises slowly. Thus, any indicator having pH range in between 3 and 8 will change its colour at the end point of the titration and will be suitable for it. Methyl orange and methyl red are this type of indicators and can be conveniently used for such titrations. It is to be noted that phenolphthalein having pH range 8.0 to 9.8 cannot be used for this type of titrations because it will continue to change colour even beyond the end point.
4. Titation of a weak acid against a weak base
The pH curve for the titration of a weak acid (CH3COOH) against a weak base (NH4OH) is shown in Fig.
The pH curve for the titration of a weak acid against
a weak base.
In this case , pH of the solution does not change sharply near the end point of the titration and there is no vertical portion in the curve. Thus in this case, the end point is not sharp. Such titrations cannot be carried out accurately by using indicators and should be avoided.
FORMULA FOR CALCULATION OF VOLUMETRIC ANALYSIS
The basic principle is that the reactants react with each other in equivalent amounts.
Suppose the two reactants involved are A and B . Further suppose V1 cm3 of the solution of reactant A having normality N1 react exactly with V2 cm3 of the solution of reactant B having normality N2.
In acid- base neutralisation , if na and nb are the basicity of acid and acidity of base respectively, then
na Ma Va = nB Mb Vb
where Ma the molality of acid and Mb the molaity of base .
BUFFER SOLUTIONS
Generally pH of the solution changes on addition of small amount of acid or base to it. But if the solution contains a weak acid and its conjugate base or a weak base and its conjugate acid, such a solution can resist change in pH and is called Buffer solution.
A buffer solution is the one which can resist the change in pH on addition of a small amount of acid or base. The ability of a buffer solution to resist the change in pH on addition of acid or base is called buffer action.
Depending upon pH values , buffer solutions are divided into two classes. If the pH of the buffer is less than 7, it is called acidic buffer and if it is more than 7, it is called basic buffer.
An acidic buffer solution contains equimolar quantities of a weak acid and its salt with a strong base, for example, a solution containing equimolar quantities of acetic acid and sodium acetate . Similarly, a basic buffer solution contains equimolar amount of weak base and its salt with a strong acid, as for instance, a solution having equimolar amount of ammonium hydroxide and ammonium chloride.
Some Common buffer solutions
Buffer pair pH
Acetic acid + sodium acetate 4.74
Formic acid + sodium formate 3.7
Ammonium hydroxide + ammonium chloride 9.25
It may be noted that an aqueous solution of a salt of weak acid and weak base can also act as buffer. Examples, CH3COONH4, (NH4)2CO3, (NH4)3PO4 etc.
Blood and sea water are well known examples of buffers. Normal blood has a pH of 7.4. The pH is maintained by the buffer action of carbonic acid (H2CO3), bicarbonate ion (HCO3) and carbon dioxide.
Sea water has a pH around 8.2. This pH is maintained by complex buffer action of various salts present in sea water.
pH OF A BUFFER SOLUTION HENDERSON HASSELBALCH EQUATION
The pH of a buffer solution can be calculated with the help of Henderson’s equation. The equation can be derived as follows.
For acidic Buffer
Suppose we have an acidic buffer containing CH3COOH and CH3COONa.
According to the law of equilibrium,
Due to common ion effect , the dissociation of the acid is negligible. Therefore [CH3COOH] can be regarded as equal to the concentration of the salt initially taken. Hence, the above equation can be written in a general form as follows.
where pKa = log Ka
The equation is called Henderson’s equation.
For Basic solutions
Similarly , it can be shown for a basic buffer,
where pKb = log Kb
APPLICATIONS OF BUFFER SOLUTIONS
1. In biological processes : The pH of our blood is maintained constant (at about 7.4) inspite of various acid and base -producing reactions going on in our body. In the absence of its buffer nature, we could not eat a vaiety of foods and spices.
2. In Industrial processes : The use of buffers is an important part of many industrial processes. E.g.,
(i) In electroplating.
(ii) In the manufacture of leather, photographic materials and dyes.
(iii) In analytical chemistry.
(iv) To calibrate pH meters.
(v) In bacteriological research , culture media are generally buffered to maintain the pH required for the growth of bacteria being studied.
Problems
28. Calculate the concentrations of H3O+ions and OH ions in :
(a) 0.01 M solution of HCl.
(b) 0.01 M solution of NaOH at 298 K, assuming that HCl and NaOH are completely ionised in the given conditions.
29. Calculate the concentration of H3O+ ions in 0.005 M solution of Ba(OH)2 at 298 K assuming that Ba(OH)2 is completely ionised under the given conditions.
30. Calculate the pH 0.01 N nitric acid, assuming complete ionisation.
31. Calculate the pH a solution of H3O+ concentration of 0.001 mol/L.
32. Calculate pH of a solution with OH ion concentration of 0.01 mol/L.
33. The pH of a solution 6.4. What is the hydrogen ion concentration ?
34. Calculate the pH of a 108 M HCl solution.
35. What is the pH of an aqueous solution with hydrogen ion concentration equal to 3 x 105 mol/L ? Is the solution acidic or basic ?
36. Find the pH of the following solutions :
(i) 3.2 g of hydrogen chloride dissolved in 1.00 L of water.
(ii) 0.28 g of potassium hydroxide dissolved in 1.00 L of water.
37. A base dissolved in water yields a solution with a hydroxide ion concentration of 0.05 mol/L. What is the concentration of hydrogen ions in the solution ?
What is the pH of the solution ?
Is the solution acidic, basic or neutral ?
38. At 298 K, calculate the pH of :
(a) 0.2 M solution of methyl amine ; Kb = 4.4 x 105 .
(b) 0.23 M weak acid HX ; Ka = 7.3 x 106.
39. At 298 K pH of lemon juice is 2.32. What are the [H3O+]and [OH] in the solution ?
40. Calculate the pH of 0.05 M sulphuric acid solution.
41. Calculate the pH of 0.5 M HCl solution.
42. Calculate the pH of 0.1 M NaOH solution.
43. Equal volumes of 0.2 M HCl and 0.1 M NaOH solutions are mixed together. Calculate the pH of the resulting solution.
44. Calculate the hydrogen ion concentration of a solution of pH = 5 .
45. The ionisation constant of propionic acid is 1.32 x 105. Calculate the degree of ionisation in its 0.05 M solution and also its pH.
46. The pH of 0.1 M solution of cyanic acid (HCNO) is 2. 34 . Calculate the [H3O+] of the acid and its degree of ionisation in the solution.
47. The ionisation constant of nitrous acid is 4.5 x 104 . Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
48. A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionisation constant of pyridine.
49. The degree of ionisation a 0.1 M Bromoacetic acid solution is 0.132. Calculate the pH of the solution and pKa of bromoacetic acid.
50. Ionic product of water at 310 K is 2.7 x 1014. What is the neutal pH of water at this temperature ?
51. Calculate the pH of a solution obtained by mixing 10 ml of 0.1 M HCl and 40 ml 0.2 M H2SO4.
52. What will be the resultant pH when 200 ml of an aqueous solution of HCl (pH = 2 ) is mixed with 300 ml of an aqueous solution of NaOH( pH= 12) ?
53. The pH of a soft drink is 4.4. Calculate [H3O+] and [OH].
54. Calculate the pH of a buffer solution containing 0.2 M acetic acid and 0.02 M sodium acetate. Ka of acetic acid is 1.85 x 105.
55. Calculate the hydrolytic constant, degree of hydrolysis and pH of a 0.01 M aqueous NH4Cl. Kb = 1.8 x 105.
56. Calculate the hydrolytic constant, degree of hydrolysis and pH of an aqueous solution of 0.01 M sodium acetate solution ( Ka =1.85 x 105 )
57. Calculate the pH of a 0.1 M ammonium acetate solution ( Ka =1.85 x 105 and Kb = 1.85 x 105)
58. What would be the pH of a 0.1 M aqueous solution of ammonium cyanide solution ? pKa = 9.04 and pKb = 4.73,
59. How much of 0.3 M ammonium hydroxide should be mixed with 30 ml of 0.2 M solution of ammonium chloride to give buffer solutions of pH 8.65 and 10.
60. How many grams of KBr can be added to 1 litre of 0.05 M solution of silver nitrate just to start the precipitation of silver bromide ? (Ksp = 5 x 1013)
QUESTIONS
1. Explain what is meant by strong and weak electrolytes ?
2. Deduce an expression for the ionisation constant for ionisation of a weak electrolyte.
3. What ia Arrhenius concept of acids and bases ?
4. What is Arrhenius concept of strength of acids and bases.
5. What are the drawbacks of Arrhenius concept of acids and bases ?
6. What is Lowry Bronsted concept of acids and bases ?
7. Discuss the Lewis definition of acid and bases. How is it more useful than Bronsted definition ?
8. What is meant by ionic product of water ?
9. Explain the term pH value.
10. Explain the pOH of a solution. How is it related to pH value ?
11. Write a note on pH scale.
12. Define the term solubility product.
13. How is solubility of a salt related to solubility product ?
14. When does a salt get precipitated in solution ?
15. Lewis acids are called electrophiles. Why ?
16. The Ksp of a salt is high. What does it indicate ?
17. Out of water and 0.1 M KCl in which silver chloride will dissolve more ?
18. Give two examples of cations which can act as Lewis acids.
19. What is the difference between a conjugate acid and a conjugate base ?
20. Whether the pH value of an aqueous solution of sodium acetate will be 7 or greater than 7 ?
21. What happens when HCl gas is passed through saturated solution of NaCl solution ?
22. What happens to the pH if a few drops of acid are added to CH3COONH4 solution ?
23. NaCl is added to a saturated solution of PbCl2 . What will happen to to the concentration of Pb2+ ions in solution ?
24. pH scale has a range from 0 to 14. Can the pH of a solution be greater than 14 or have negative value ?
25. What is the concentration of H3O+ and OH ions in water at 298 K ?
26. Will ionic product of water increase or decrease if temperature is increased ?
27. Define common ion effect.
28. On the basis of pH values, classify the following as strong acids , strong bases, weak acids and weak bases.
Solution A B C D E
PH 8.8 0.1 14.0 6.0 8.4
29. Give an example of acidic buffer.
30. Give an example of basic buffer.
31. What is meant by conjugate acid and base pair ? Find the conjugate acid / base for the following species :
HNO2, CN , HClO4, F , OH , Cr3+ and S2
32. Justify the statement that water behaves like an acid and also like a base on the basis of protonic concept.
33. Fear or excitement leads to fast breathing and thus, in the decrease of CO2 concentration in blood. In what way will it change the pH of the blood ?
QUESTIONS
1. Explain the term : (i) irreversible reaction (ii) reversible reaction. Give an example each.
2. What is meant by chemical equilibrium ?
3. What are the characteristics of a chemical equilibrium ?
4. What is physical equilibrium ? Give an example to illustrate it.
5. Write a note on ‘’dynamic nature ‘’ of chemical equilibrium
6. Give an example of :
(i) Solid-liquid physical equilibria.
(ii) Liquid gas physical equilibria
(iii) Solid in physical equilibrium.
(iv) Gas in liquid physical equilibrium.
7. Explain the term :
(i) Homogeneous equilibrium.
(ii) Heterogeneous equilibrium.
8. Can equilibrium be achieved between water and its vapour in an open vessel ? Explain your answer and say what happens eventually.
9. A liquid in equilibrium with vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(i) What is the initial effect of change on the vapour pressure ?
(ii) How do the rates of evaporation and condensation change initially ?
(iii) What happens when equilibrium is restored finally and what will be the final vapour pressure ?
10. A vessel has two compartments A and B . In compartment B is placed radioactive methyliodide(CH3I*) and in A the normal methyl iodide (CH3I)
(i) Will the vapour over A and B become radioactive ?
(ii) Will the radioactivity spread to the liquid in compartment A ? Discuss in terms of the dynamic nature of the equilibrium between the vapour and its liquid.
11. (i) Which of the following liquids will have the lowest and highest boiling points ?
Liquid Water Acetone ethanol
Equilibrium vapour pressure at 293 K (kPa) 2.34 12.36 5.15
(iii) Which of these will evaporate least at 293 K in a sealed container, before equilibrium is attained ?
12. A vessel contains a saturated solution of normal sugar. If we add radioactive sugar to it, what will you observe ? Explain your observation.
13. What are general characteristics of equilibria involving physical processes ?
14. What are common characteristics of equilibria involving physical changes ?
15. Write a note on Henry’s law. What are its limitations ?
16. A lump of common salt of a given mass is kept in an aqueous solution of sodium chloride. A fter 24 hours , its mass was found to remain the same. Is the crystal in equilibrium with the solution ?
17. When 1 kg of sugar is added into 1 L of water at 298 K , it was found that sugar is only partly dissolved. What inference can be derived about the state of equilibrium ?
18. What is law of equilibrium ?
19. State and explain the law of mass action.
20. What are the characteristics of equilibrium constant ?
21. How can you express the equilibrium constant for gaseous reversible reaction ?
22. How is equilibrium constant KP related to equilibrium constant KC ?
23. Derive the expression for KC, the equilibrium constant , for reaction :
A + B ⇌C + D
24. For which of the following cases does the reaction go to maximum completion ?
(i) K = 1010 (ii) K = 1010 (iii) K = 1
25. What does the magnitude of equilibrium constant indicate ?
26. The value of equilibrium constant depends on what ?
27. Explain the effect of following factors on equilibrium concentration of a reaction:
(i) Change of concentration of reactants or products.
(ii) Change of pressure, if gas (or gases) is present
(iii) Change of temperature.
28. State Le-Chatleir Braun principle.
29. In the reaction : A + B ⇌ C + D , what will happen to the concentration of A , B and D , if the concentration of C is increased ?
30. Discuss Le-Chatelier’s principle in the manufacture of ammonia by Haber process.
31. Discuss the application of Le-Chatelier Principle in the :
(i) oxidation of SO2 to SO3.
(ii) Oxidation of N2 to NO
32. In the light of Le Chatelier’s Principle, discuss the influence of : (i) Temperature (ii) Concentration change (iii) Pressure change of reversible reaction.
33. Using Le Chatelier’s principle, predict the influence of :
(i) Temperature and pressure on melting.
(ii) Temperature and pressure on vaporisation of water.
(iii) Temperature on solubility of salts
34. What is the effect of catalyst on equilibrium of a reversible reaction ?