UNIT 7 ( PAGE 2)
CONCENTRATIONS OF [H3O+] AND [OH-] IN AQUEOUS SOLUTIONS OF ACIDS AND BASES
In pure water, the [H3O+]equal to [OH-]. But on addition of some acid or base to water, these concentrations no longer remain equal. However, the value of ionic product of water Kw at a particular temperature always remain constant irrespective of the fact that whether water is pure or some acid or base has been added to it. For example, if an acid (say HCl) is added to water, the concentration of hydronium ion becomes quite high. Consequently, dissociation equilibrium of water shifts in the reverse direction, i.e., H3O+ ions would combine with OH- ions to form undissociated water molecules, so that the value of Kw in the solution may remain the same as that in pure water. Thus, addition of an acid in water decreases the [OH-] according to the relation,
Similarly, the addition of a base such as NaOH increases the [OH-] and decreases the [H3O+] according to the relation :
From the above discussion, it is clear that H3O+ and OH- ions are always present in aqueous solution whether it is acidic or basic. However, the relative concentrations of these ions vary in different solutions.
In general,
In neutral solution [H3O+] = [OH-]
In acidic solutions [H3O+] > [OH-]
In basic solutions [H3O+] < [OH-]
EXPRESSING HYDRONIUM ION CONCENTRATION - pH SCALE
Sorensen (1909) suggested a new term for expressing the concentration of hydrogen ion known as pH scale. The symbol pH is derived from the Danish word Potenz meaning power. Thus pH means power of hydrogen ion.
The pH is defined as the negative logarithm of H3O+ ion concentration in moles per litre. Mathematically, it may be expressed as :
For acidic solutions, [H3O+] concentration is more than 1 x 10-7 mol L-1. Therefore, pH of acidic solution is less than 7. For basic solutions, the pH value is greater than 7. Corresponding range of pH is from 0 to 14. The solution having pH between 0 and 2 are strongly acidic, those with pH between 2 to 4 are moderately acidic, while others having pH between 4 to 7 are weakly acidic. Similarly, the solutions having pH value between 7 to 10 are weakly basic, those having pH 10 to 12 are moderately basic whereas others which have pH range between 12 to 14 are strongly basic.
pOH OF A SOLUTION
pOH of a solution is the negative logarithm to the base 10 of OH- ion concentration expressed in moles per litre.
Thus :
pOH = - log[OH-]
Relation between pH and pOH
In any solution :
[H3O+] [OH-] = Kw = 10-14
Taking logarithms on both sides:
log[H3O+] + log [OH-] = log Kw = log [10-14]
Reversing the signs throughout we get :
-log[H3O+] - log [OH-] = -log Kw = -log [10-14]
pH + pOH = pKw = 14
pH of Some common substances
The name of fluid | pH | Name of fluid | pH |
Saturated soln of NaOH | » 15 | Blank coffee | 5 |
0.1 M NaOH solution | 13 | Tomato juice | » 4.2 |
Lime water | 10.5 | Soft drinks & vinegar | » 3.0 |
Milk of magnesia | 10 | Lemon juice | » 2.2 |
Egg white, sea water | 7.8 | Gastric juice | » 2.2 |
Human blood | 7.4 | 1 M HCl | » 0 |
Milk | 6.8 | Concentrated HCl | » -1.0 |
Human saliva | 6.4 |
Ionisation of Weak Electrolytes –
Degree of ionisation
Weak electrolytes are ionised partially. The ions produced as a result of dissociation of weak electrolytes are present in dynamic equilibrium with the undissociated molecules. The fraction of total number of molecules of the electrolyte dissolved, that ionises at equilibrium is called degree of dissociation. Let us consider the ionisation of some weak electrolyte AB in water. Let C be the concentration of the electrolyte in the solution and a the degree of ionisation. The concentration of various species at equilibrium would be as given under:
According to Law of chemical equilibrium:
For weak electrolytes under normal concentrations, a is very small as compared with unity and hence ( 1 - a ) can be taken as 1 in the denominator.
So the degree of ionisation is inversely proportional to the square root of concentration. Thus, as the concentration decreases, the degree of ionisation increases. It can also be followed from the above relationship that, as C approaches zero or dilution approaches infinity, the degree of dissociation approaches unity, i.e., its maximum value. In other words, in this dilution , the whole of electrolyte is present in the dissociated form as ions. This important generalisation is known as Ostwald dilution Law.
SOLUBILITY EQUILIBRIA OF SPARINGLY SOLUBLE SALTS
The solubility of ionic solids in water varies a great deal. Some are so soluble that they are hygroscopic in nature and even absorb water vapours from atmosphere. Others have so little solubility and are called insoluble. The solubility depends on a number of factors, important among which are the lattice energy of the salt and how well the ions are hydrated in aqueous solution. Each salt has its characteristic solubility and its dependence on temperature. We classify salts on the basis of their solubility in the following three categories.
· Category I Soluble Solubility > 0.1 M
· Category II Slightly soluble Solubility < 0.1 M
· Category III Sparingly soluble Soubility < 0.01 M
SOLUBILITY PRODUCT CONSTANT
Certain electrolytes such as BaSO4 and AgCl are sparingly soluble in water. Even in their saturated solutions, the concentration of the electrolytes is very low. So , whatever little of electrolyte goes into solution, undergoes complete dissociation (due to low concentration). Therefore , in saturated solutions of such electrolytes solid electrolyte is in equilibrium with the ions as represented below :
Consider a saturated solution of a salt containing the solid salt. There are two equilibria, one between solid salt and dissolved salt and second between the dissolved salt and its ions.
Applying the Law of mass action to the second equilibrium,
where K is the equilibrium constant and [AB] is the concentration of the dissolved salt. Cross multiplying we get
Since the solution is saturated , the concentration of the dissolved salt remains constant at a fixed temperature.
Hence . [A+][B-] = K x Constant = Ksp where Ksp is another constant. This constant Ksp is known as the solubility product of the electrolyte. It is the maximum value of product of concentrations of the ions of the electrolyte.
In the case of silver chloride, we have :
Ksp = [Ag+] [Cl-]
For lead chloride :
Ksp = [Pb2+] [Cl-]2
In general , for any sparingly soluble salt AxBy
which dissociates to set up the equilibrium :
where AY+ and BX- denote the positive and negative ions , x and y represent the number of these ions in the formula of the electrolyte. The solubility product constant may be expressed as :
Ksp = [AY+]x [BX -]y
Thus solubility product of a sparingly soluble salt at a given temperature may be defined as the product of the concentrations of its ions in the saturated solution, with each concentration term raised to the power equal to the number of times the ion occurs in the equation representing the dissociation of the electrolyte.
Problem
40. Express the solubility products of :
i) PbCl2 ii) Mg(OH) 2 iii) Ca3(PO4) 2 iv) ZnS
Calculation of Solubility product
Knowing the solubility of the salt, its solubility product can be calculated. Consider the salt AB. Suppose at a particular temperature its solubility is S mol L-1. S moles of salt on ionisation give S moles of A+ and S moles of B- ions.
Knowing the solubility S, Ksp can be calculated.
In general , for any sparingly soluble salt AxBy which dissociates to set up the equilibrium :
Problems
41. The solubility of AgCl is 1.06 x 10-5 mol L-1at 298 K. Find
out the solubility product of AgCl at this temperature.
42. The solubility of PbCl2 at 298 K is 2 x 10-2 mol L-1. Find out
the solubility product at this temperature.
43. Which of the following is more soluble ?
AgCl (Ksp = 1.7 x 10-10)
Ag2CrO4 (Ksp = 4 x 10-12 )
44. The ionisation constant of phenol is 1.0 x 10-10 . What is the concentration of phenate ion in 0.05 M solution of phenol ? What will be its degree of ionisation if the solution is also 0.01 M in sodium phenate.
Applications of Solubility Products
1. Calculation of Solubility
If solubility product of a sparingly soluble salt at a particular temperature is known, its solubility at that temperature can be calculated.
2. Predicting the precipitation of a salt
The concept of solubility product principle helps us to predict whether a salt will precipitate or not, on mixing the solutions containing its ions under particular conditions. In the solution, the ionic product, i.e., product of concentration of the ions of the salt(raised to appropriate power) cannot exceed the value of its solubility product because solubility product is the highest limit of ionic product at a particular temperature. Thus, if ionic product exceeds solubility product, excess ions combine each other to form precipitate of the salt. So in order to predict whether a salt will precipitate or not, ionic product of the salt is calculated. If it exceeds the value of solubility product, the salt will be precipitated , otherwise not. Thus, it can be concluded that :
Precipitation occurs :
if calculated ionic product > Ksp
No precipitation :
if calculated ionic product < Ksp.
Problems
45. The solubility product of Ag2CrO4 at 298 K is 4 x 10-12. Find out the solubility at this temperature.
46. The solubility product of AgBr at a certain temperature is 2.5 x 10-13. Find out the solubility of AgBr in gram per litre at this temperature.
47. How many moles of AgBr (Ksp = 5 x 10-13 mol2L-2) will dissolve in 0.01M NaBr solution.
48. What [H3O+] must be maintained in a saturated solution to precipitate Pb2+ , but not Zn2+ from a solution in which each ion is present at a concentration of 0.01 M ? [Ksp (H2S) = 1.1 x 10-22, Ksp (ZnS) = 1.0 x 10-21 )
49. Equal volumes of 0.02 M sodium sulphate and 0.02 M Barium chloride solutions are mixed together. Predict the precipitation will occur or not (Ksp of BaSO4 is 1.5 x 10-9)
50. 50 ml of 0.01 M solution of calcium nitrate is added to 150 ml of 0.08 M solution of ammonium sulphate. Predict whether CaSO4 will be precipitated or not. (Ksp of CasO4 = 4 x 10-5.
3. Precipitation of Soluble Salts
The principle of solubility product is also applied in the precipitation of soluble salts from their saturated solution, in pure state. The phenomenon known as salting out, is used in the purification of sodium chloride. This is done by preparing a saturated solution of impure sodium chloride in water when the following equilibrium exists :
HCl gas is passed through this solution. The concentration of chloride ions increases considerably. The ionic product exceeds the solubility product of NaCl and therefore , it precipitates out from the solution in pure state. The impurities remain in solution.
4. Inorganic Qualitative Analysis
The classification of basic radicals into different groups in the inorganic qualitative analysis is based upon the knowledge of solubility products of salts of these basic radicals. For example , chlorides of Hg22+, Pb2+and Ag+ have very low solubility products. On the basis of this knowledge these radicals are grouped together in Group-I and are precipitated as their chlorides by adding dil. HCl to their solutions. For adjusting the condition for precipitation, another concept called Common Ion Effect plays a very important role.
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)
1. 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.
2. 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.
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
The pH of a buffer solution can be calculated with the help of Henderson
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
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
51. 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.
52. 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.
53. Calculate the pH 0.01 N nitric acid, assuming complete ionisation.
54. Calculate the pH a solution of H3O+ concentration of 0.001 mol/L.
55. Calculate pH of a solution with OH- ion concentration of 0.01 mol/L.
56. The pH of a solution 6.4. What is the hydrogen ion concentration ?
57. Calculate the pH of a 10-8 M HCl solution.
58. What is the pH of an aqueous solution with hydrogen ion concentration equal to 3 x 10-5 mol/L ? Is the solution acidic or basic ?
60. 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.
61. 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 ?
62. At 298 K, calculate the pH of :
(a) 0.2 M solution of methyl amine ; Kb = 4.4 x 10-5 .
(b) 0.23 M weak acid HX ; Ka = 7.3 x 10-6.
63. At 298 K pH of lemon juice is 2.32. What are the [H3O+]and
[OH-] in the solution ?
64. Calculate the pH of 0.05 M sulphuric acid solution.
65. Calculate the pH of 0.5 M HCl solution.
66. Calculate the pH of 0.1 M NaOH solution.
67. Equal volumes of 0.2 M HCl and 0.1 M NaOH solutions are mixed together. Calculate the pH of the resulting solution.
68. Calculate the hydrogen ion concentration of a solution of pH = 5 .
69. The ionisation constant of propionic acid is 1.32 x 10-5. Calculate the degree of ionisation in its 0.05 M solution and also its pH.
70. 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.
71. The ionisation constant of nitrous acid is 4.5 x 10-4 . Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
72. A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionisation constant of pyridine.
73. 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.
74. Ionic product of water at 310 K is 2.7 x 10-14. What is the neutal pH of water at this temperature ?
75. Calculate the pH of a solution obtained by mixing 10 ml of 0.1 M HCl and 40 ml 0.2 M H2SO4.
76. 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) ?
77. The pH of a soft drink is 4.4. Calculate [H3O+] and [OH-].
78. 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 10-5.
79. Calculate the hydrolytic constant, degree of hydrolysis and pH of a 0.01 M aqueous NH4Cl. Kb = 1.8 x 10-5.
80. Calculate the hydrolytic constant, degree of hydrolysis and pH of an aqueous solution of 0.01 M sodium acetate solution ( Ka =1.85 x 10-5 )
81. Calculate the pH of a 0.1 M ammonium acetate solution ( Ka =1.85 x 10-5 and Kb = 1.85 x 10-5)
82. What would be the pH of a 0.1 M aqueous solution of ammonium cyanide solution ? pKa = 9.04 and pKb = 4.73.
83. 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.
84. ow 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 10-13)
85. Calculate the pH of solution formed on mixing 0.2 M NH4Cl and 0.1 M NH3, The pOH of ammonia solution is 4.75. (N 215 P 22)
86. The pKa of acetic acid and pKb of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution. (N 219 P 7.25)
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 ?
34. Explain the term : (i) irreversible reaction (ii) reversible reaction. Give an example each.
35. What is meant by chemical equilibrium ?
36. What are the characteristics of a chemical equilibrium ?
37. What is physical equilibrium ? Give an example to illustrate it.
38. Write a note on ‘’dynamic nature ‘’ of chemical equilibrium
39. Give an example of :
40. Explain the term :
(i) Homogeneous equilibrium.
(ii) Heterogeneous equilibrium.
41. Can equilibrium be achieved between water and its vapour in an open vessel ? Explain your answer and say what happens eventually.
42. 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 ?
43. 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 ?
(iii) Discuss in terms of the dynamic nature of the
equilibrium between the vapour and its liquid.
44. A vessel contains a saturated solution of normal sugar. If we add adioactive sugar to it, what will you observe ? Explain your observation.
45. What are general characteristics of equilibria involving physical processes ?
46. What are common characteristics of equilibria involving physical changes ?
47. Write a note on Henry’s law. What are its limitations ?
48. A lump of common salt of a given mass is kept in an aqueous solution of sodium chloride. After 24 hours , its mass was found to remain the same. Is the crystal in equilibrium with the solution ?
49. 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 ?
50. What is law of equilibrium ?
51. State and explain the law of mass action.
52. What are the characteristics of equilibrium constant ?
53. How can you express the equilibrium constant for gaseous
reversible reaction ?
54. How is equilibrium constant KP related to equilibrium
constant KC ?
55. Derive the expression for KC, the equilibrium constant , for
reaction:
A + B ⇋C + D
56. For which of the following cases does the reaction go to maximum completion ?
i) K = 10-10 (ii) K = 1010 (iii) K = 1
57. What does the magnitude of equilibrium constant indicate ?
58. The value of equilibrium constant depends on what ?
59. 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.
60. State Le-Chatleir Braun principle.
61. 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 ?
62. Discuss Le-Chatelier’s principle in the manufacture of ammonia by Haber process.
63. Discuss the application of Le-Chatelier Principle in the :
(i) oxidation of SO2 to SO3.
(ii) Oxidation of N2 to NO
64. In the light of Le Chatelier’s Principle, discuss the influence of : (i) Temperature (ii) Concentration change (iii) Pressure change of reversible reaction.
65. 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
66. What is the effect of catalyst on equilibrium of a reversible reaction ?