Ionic Equilibrium Notes

Page # 1 IONIC EQUILIBRIUM IONIC EQUILIBRIUM 1. ELECTRIC CONDUCTIVITY Those substance which allow the electric current to pass through them are called electric conductors and property is called electric conductivity. On the basis of Electric conductivity, substances are of two types 1.1 Nonconductors : Those substance which do not allow the electric current to pass through them are called nonconductors. eg. All covalent compounds & nonmetals. 1.2 Conductors : Those substance which allow the electric current to pass through them are called conductors. eg. all metals, alloys, all acid and bases, salt and graphite etc. On the basis of conducting units conductors are of two types 1.2.1 Metallic or Electric Conductors : Electricity conduct them due to the presence of free and mobile electron which act as electricity conducting unit called metallic or electric conductors. eg. Metals, Alloys, Graphite, Gas, Carbon etc. 1.2.2 Ionic Conductors or Electrolytes : Conductors in which the current is passes through them due to the presence of free ions are called Ionic Conductors or Electrolyte or Electrolytic conductors. Ionic conductors are further divided into two types on the basis of their strengths (a) Strong electrolytes : (i) Th ose substance whi ch are almost completely ionize into ions in their aqueous solution are called strong electrolytes. (ii) Degree of ionization for thi s type of electrolyte is one i.e.   1. eg. HCl, H2SO4, NaCl.HNO3, KOH, NaOH, HNO3, AgNO3, CuSO4, etc. Means all strong acids and bases and all types of salts. (b) Weak electrolytes : (i) Those substance which are ionize to a small extent in their aqueous solution are known weak electrolytes.eg. H2O, CH3COOH, NH4OH, HCN, HCOOH, Liq. SO2 etc. Means all weak acids and bases. (ii) Degree of ionization for this types of electrolytes in  <<< 1. 2. IONIZATION OF WEAK ELECTROLYTES 2.1 Ionisation : The process in which molecules of acids bases and salts when melted or dissolve in water dissociate into ions is called ionization. 2.1.1 Ionic Equilibrium : Weak electrolytes are partially ionised in aqueous solution and an equilibrium is situated between the ionized and unionised electrolyte. This type of equilibrium is known as ionic equilibrium. Let us consider an acid HA which when dissolved in water, an equilibrium will setup between ionized and unionized acid molecule as below HA + H2O H+ + A – On applying the law of mass action, the equilibrium constant K is written as K = [ H  ] [A  ] [ HA ] [ H 2 O ] Since in an aqueous solution, the molar concentration of water is constant i.e. [H2O] = constant. Therefore K [H2O] replaced by a new constant Ka which is known as ionization acid constant or dissociation constant of acid. Thus, [ H  ] [A  ] Ka = [ HA ] ‘Ka’ is the characteristic of the individual acid. From the above equation K a i s di rectl y proportional to [H + ] concentration hence, greater the Ka of an acid, more will be acidity. Assume that the initial concentration of C is moles/lit. and  is the degree of ionization then HA + H2O Initial concentration C Equilibrium (1 – )C C H+ + A – 0 0 C C  C  2C concentration Ka = = C(1    (1    The degree of ionisation  is very small as compared to 1 means  <<< 1. So, 1 –  = 1 = 2c ; Thus, Ka or  = Ka C So,  = Ka  V 2  = Ka C 1 V C Page # 2 IONIC EQUILIBRIUM This equation is known as ostwald dilution law equation. Similarly, for a weak base we have  = So,  = Kb C  1 V C (g) Similarly basic strength of bases is directly proportional to Kb. kb = – log kb = pkb Basic strength of base  kb   pka Kb  V Where, Kb is the dissociation constant of weak base. C is the initial concentration  is the degree of dissociation of weak base. Hence, according to Ostwald dilution law, the degree of ionization of weak electrolyte is directly proportional to the square root of volume containing one mole of electrolyte or Inversely proportional to the square root of its molar concentration. 3. ARRHENIUS THEORY OF ELECTROLYTIC DISSOCIATION OR IONIZATION (a) When an electrolyte dissociates into water, it gives two types of charged particles called ions. (b) Ions which carry (+) ve charge and move towards cathode are called as ‘Cations’ while ions carrying (–) ve charge and moving towards anode called as ‘anion’. (c) Every electrolytic solution is always neutral in nature. (d) Quantity or part of electrolyte which is ionized or decomposed or dissociate called as “Degree of Ionisation”. H+ (e) Electrolyte which gives dissociation in the aqueous solution aci d wh ile that which gi ves dissociation in the aqueous solution base. ions after is called as OH – after is called as (f) Aci dic strength of aci ds i s di rectly proporti on al to the di ssoci ati on constant ka. ka = – log ka = pka Thus, Acidic strength  ka 1  pk a  pkb  1 kb 1 pk b  1 ka (h) Conductivity of solution depends upon the number of ions produced by the electrolyte, such as Conduction of solution  number of ions produced by the electrolyte (i) Solution of strong electrolyte has more electric conductivity property as compared to weak electrolyte. (j) Only weak electrolyte followed the law of mass action and Ostwalds dilution law. (k) When electricity passed through in the electrolytic solution, it gives only direction to movement of ions towards the electrodes. (l) Movement of ions is inversely proportional to the molecular mass or atomic mass of ions. 3.1 Limitations of Arrhenius Concept : – (a) H+ and OH– ions exist as hydrated ions. (b) He was unable to explain the acidic nature of CO2, SO2 etc. and basic nature of NH3, CaO, Na2CO3 etc. (c) He could not explain the acid base reaction in the absence of water. SO3(g) + CaO(s)  CaSO4(g) 3.2 Factors affecting the degree of ionization : (a) Temperature – Wi th the ri se i n temperature, the degree of dissociation of an electrolyte in solution increased. Thus, Degree of dissociation  Temperature (b) Dilution : – On the increasing of dilution, the degree of dissociation increases. But at infinite dilution, their is no effect on the degree of dissociation. (c) Concentration of the solution : – Degree of dissociation    1 Concentration of solution 1 Amount of solute in given volume or wt. of solution Amount of solvent Page # 3 IONIC EQUILIBRIUM (d) Nature of Solvent :– Higher the dielectric constant of a solvent, more is its dissociation power or ionising power. Thus Degree of ionization or dissociation of an electrolyte  dielectric constant of solvent. Note:- A CABP is different from each other only by single proton. e.g. HSO–4 is the conjugate base of H2SO4 but SO24– is not. Dielectric constant :– The dielectric constant of solvent is a measure of its tendency to weaken the forces of attraction between oppositely charged ions of the given electrolyte or the force of attraction applied by solvent molecules or solute molecule is defined as Dielectric constant of solvent. 4.3 Relative strength of Acids/Bases :Any Species and its conjugate species are opposite of each other in terms of strength. e.g. Acid (or Base) Conjugate Base (or Acid) (i) Weak Strong (ii) Strong Weak Note : – Water is the most powerful ionizing solvent as its dielectric constant is highest. (e) Presence of Common Ion :– In the presence of a strong electrolyte having common ion, the degree of dissociation of an electrolyte decreases. eg. Ionisation of CH 3 COOH is suppressed in the presence of HCl due to common H+ ions. (f) Nature of Electrolyte :– At constant temperature, electrolytes ionize to a different extent in their solutions of same concentration. 4. BRONSTED AND LOWRY CONCEPT OF ACID & BASES : 4.1 Postulates :(1) Acid - Proton (H+) donor (2) Base - Proton (H+) acceptor e.g. HCl(aq) + H2O() Acid Base H2O+(aq) + Cl–(aq) NH4+(aq) + Cl–(aq) HCl(aq) + NH3(aq) Acid Base HCl(aq) + CH3COOH(aq) Acid Base+ Cl–(aq) CH3COOH+2(aq) Note :- Here CH3COOH has a less tendency to donate H+ than HCl, therefore CH3COOH acts as a weak base. 4.2 Conjugate Acid-Base Pair(CABP) :In an acid-base reaction Acid  H+ + conjugate base Base + H+  conjugate acid. e.g. NH4+(aq) + Cl–(aq) HCl(aq) + NH3(aq) Acid1 Base2 CABP Acid2 CABP 1 Base1 e.g. Strength order of acids. HClO4 > H2SO4 > HCl > CH3COOH strength order of conjugate bases ClO4– < HSO4– < Cl– < CH3COO– 5. IONIC PRODUCT WATER : Pure water is a very weak electrolyte H 2O H+ + OH– on appl yi ng the law of mass acti on at equilibrium, [H ] [OH ] [H2O] Then K = or [H+] [OH–] = K[H2O] Since, ionization takes places to a very small extent, so the concentration of unionized water molecule is regarded as constant. Thus the product of K[H2O] gives another constant Kw. So, [H+] [OH–] = KW The product of concentration of H+ and OH– ion in water at a particular temperature is known as ionic product of water. KW = 10–7 × 10–7 KW = 10–14 Note : – The value of KW is increases with the increase in temperature i.e. the ionisation of water increases with increase in temperature and finally the concentration of H+ and OH– ion increases. (a) Values of KW at various temperature :– Temperature (ºC) Value of KW (1)0 0.11 × 10–14 (2)10 0.31 × 10–14 (3)20-35 or 25º 1 × 10–14 (Room temp.) (4)60 2.9 × 10–14 (5)80 5.6 × 10–14 (6)90 1× 10–12 Page # 4 IONIC EQUILIBRIUM KW = [H+] [OH–] – log KW = – log [H+] + (–) log [OH–] pKW = pH + pOH 14 = pH + pOH The acidity and basicity depend upon the concentration of H+ and OH– ions. If, [H+] > [OH–] = Acidic solution [OH–] > [H+] = Basic solution [OH–] = [H+] = Neutral solution Some Important Points to Remember : – (a) Mass of 1 litre of water = 997 gm. (b) Molar concentration of water = 55.5 gm-mole / litre. (c) Number of water molecule in 1 litre of water = 55.5×6.023×1023 = 3.34×1025. (d) Concentration of H+ ion in litre of neutral water = 10–7 moles / litre. (e) Concentration of OH– ion in litre of neutral water = 10–7 moles / litre. (f) Number of H+ ion in one litre of neutral water = 6.023 × 1016 . (g) Number of OH– ion in one litre of neutral water = 6.023 ×1016. (h) Hydrolysis constant of water H2O H+ + OH– K = = [H ] [OH ] [H2 O] 10 7  10 7 = 1.8 × 10–16 Ans. 55.5 6. NEUTRALISATION : A reaction between acid and base to form salt and water molecule is known as neutralisation. In this type of reaction acid gives H+ ion and base gives OH– ion. 6.1 Salts : (a) Speci es formed by t he reaction of neutralization of acid and base is called as salt. (b) salt formation is the exothermic process and released energy by this reaction is called as the heat of neutral ization for 1 eq. monoacidic base and 1 eq. monobasic acid. (c) The value of heat of neutralization is equal to 13.6 kcal. (d) Salts are generally crystalline solid. (e) Salts are classified into following four types. 6.1.1 Simple Salts : These salts formed by the neutralisation process, which are of three types. (a) Normal Salt : These salts are formed by the neutralization reaction of simple acid and simple base, like - NaCl, NH4Cl, CH3COONa, KNO3 etc. (b) Aci d Salt : It is formed by the incomplete neutralization reaction of acid with the base is called the acidic salt and gives proton in aqueous solution. Like - NaHCO 3, NaHSO 4 , Na2HPO4. eg. H2SO4 + NaOH — NaHSO4 (Dibasic) 1 mole  H (c) Basic salt : Salt which is formed by the incomplete neutralization reaction of base with acid called the basic salt and gives OH– ion in aqueous solution. Like - Zn(OH)Cl, Mg(OH)Cl, Fe(OH)2Cl etc. eg. Ca(OH)2 + HCl — Ca(OH)Cl + OH– 1mol Basic salt eg. Mg(OH)2 + HNO3 — Mg(OH)NO3 1mol Basic salt eg. Al(OH)3+HCl—Al(OH)2Cl  Al(OH)Cl2 1mol Basic salt 6.1.2 Mixed Salts : Salt which are formed by the neutralization reaction of more than two different acids and bases are called mixed salt. Or Salts which furnish more than one cation or more than one anion when dissolved in water are also as mixed salt. Like Rochelle salt, eg. HO  CH  COOH | HO  CH  COOH + NaOH KOH (Two bases) — OH  CH  COONa | OH  CH  COOK eg. Ca(OH)2 + HOCl / HCl — CaOCl2 Two acids 6.1.3 Double salts : Salts which are formed by the addition of two or more simple salts are called as double salt. eg. K2SO4.Al2(SO4)3.24H2O  Potash alum eg. FeSO4.(NH4)2SO4.6H2O  Mohr’s salt etc. 6.1.4 Complex salts : – (a) Salt which are formed by the donation of electron pair by the legand molecule with the metal ion are called as complex salts. Or salts which are formed by the combination of simple salts or molecular compounds. (b) Complex salts are stable in solid state. (c) On dissolving the water, complex salts. Give minimum one complex ion eg. K4[Fe(CN)6, [Cu(NH3)4 SO4 or all coordination compound K4 [Fe(CN)6 — 4K+ + [Fe(CN)6]4 Page # 5 IONIC EQUILIBRIUM 7. pH SCALE : (i) H + concentration i n any solution can vary within a wide range from 1 mol dm–3 to 10– 14 mol dm–3. So a logarithmic notation has been devised by Sorensen in 1909 to simplify the expression of these quantities. (ii) The above notation is termed is pH scale. According to the scale the hydrogen ion concentrations [H+] are expressed in terms of the numerical value of negative power to which 10 must be raised. i.e. [H+] = 10–pH 1 or pH =– log [H+] = log [H ] FG H IJ K Note : (i) [H+] should be taken always in mol dm–3 or mol L–1. (ii) Similarly [OH–] is also expressed i.e. pOH = – log[OH–] since we know [H+][OH–] = Kw = 10–14 or pH + pOH = 14 at 25ºC (iii) [H+] [OH–] pH pOH Nature of solution 10º=1 10 –14 0 14 Strongly Acidic 10 –2 10 –12 2 12 Acidic –5 –9 10 10 5 9 Weakly acidic 10 –7 10 –7 7 7 Neutral –9 –5 10 10 9 5 Weakly basic 10 –11 10 –3 11 3 Basic 10 –14 10º=1 14 0 Strongly basic (iv) pH values of the solutions do not give immediate idea of the relative strength of the solution. e.g. (a) For a solution of pH = 1 has a [H+] 100 ti mes that of a sol uti on of p H = 3. (not three times) (b) A 4 × 10–5 N HCl is twice concentrated as a 2 × 10–5 N HCl solution but their pH values will be 4.4 and 4.7. (not double) (v) A solution of an acid having very low concentration, say 10 –8 N HCl, can not have pH = 8 as shown by definition but the pH will be less than 7. 7.1 Applications of pH (a) pH of strong acid or strong base. (i) pH of a strong acid or strong base can be calculated directly by its normality, since they are completely ionised. e.g. (a) For 10–2M HCl  10–2 N HCl + –2 [H ] = 10 M or pH = –log(10–2) = 2 (b) For 10–3 M NaOH  10–3 N NaoH [H+] = or pH = 10 14 = 10–11 [OH ] –log (10–11) = 11 (c) For 10–2M H2SO4  2×10–2 N H2SO4 [H+] = 2×10–2M or pH = –log (2 × 10–2) = 1.699 (d) For 10–3M Mg(OH)2  2 × 10–3 N Mg(OH)2 [H+] or = pH = 10 4 2  10 3 = 5 × 10–12 M – log (5 × 10–12) = 11.3 Note: Normality = Acidity or Basicity × Molarity (b) pH of weak acid or weak base (i) For weak acid or base [H+] can not be calculated by its concentration only since they are partially ionised. (ii) For [H+] calculation either Ka (or Kb) or degree of dissociation will be required in addition to its concentrations (C). (iii) As we know for weak acids [H+] = K a . C or so pH = 1 (pKa – log C) 2 (iv) For weak bases, [OH–] so pOH = [H+] = C = [OH–] = K b . C or C  1 (pkb – log C) 2 8. HYDROLYSIS : It is defined as a process involving the reaction of water on a salt to form mixture of acid and base. (a) It i s the just reverse process of neutralization Salt + water Acid + Base (b) In this reaction the solution is always neutral when both acid and base are strong. (c) If acid is stronger than base, the solution is acidic and if base is stronger than acid, the solution is basic. (d) Depending upon the nature of an acid or a base, there can be four types of salt - Page # 6 IONIC EQUILIBRIUM (i) Salt (ii) Salt (iii) Salt (iv) Salt Strong acids HCl H2SO4 HNO3 HClO4 HI of of of of strong acid and strong base. strong acid and weak base. weak acid and weak base. weak acid and strong base. Weak acids CH3COOH HCN KOH H2CO3 H3PO4 H3PO3 Strong bases NaOH LiOH RbOH CsOH Ba(OH)2 Weak bases NH4OH Ca(OH)2 Be(OH)2 Zn(OH)2 COOH Fe(OH)3 RNH2 | NH3 concentration of [H2O] remains almost constant means only one H2O molecule is ionized out of 55 crore molecule of H2O. [ BOH ] [ H  ] So, K[H2O] = [ B ] [ BOH ] [ H  ] Kh = [ B ] Some part of H2O would be ionized otherwise reaction will not be reversible.  COOH Result: H2O — H+ + 8.2 Hydrolysis of Salt of Strong acid and Weak base : HCl + NH4OH — NH4Cl + H2O strong acid strong base salt on hydrolysis, NH4Cl + H2O — NH4OH + HCl Salt weak base strong acid NH4+ + Cl– + H2O — NH4OH + H+ + Cl– Final equation NH4+ + H2O — NH4OH + H+ A. Relation between hydrolysis constant of water (Kh), weak base dissociation constant (Kb) and ionic product of water (Kw) : Let us consider a salt – BA H2O + BA BOH + HA Salt weak base strong acid B+ + A– BOH + H+ + A– Resultant eq. B+ + H2O BOH + H+ on applying the law of mass action K = [ BOH ] [ H  ] [ B ] [ H 2 O] + OH– [ H  ] [OH  ] [ H 2 O] = K [H2O] = [H+] [OH–] KW = [H+] [OH–] [H+] = KW [OH  ] ......(2) Similarly some part of BOH will be also ionized  BOH B+ + OH– OH– Thus, hydrolysis does not occur. So, the solution is neutral in nature i.e. salt of strong acid and strong base does not hydrolysed. Note : No hydrolysis reaction takes place due to conjugate acid and base of this type of salt, which are weak. H+ H 2O K 8.1 Hydrolysis of Salt of Strong acid and Strong base : HCl + NaOH — NaCl + H2O strong acid strong base salt on Hydrolysis, NaCl + H2O — NaOH + HCl Na+ + Cl– + H2O — Na+ + OH– + H+ + Cl– .....(1) [ B ] [OH  ] Kb = [ BOH ] [B+] = K b  [ BOH ] [OH  ] ........(3) On putting the value of [H+] and [B+] on the equation (1) by the equation (2) and (3), we get - Kh  KW Kb B. Calculation of degree of hydrolysis : – Assume the original concentration of the salt in the solution is C moles / lit. and h is its degree of hydrolysis at this concentration. Then we have B+ + H2O BOH + H+ Initial molar C 0 0 concentration Molar concentration (1 – h)C hC hC at equilibrium  1 h V h V h V [ BOH ] [ H  ] K = [ B ] [ H 2 O] [ BOH ] [ H  ] K[H2O] = [ B ] Page # 7 IONIC EQUILIBRIUM h h  h2 V V Kh = = 1 h (1  h) V V Since, h << 1, therefore pH 1 – h = 1, Kh = h2 = h = Kh = Hydrolysis constant Then, h   Kh × CH3COOH + NaOH on hydrolysis V KhV h = = = V Kh C Since,  But KW Kb  C KW = [OH  ] and [OH–] pH = – = hC log FG K C IJ HK K 1/ 2 W A. Relation between Kh, KW and Ka : Let us consider a salt - BA H2O + BA BOH + HA Salt Strong base Weak acid B+ + A– B+ + OH– + HA [ HA ] [ OH  ] [A  ] [ H 2 O] b [ HA ] [OH  ] So, K[H2O] = [A  ] [H+] Kh 1/ 2 b = (iv) As solution is basic, the pH will be greater than 7. Concentration of [H2O] remains almost constant means only one H2O molecule is ionized out of 55 crore molecule of H2O. W pH (ii) Solution of this type of salt is basic in nature as forming of OH – ion in resultant reaction. K = F K C IJ pH = – log G HK K pH = – Result : – (i) Hydrolysis of salt of weak acid and strong base is called anionic hydrolysis. Resultant equation :– A– + H2O OH– + HA on applying the law of mass action, KW [H+] = hC KW h = Kb .C KW KWC [H+] = C = Kb .C Kb = Since, CH3COONa + H2O (iii) This type of solution turns red litmus to blue. Kh C C. Calculation of hydrogen ion concentration or calculation of pH : – [H+] 1 log C 2 CH3COONa+H2O CH3COOH + NaOH Salt Weak acid strong base CH3COO— + Na+ + H2O CH3COOH+ Na++OH– on putting the value of Kh in the above equation h 1 pK – 2 b 8.3 Hydrolysis of salt of Weak acid and Strong base : Hence, the degree of hydrolysis is directly proportional to the square root of volume, inversely proportional to the square root of C.  V = 1/C  h2 = Kh×V h2 7 – Here, on the study of above equation, we can say that the pH of the strong acid weak base is less than 7. 2 h V = 1 1 1 log KW + log Kb – log C 2 2 2 1 pK – 2 W 1 pK – 2 b 1 log C 2 = [ HA ] [OH  ] [A  ] ........(1) Some part of H2O would be ionized otherwise reaction will not be reversible  H 2O H+ + OH– [ H  ] [OH  ] K = [ H 2 O] K[H2O] = [H+] [OH–] Page # 8 IONIC EQUILIBRIUM KW = [H+] [OH–] C. Calculation of hydrogen ion concentration or calculation of pH : KW [H  ] [OH–] = .......(2) Since, [H+] = KW [OH  ]  [H+] = KW hC but h = KW KaC  [H+] = KW × C Since, pH Similarly some part of HA will be also ionized.  HA H+ + A–   [ H ] [A ] [ HA ] Ka = [A–] = K a  [ HA ] [H  ] .......(3) on putting the value of OH– and A– on the equation (1) by the equation (2) and (3), we get - Kh  KW Ka = B. Calculation of degree of hydrolysis : Assume the original concentration of salt in the solution is C moles / lit. and h is its degree of hydrolysis at this concentration. Then we have A – + H2O C Initial molar 0 concentration Molar concentration (1 – h)C at equilibrium  Kh OH– + HA 0 hC hC hC  hC [OH  ] [ HA ] = =  (1  h)C [A ] = It h is very small as compared to one. Then (1 – h) can be equal to 1. Hence, the above equation becomes Kh = h2C  h = Kh C h = Kh  V  C = 1 V Thus, the degree of hydrolysis of a salt of weak acid and strong base is inversely proportional to the square root of the molar concentration and directly proportional to the square root of the volume of the solution. on putting the value of Kh , h = KW Ka C [OH–] = hC KaC KW = KW Ka C = – log [H+] – log K W . Ka C F K W . K a I 1/2 GH C JK = – log = – = 1 1 pKW + pKa 2 2 1 1 1 log KW – log Ka + log C 2 2 2 + 1 log C 2 1 1 pKa + log C 2 2 Here, on the study of above equation, we can say that the pH of strong base weak acid salt is greater than 7 and therefore solution of salt of strong base weak acid be always alkaline. pH = h2C (1  h) and 7 + 8.4 Hydrolysis of salt of weak acid and weak base : CH3COOH + NH4OH CH3COONH4 + H2O on hydrolysis CH3COONH4+H2O CH3COOH + NH4OH Salt Weak acid Weak base CH3COO– + NH4+ + H2O CH3COOH + NH4OH Result : – (i) CH3COO–+NH4++H2O CH3COOH+ NH4OH This type of salt hydrolysis is called isoionic salt hydrolysis. Conditions : (a) If pKa = pKb, the pH = 7 and solution is neutral. (b) If pKa > pKb, the pH > 7 and solution is alkaline because here the acid is relatively weaker than base. Page # 9 IONIC EQUILIBRIUM (c) If pKa < pKb, the pH < 7 and solution is acidic because here the base is relatively weaker than acid. A. Relation between Kh, Kw, Ka and Kb : Let us consider a salt - BA BA + H2O HA + BOH Weak acid Weak base B+ A– + + H2 O HA + BOH B+ + A – + H2 O HA + BOH C 0 0 Initial molar C concentration Molar (1 – h)C (1 – h)C hC hC concentration at equilibrium  Note :– Here is this solution the HA and BOH are weak acid and bases. So, They remain almost undissociated. on applying the law os mass action, [HA] [BOH] K = [B  ] [A  ] [H O] 2 Concentration of [H2O] remains almost constant. So, K[H2O] = Kh = [HA ] [BOH] [B  ] [ A  ]  .......(1)  [B ] [ A ] Now, let us consider the dissociation of weak species HA and BOH HA H+ + A – BOH B+ + OH– [ H  ] [A  ]  Ka = [ HA ] .......(2) [ B ] [OH  ] and Kb = [ BOH ] ........(3) We know that, KW = [H+] [OH–] Dividing (4) by (2) and (3) - .........(4) KW Ka  K b [ H  ] [OH  ] [ HA ] [ BOH ] = [ H  ] [A  ] [ B ] [OH  ] = [ HA ] [ BOH ] [ B ] [A  ] Kh = KW Ka  K b h2 Kh = (1  h) 2 If h is very small as compared to 1 then (1 – h) is equal to 1. Hence the above equation becomes, Kh = h2 h B. Calculation of degree of hydrolysis : Assume again, the original concentration of salt in the solution is C moles / litre and h is its degree of hydrolysis at this concentration. Then we have Kh KW Ka  K b h = C. Calculation of hydrogen ion concentration or calculation of pH : In this type of cases, the hydrogen ion concentration may be calculated from the following equation of weak acid HA. HA H+ + A– Thus, Ka or [H+] = = [H+] = Since, .........(6) = Thus, the degree of hydrolysis in this case is independent of concentration or dilution of the salt. on putting the value of Kh from equation (6) ........(5) from equation (1) and (5)- [ HA ] [ BOH ] [ B ] [A  ] hC  hC (1  h)C  (1  h)C = or [HA ] [BOH] Kh = [A  ] [ H  ] [ HA ] K a [ HA ] [A  ] Ka hC (1 h)C = Ka h (1 h) h << 1 then (1 – h) = 1 [H+] = Ka × h on putting the value of h [H+] = Ka We know that pH = KW Ka . K b – log [H+] = KW . Ka Kb Page # 10 IONIC EQUILIBRIUM = = – log – log KW . Ka Kb FG K . K IJ H K K W CH3COO– + H+ (from acid) equilibria wi ll shift in forward directi on. Similarly on addition of small amount of base, the [OH–] in solution will increase 1/ 2 a b = = pH = – 1 1 1 log KW – log Ka + pK 2 2 2 B 1 pK + 2 w 7 + 1 pKA – 2 1 pKa – 2 1 pKb 2 1 pKb 2 9. BUFFER SOLUTIONS : (i) In certain applications of chemistry and biochemistry we require solutions of constant pH. Such solution are called buffer solution. (ii) A solution whose pH any great extent by the quantities of either an acid (OH– ions) is called buffer CH3COOH (very less ionised) is not altered to addition of small (H+ ions) or a base solution. NH4+ + OH– (from base) NH4OH (very less ionised) equilibria will shift in forward direction Hence one can conclude that there will be no change (almost) in the pH of the solution. (iii) Simple buffer solutions have very little 1 1 pKa – pKb so 2 2 pH can not be significance since pH = 7 + sol ution prepared. of desi red 9.2 Mixed buffer Mixed buffers are solutions of more than one compounds.They can be further classified as (a) Acidic buffer (b) Basic buffer (c) A mixture of an polyprotic acid and its acidic salt (iii) Buffer solutions are also called solutions of reverse acidity or alkalinity. (d) Soluti on of amphol yte or amph oteric electrolyte viz. Proteins and Amino acids. (iv) Following are the characteristics of buffer solutions Note : (i) Ampholyte or amphoteric electrolytes are those which can show properties of both acid and base. (a) It must have constant pH. (b) Its pH should not be changed on long standing (c) Its pH should not be changed on dilution. (d) It pH should not be changed to any great extent on addition of small quantity of acid or base. (v) Buffer solutions can be classified as follows. (A) Simple buffer (B) Mixed buffer 9.1 Simple buffer (i) It is a solution of one compound i.e. salts of WA + WB (Refer salt hydrolysis.) (ii) Buffer action of such solution can be explained as follows, In the salt solution of CH3COONH4 following equilibria will be there, (a) CH3COO+(aq.) + NH4+ (aq.)+ H2O CH3COOH(aq.)+ NH4OH(aq.) (Salt hydrolysis) (b) NH4OH NH4+ + OH– (weak base) (c) CH3COOH CH3COO–+H+(weak acid) (d) H2O+H2O H3O+ + OH– (feebly ionised) on addition of small amount of acid the [H+] in solution will increase and 9.2.1 Acidic buffer (i) These are the mixture of a weak acid and its salt with strong base. e.g. (a) CH3COOH + CH3COONa (b) Boric acid (H3BO3) + Borax (Na2 B 4O7) (ii) Buffer action of acidic buffer can be explained with following equilibria, (a) CH3COOH H+ + CH3COO– (Feebly ionised.) (b) CH3COONa   Na+ + CH3COO– (completely ionised.) (c) H2O H+ + OH– When a strong acid is added in the solution [H+] increases, which will combine with CH3COO– to form feebly ionised CH3COOH, which is also suppressed by common ion effect, CH3COO– + H+ (from acid) CH3COOH (Weak Acid) Again when strong base is added in the solution it will attack on unionised acid CH3COOH to form feebly ionised H2O molecules Page # 11 IONIC EQUILIBRIUM CH3COOH +OH– (from base) CH3COO– + H2O (iii) One can remember the buffer action of acidic buffer with the help of following figure - (iv) pH of such acidic buffer can be calculated as follows. (a) CH3COOH CH3COO– + H+ (Ka) – 2O (b) CH3COONa H  CH3COO (aq.) + Na+ (aq.) CH3COOH is feebly ionised and its ionisation is also suppressed by presence of common ion (CH3COO–). So one can fairly assume [CH3CO O–]=[salt] and [CH3COOH]=[Acid] taken initially in buffer solution. so Ka or = [H+] [H ][CH3 COO  ] [CH3 COOH] = Ka [CH3 COOH] [CH3 COO  ] log[H+]=logKa+log[CH3COOH]–log[CH3COO–] –log[H+] = – logKa – log[CH3COOH] + log [CH3COO–] Note : Mixture of weak acid and strong base solution can also act as an acidic buffer, if value of N1V1 of weak acid is greater than the value of N2 V2 of strong base. 9.2.2Basic Buffer (i) These are the mixture of a weak base and its salt with strong acid. e.g. (a) NH4OH + NH4Cl (b) Glycine (NH2CH2COOH) + +Glycine hydrochloride (Cl NH3CH2COOH) (ii) Buffer action of basic buffer can be explained with the help of following equilibria (a) NH4OH (b) NH4Cl NH4+ + OH– + –  NH4 + Cl (Kb) (c) H2O H+ + OH– (Kw) When a strong base is added in the solution [OH–] increases, which will combine with NH4+ to form feebly ionised NH4OH, which is also suppressed by common ion effect. NH4++OH–(From base) NH4OH (Weak base) Agai n when strong acid is added in the solution it will attack on unionised base NH4OH to form feebly ionised H2O molecules. NH4OH + H+ (From acid) NH4+ + H2O (iii) One can remember the buffer action of basic buffer with the help of following figure - [CH3 COO  ] pH = pKa + log [CH3 COOH] pH = pKa + log or pH = [Conjugate base] [ Acid] pKa + log [Salt] [Acid] (iv) pH of such basic buffer can be calculated from Henderson - Hasselbalch equation as follows. (a) NH4OH NH4+ + OH– (Kb) + – 2O (b) NH4Cl H  NH4 (aq.) + Cl (aq.) Above equation called Henderson's equation. * If we increase the concentration of given salt in acidic buffer, pH will also increases. * If we increase the concentration of acid in acidic buffer, pH will decreases. * If conc. and volume are given for salt and acid then the pH is given by the following formula- NH4OH is feebly ionised and its ionisation is also suppressed by presence of common ion (NH4+). So one can fairly assume [NH4+] = [Salt] and [NH4OH] = [Base] taken initially in buffer solution. so Kb [OH ] [NH4 ] = [NH4 OH] [N2 V2 ] pH = pKa + log [N V ] 1 1 or = Where N2 V2 = conc. & volume of salt N1V1 = conc. & volume of acid [OH–] = K b [NH4 OH] [NH4 ] K a [Base] [Salt] Page # 12 IONIC EQUILIBRIUM or pOH = pKb + log or pH = 14 – [Salt] [Base] AB Insoluble Solute pOH 9.3 Buffer Capacity (i) The property of a buffer solution to resist alteration in its pH value is known as buffer capacity. (ii) Buffer capacity is number moles of acid or base added in one litre of solution as to change the pH by unity, i.e. Buffer capacity ()  = b  (p H) where b is number of moles of acid or base added and  (pH ) is change in pH. 9.4 Applications of Buffer solutions in analytical chemistry. (i) For the removal of PO43– ion after second group using CH3COOH + CH3COONa buffer. (ii) For the precipitation of lead chromate quantitatively in gravimetric analysis using CH3COOH + CH3COONa buffer. (iii) For the precipitation of hydroxides of group using NH4Cl + NH4OH buffer. IIIrd (iv) A buffer solution of NH 4Cl, NH 4OH and (NH 4 ) 2 CO 3 i s us ed for preci pi tat ion of carbonates of Vth group. 10. SOLUBILITY (s) : At a constant temperature, the mass of a solute or electrolyte dissolved in the 100 gm of solvent in its saturated solution is called as solubility. Or number of gm mole of a solute dissolved in one litre of water at constant temperature is called as solubility of that solute. Solubility of a solute in moles / litre = Solubility of solute in gm / litre molecular weight of the solute 10.1 Solubility product (KSP) : nic concentration of the ions of binary solid electrolyte in saturated state at constant temperature. (a) Let solubility of a compound A x B y be s moles L –1 it means that if more than s moles are dissolved in solvent (one litre) only s moles will be soluble, rest will be insoluble, following equilibrium is established, 1 Note : – In the solubility s moles L–1, moles of only unionised are counted moles of ions and insoluble solute do not have anything to do. (b) According to law of mass action - [ A  ] [ B ] K1 = [AB] or K1 [AB] = [A+] [B–] = KSP This KSP is called solubility product. Number of moles acid or base added to 1l sol. = Changein pH or AB A + + B– Unionised Ions (Saturated solution) (c) At a certain temperature solubility product of a compound is constant, it means that ions are formed in the manner that product of their concentration is always a constant. However, it becomes clear that if one of ions (A+ or B–) is added from outside, it would tend to increase KSP because [A+] or [B–] has get increased, so that extra ions will react with other ions to convert in insoluble part and this will get precipitated. (d) KSP increases with increase in temperature. (e) In a saturated solution. KSP = [A+] [B–] (f) In an unsaturated solution of AB KSP > [A+] [B–] i.e. more solute can be dissolved. (g) In a supersaturated solution KSP < [A+] [B–] i.e. precipitation will start to occur. 10.2Relationship between Solubility and Solubility Product : The equilibrium for a saturated solution of a salt Ax B y may be expressed as, Ax By xAy+ + yBx– Thus, solubility product KSP = [Ay+]x [Bx–]y Let the solubility of the salt Ax By in water at a particular temperature be ‘s’ moles per litre then Ax By xAy+ + yBx– xs ys x So, KSP = [xs] [ys]y KSP = xx . yy(s)x + y Page # 13 IONIC EQUILIBRIUM (a) 1 : 1 types salts or AB type of salts : eg. AgCl, AgI, BaSO4, PbSO4, etc. AB A+ + B– let the solubility of AB is s moles per litre. So, KSP = [A+] [B–] = s × s = s2 s = KSP (b) 1 : 2 or 2 : 1 type of salts or AB2 or A2B type of salts : eg. Ag2CrO4, PbI2, Ag2CO3, CaF2, CaCl2 etc. (i) AB2 A2+ + 2B– s 2s let the solubility of AB2 is ‘s’ moles per litre So, KSP = [A2+][B–]2 = s × (2s)2 = 4s3 s = 3 K SP 4 (ii) A2B 2A+ + B2– let s the solubility of A2B So, A2B 2A+ + B2– 2s s + 2 KSP = [A ] [B–2] = (2s)2 (s) = 4s3 s = 3 K SP 4 (c) 1 : 3 type of salts or salts of AB3 or A B type of salt AB3 = Valency of A = 3 × Valency of B eg. FeCl3, AlCl 3, PCl3, Al(OH)3, Fe(OH)3 etc. A 3B = 3 × Valency of A = Valency of B eg. Na3BO3, Na3PO4, H3PO4 etc. (i) AB3 A3+ + 3B– let the solubility of A3B is ‘s’ mole / litre. AB3 KSP A3+ + 3B– s 3s = [A3+] [B–]3 = s × (3s)3 = 27s4 s = 4 K SP 27 (ii) A3B 3A+ + B–3 let the solubility of A3B is ‘s’ moles/litre. A3B 3A+ + B–3 3s s + 3 KSP = [A ] [B–3] = (3s)3 × s = 27s4 (d) 2 : 3 or A2B3 type of salts : eg. Al 2(SO4) 3 A2B2 2A+ 3 + 3B– 2 let the solubility of salt A2B3 is ‘s’ So, A2B3 2A+3 + 3B–2 2s 3s KSP = [A+3]2 [B–2]3 = (2s)2 × (3s)3 = 4s2 × 27 s3 5 KSP = 108 s s = 5 K SP 108 10.3 Applications of KSP : A. In purification of common salt : In a saturated solution of NaCl & impurities, by passing HCl gas through it, increase the Cl – ion concentration which shifts the equilibrium NaCl Na+ + Cl – to left & causes the precipitation of NaCl. B.In preparation of NaHCO 3 by solvay method : Precipitation of NaHCO3 from its saturated solution is done by addition of NH4HCO3, HCO3– as common ion. C. Predicting precipitation in Ionic reactions Precipitation in an ionic reactions could be predicted by compari ng K SP to the ionic concentration product of ions. D. Salting out action of Soap : When NaCl is added to saturated solution of soap (RCOONa) concentration of Na+ increases & causes to precipitation of soap, it is due to [Na+] [RCOO–] > KSP. E. In qualitative analysis : Qualitative analysis of mixtures is based on the principle of solubility product. Some important applications are as follows (a) Precipitation of I group radicals : Group reagent is dilute HCl. In the presence of dilute HCl, the ionic product of I group radicals as there chlorides becomes more than the solubility product. Thus I group radicals get precipitated as their chlorides. (b) Precipitation of II group radicals : Group reagent is H2S. In presence of dilute HCl, the dissociation of H2S is suppressed due to common ion (H+) and only ionic product of the sulphides of II group radicals exceeds their solubility product and get precipitated. (c) Precipitation of III group radicals : Group reagent is NH4OH in presence of NH4Cl. The K SP s = 4 27 presence of NH4Cl suppresses the ionisation of 1 Page # 14 IONIC EQUILIBRIUM NH4OH, NH+4 as common ion. Thus, the ionic (d) This is called Common ion Effect that if in product of hydroxides of III group radicals a solution of weak electrolyte solution of strong exceeds corresponding solubility products and get precipitated. electrolyte with one of its common ion is added ionisation of weak electrolyte is suppressed. (d) Precipitation of IV group radicals : Group reagent is H2S in presence of NH4OH. In the presence of NH4OH enhances the dissociation of H2S. Thus, due to high ionisation of H2S the 11.1Effect of Common ion on Solubility : As we saw that for a saturated solution product of concentration of ions should not exceed a concentration of S–2 ions increases and ionic constant called solubility product. Now suppose, product of sulphides of IV group radicals exceeds we some how added extra common ions to correspondi ng solubility product and get increase concentration of ions it will tend product precipitated. to exceed the constant but it cannot so ions (e) Precipitation of V group radicals : Group will react to give molecule and get precipitated reagent is ammonium carbonate in presence of e.g. let solubility of AB in water be s moles L– NH4Cl. Common ion NH4 + suppresses ionisation of (NH4) 2CO 3 thus only V group carbonates having low solubility product get precipitated. 1. ion B– let concentration of CB be C. 11. COMMON ION EFFECT : (a) If we consi der ion isa ti on of a weak electrolyte say CH3COOH, it ionises as, CH3COOH CH3COO– + H+ [CH 3COO  ] [ H  ] Ka = [CH 3COOH ] It is dissolved in solution CB having a common AB A+ + B– s s s KSP = s2 [A+] [B–] = (Product should not exceed in CB be S1. CB C+ + C C AB A+ + and H + will not get changed further, their s1 s1 concentration on a definite mathematical [A+] = s1 calculation will give a constant called equilibrium [B–] constant which will remain constant at same s1 + C K SP = this equilibrium constant ? A stage will come such that concentration of CH3COOH, CH3COO– has stopped only concentration will remain same, If ions will continue to change in molecule and = (c) Now suppose some how CH3COO– ions are added to the solution to increase CH3COO– ions concentration since mathematical calculation will tend to change itself to change that equilibrium constant, electrolyte (CH3COOH) will do same thing in order to keep Ka constant because it has to be a constant, it will reduce its ionisation. Thus [CH 3 COOH] wi ll increase, [H + ] will decreases and Ka will remain constant. or B– C B– s1 (s1) (s1 + C) s1 <<< C K SP = s 1C s1 s2 C molecule will continue to get ionised this is said to be Dynamic Equilibrium. otherwise precipitation will start) Now. Let solubility of AB (b) What do we mean by this equilibrium and temperature. It does not mean that reaction ......(1) s2 = = s2, from equation (1) So, solubility gets reduced.