Chapter 09: Acid Base Chemistry

General properties of acids, bases and indicators • Scientists have recognized that acids and bases have distinct characteristic properties, since the earliest days of experimental chemistry. • Acids have a sour taste while bases are bitter. • Acids and bases change the colour of certain dyes called indicators, such as litmus and phenolphthalein. • Acids change litmus from blue to red and phenolphthalein from pink to colourless. • Bases change litmus from red to blue and phenolphthalein from colourless to pink. From these colour changes, acids and bases neutralize each other • During neutralization, acids and bases react with each other to produce ionic substances called salts. BRONSTED-LOWRY CONCEPT

Bronsted-Lowry Concept Introduction: In 1923, the Danish chemist J.N. Bronsted and the English chemist T.M. Lowry independently expanded the Arrhenius theory of acid-base. Need of Bronsted-Lowry concept The limitations of Arrhenius theory are that it describes the reaction of an acid and base only in the aqueous medium. There are many reactions that occur in solvents other than water or in the absence of any solvent. Bronsted-Lowry pointed out that acid-base reactions can be seen as proton-transfer reactions and that acids and bases can be defined in terms of this proton (Ht) transfer. Definition of Acid: An acid is the specie donating a proton in a proton-transfer reaction. Definition of Base: A base is a specie that accept proton from the other species in a proton transfer reaction. When HCl dissolves in water, an H* ion (a proton) is transferred from HCl to water, where it Interesting Information! S.Q. How acidity of stomach is treated? Ans. Acidity in our stomach is due to excess of HC1. It is treated by taking mild bases such as baking soda (NaHCO3), milk of magnesia Mg(OH)2 and aluminum hydroxide AI(OH)3. becomes attached to a lone pair of electrons on the O atom and forms H30+ (the acid) has donated the Ht and H2O (the base) has accepted it. HCl (g) Base Acid Conjugate Acid: A specie formed after a Bronsted base accepts a proton from the acid is called the conjugate acid. Example: The hydronium ion (H3O*) is the conjugate acid of water. Conjugate Base: A specie formed when an acid donates a proton to a base is called the conjugate base. Cl is conjugate base of Hcl. Hcl-CI and H20-H3O* are conjugate acid-base pairs. molecule becomes OHion: NH3(g) + H2O Acid Base Conjugate base H20 (the acid) has donated the proton and NH3 (the base) has accepted it. NH3-NH4* and H2O- OH are conjugate acid-base pair Definition of Amphoteric: An Amphoteric substance is a species that can act as either an acid and a base (it can lose or gain a proton), depending on the other reactant H2O is amphoteric in nature because it acts as a base in one case and as an acid in the other. Application of Bronsted-Lowry theory It can be applied to acids in solvents other than water or even solventless reactions. The reaction between gaseous ammonia and HC gives solid NH4CL. Hcl is an acid because in the reaction it donates a proton to the NH3 molecule. The NH3 acts as a base, even though hydroxide ion OH is not present, and accepts a proton from the Hcl molecule. Table 9.1 Conjugate acid-base pairs of common species Acid Base + H20 HNO3 + H2O H2O + CO3- LEWIS CONCEPT OF ACIDS AND BASES

Lewis concept of acids and bases Introduction: In 1923, G.N. Lewis proposed a generalized definition of acid-base behavior in which acids and bases are identified by their ability to accept or to donate a pair of electrons and form a coordinate covalent bond. Lewis Acid: A Lewis acid is any species (molecule or ion) that can accept a pair of electrons. Lewis Base: A Lewis base is any species (molecule or ion) that can donate a pair of electrons. Reaction of NH3 and Ag+ A Lewis acid-base reaction occurs when a base donates a pair of electrons to an acid. The two ammonia molecules act as Lewis bases, donates a pair of electrons to a positively charged silver ion (the Lewis acid). In effect, HCl HO* Cl (aq) (aq) Conjugate acid Conjugate base + NH 4(ag) Conjugate acid → NH,CC (s) Conjugate Acid Conjugate Base H3O+ + NO3 H3O+ HSO4 HCO3 Adduct: The sum of charges on the left side is +1, so the acid-base adduct on the right hand side must carry a charge of +1: H + H - Nº Ag* H Lewis acid Lewis base Reaction of BF3 and F The boron atom in boron trifluoride, BF3, has only six electrons in its valence shell. The boron atom has an incomplete octet, it can behave as an electron pair acceptor. As a result, BF3 1s a very good Lewis acid and reacts with many Lewis base; a fluoride on in the Lewis base in this reaction, donating one of its lone pairs: • • • • B + • • Lewis base Lewis acid Adduct: The negative charge on the adduct is the sum of charges on the left hand side of the equation is -1, the sum of charges on the right hand side must also be -1. Quick Check 9.11000 (a) Identify the conjugate acid-base pairs in the following reactions: (ag) + CN (aq) Ans. In this reaction, the conjugate acid-base pairs are HCN-CN and H2O-H30+ (aq) + CH, COO: (ii) CH, COOH(., + H2O() = HO* Ans. In this reaction, the conjugate acid-base pairs are CHCOOH - CH300 and H20- H3O+ H30(ag) + HS (ag) Ans. In this reaction, the conjugate acid-base pairs are H2S-HS and H2O-H30+ (iv) HC(g) + HCO3(ag) H, CO 3(ag. + C(ag) Ans. In this reaction, the conjugate acid-base pairs are Hcl-CI and HCO, -H2СО3. (b) Identify the Lewis acid and Lewis base in the reaction between: SO and O2- (ii) Hcl and H20 (iii) BF3 and NH3 Àlso write down the balance chemical equation for each reaction and explain. Ans. (i) SO, +0^ _→SO?- In this reaction, SO3 is Lewis acid and O-2 is Lewis base. In this reaction, HCl is Lewis acid and H20 is Lewis base. (iii HN: + BF →HN:→BF] In this reaction, NH3 is Lewis base and BF3 is Lewis acid. + H H H-N:→Ag<:N-H H H Acid base adduct •F: • • Acid base adduct + Cl(ag) IONIC PRODUCT OF WATER

Definition: The product of concentration of Ht and OH ions in water is called ionic product of water or dissociation constant of water (Kw). Explanation • Pure water is a poor conductor of electricity but its conductance is measurable. • Water undergoes self-ionization reversibly as follows, Net reaction The equilibrium constant for this reaction can be written as follows: [H2O] • The concentration of H2O, [H20] in pure water may be calculated to be 1000g dm3 divided by 18g mol-1 giving 55.5 mol dm • Water is present in very large excess and very few of its molecules undergo ionization, so its concentration remains effectively constant. • Constant concentration of water is taken on left hand side and multiplied with Kc to get another constant called Kw. tree 1) 18 × 10-16 ½ 555 = 101 1014[10 This value, 1.01 × 10-14 is called Kw of water at 25°C. Kw = [Ht] [OH] = 10-14 at 25°C Effect of temperature on Kw: The effect of temperature on Table 9.2 Kw at various Kw is shown in table. The value of Kw increases almost 75 times when temperature is increased from 0°C to 100°C. For neutral water [+1=10114 [HT]2 = 10-14 (at 25°C) [Ht] = 10-7 moldm The concentration of OH is also the same in neutral water. [OH:] = 10-7 moldm3 Effect of addition acid and base on Kw Whenever some quantity of acid or base is added to water, then Kw remains the same, but [Ht] and [OH] are no more equal. Addition of acid to water: In case of addition of small amount of acid Addition of base to water: In the case of addition of few drops of a base. During both of these additions, the values of Kw will remain the same, i.e. 10-14 at 25°C. (ag) + OH (ag) temperature Kv Temp. (C) 0 0.11 × 10-14 10 0.30 × 10-14 25 1.0 x 10-14 40 3.00 × 10-14 100 7.5 × 10-14

pH Definition: The negative logarithm of molar concentration of H' ions in a mixture is called pH pOH Definition: The negative logarithm of molar concentration of OH ions in a mixture is called pOH. Explanation • In all the aqueous solutions, the concentration of Ht and OH are too low to be conveniently expressed and used in calculations. • In 1909, Sorenson, a Danish biochemist, introduced the term pH and pOH. So, the scales of pH and pOH were developed. These two quantities can be calculated as: pH = -log[H] pOH = -log[0H] For neutral water, pH=-log[10-7] = 7 pOH = -log[10-] =7 pH + pOH = 14 • The value of pH normally varies between 0 → 14 at 25°C. The pH can have a negative value or greater than 14 pH can be possible. • The pH values of some familiar aqueous solutions are shown in figure. This table helps to understand the acidic or basic nature of commonly used solutions. Acidic [H301110° 10210310410-5 10-6 10-7 10-810-9 10-10 10-11 10-12 10-13 10-14 Hdi (1:0 M) Lemon Juice Carbonated water (pH 3.9) (PH 0:0) (pH 2.2-2.4) Beer Stomach acid Vinegar. (PH1:0-3.0) (PH2:4-3.4) (pH4.0-4.5) T 0 1 5 2 4 3 6 pH Fig: pH values of some common substances Quick Check 9.2 (a) A solution is prepared by mixing equal volumes of two solutions: one with a pH of 4.0 and another with a pH of 10.0. Calculate the Kw for this mixture at 25°C: Ans. For the first solution: pH = 4.0 [=10=10 pОН = 14-4 = 10 Similarly, for second solution. As two solution have equal volume total [H*] and [OH] is pH = 10 Interesting Information! S.Q. What pH values other than 0-14 are possible? Ans. Solutions of negative pH and having values more than 14 are also known. Basic Neutral Blood Milk Baking soda Household ammor (pH 6.4) (pH 7.4) (0.1 M) (pH 8.4) (pH 11.9) Seawater Milk of magnesia NaOH (1.0M (pH 7.0-8.3) (PH 10.5) (PH 14:0 T T 9 8 10 12 11 13 14 pOH = 14-pH pOH = 14-10=4 [OH]= 10pоn = 104 M 104 + 10-10 - = 5x10-M [H" ]mix = - + 104 10-14 - = 5x10-M [OH" Imix = 2 Kw = [H]mix_OH ]mix Kw = (5x10-5) (5x10-5) Kw=2.5 × 10- (b) At a specific temperature, the ionic product of water Kw is 1.0 x 10-14. If the concentration of OH ion in a solution is 2.5 × 10-8 M. Calculate the concentration of H* ions and the pH of the solution. Ans. Kw = 1.0×10-14 [OH] = 2.5 × 10-8 M [H]=? pH=? Solution 10×10-14= [H] [2.5x10-] 10x101 1540A50M . pH = log [HT] pH = log (4.0 × 10-) pH = 6.40 (c) Copper-plate etching solutions is prepared by diluting concentrated HNO3 to 0.30 M HNO3. Calculate [H*), pH, [OH] and pOH of this solutions at 25°C. Ans. HNO, H* + NO, [H] = 0.30 M pH = -10g (0:30) = 0.5229 Similarly pH + pOH = 14 pOH = 14-0.5229 = 13.4771 Now 1.0x10-14 K [OH] =- 0.3 [H+] [OH] = 3.33 × 10-4 M IONIZATION CONSTANT OF ACIDS (Ka)

Ionization constant of acids Definition: The value of Ka called the dissociation constant of an acid, is the quantitative measure of the strength of the acid. Explanation Acids and bases, when dissolved in water, may or may not be completely dissociated. Many acids are weak electrolytes and ionize to an extent which is much less than 100%. Derivation: Suppose we have an acid HA dissolved in water, HA + H2O → H3Ot+ A Kc for the reversible reaction will be written as follows: = K [H2O]=: [HA] Let K[H2O]= K. Hence [HA] • This equation can be used to calculate Ka for any acidic solution if we know the pH or [H] of that solution and the initial concentration of the dissolved acid [HA]. • This can also be used to calculate the equilibrium concentration of H3Ot and A produced if we know the initial concentration of acid HA and its Ka Value. Effect of Ka on strength of acid (i) Ka < 10-3 acid is weak (ii) Ka = 1 to 10-3 acids are moderately strong. (iii) Ka> 1 acid is strong Table 9.3 The values of Ka for some acids. Acid Dissociation Hcl HNO3 HNO, H* + NO: H2SO4 H,SO, H* + HSO: 'HSO 4 HSO, =H* + SO?- HF CH:COOH CH COOH-H* + CH,00- H¿CO3 H2S H,S,-÷H*+HS-' NH!+ Ht + NH3 HCO3* H20 Relative Ka strength Very large (10 6) Very strong Very large (10 3) Very strong Large (10 3) Very strong 1.3 × 10-2 Strong 6.7 × 10-5 Weak 1.85 × 10-5 Weak 4.4 × 10-7 Weak 1.0 × 10-7 Weak 5.7 × 10-10 Weak 4.7 × 10-11 Weak 1.8 × 10-16 Very Weak

Calculating HOt ions from Ka • The weak acids do not completely dissociate in water. • The concentration of hydronium ions is not equal to the initial concentration of the acid. • The equilibrium expression is used to calculate the H3O* concentration. For a weak acid HA, the dissociation in water can be represented as: HA(aq) = H* (aq) + A (aq) In water, H* combines with water to form H30*: H+ (aq) + H2O = H3O (aq) The equilibrium equation in terms of H3O* (ag) is: HA(ag) H30* (ag) + A (ag) Let the initial concentration of the weak acid HA be C mol dm3 At equilibrium: [A] = x and [HA] = C-x [H30] = x; The expression for the acid dissociation constant (Ka) is: Substitute the equilibrium concentrations into this expression: XXX K; = C- x For weak acids, x is usually very small compared to C, so C-x ~C. Therefore, the expression simplifies to: x~ Kax C Taking the square root of both sides: Thus, the concentration of H3O* is given by: Quick Check 9.3 (a) The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. Ans. pH = 2.38 [Ht] = 10-2:38 = 4.17 × 10-3 M Initial concentration of formic acid = [HCOOH] = 0.10M Ka=? Solution Dissociation of HCOOH HCOOH → Ht + HCOO- [HA] C-x x' 0.10 = O + 0 initial stage 0.10-x=x + X Equilibrium stage As concentration of Ht ion and HCOO ion is same at equilibrium So, [HCOOH] Ka= (x)(x) 0.1 - x :x=4.17x103M Ka= (4.17× 103)(4.17×10-3) 0.1-4.17x10-3 1.74×10-3 - = 1.81×10- Ka = 0.09583 (b) Calculate the concentration of [HOt] in a 0.1 M solution of nitrous acid [HNOz), given that the acid dissociation constant Ka is 4 × 104. Ans. Ka = 4 x 104 pl= [H30]=? . Solution: [H30+]=VKxC (c) A vinegar sample is found to have 0.837 M CHCOOH. Its hydronium ion concentration is found to be 3.86 × 103 mol dm3. Calculate Ka for acetic acid. [CH3CO0H] = 0.837 M Ans. [HOt] = 3.86 × 10-3 mol dm3 Ka=? Solution: X C [H30]=/K Taking square on both sides [H,0+] = K C (3.86x10-3)2 K. = 0.837 1.489x10-5 K.= 0.837 Ka = 1.79 × 10-5 Answer COMMON ION EFFECT

Common Ion Effect Definition: The suppression of ionization of a weak electrolyte by adding a common ion to it is called common ion effect. Explanation Purification of sodium chloride: NaCl is purified by passing hydrogen chloride gas through brine (saturated solution of Nacl). Sodium chloride is fully ionized in the solution. Equilibrium constant expression for this process can be written as follows: NaC1(a) Na(ac) + C1(ag) [Na I[C(ag)] Hcl also ionizes in solution HC (g)H (ag) + C(aq) • On passing Hcl gas, concentration of Clion is increased. • Nacl crystallizes out of the solution to maintain the constant value of the equilibrium constant. This type of effect is called the common ion effect. • The addition of a common ion to the solution of a less soluble electrolyte suppresses its ionization and the concentration of unionized species increases, which may come out as a precipitate. Na (ag) + C(ag) NaC(s) More examples of common ion effect (i) Solubility of KClO3 The solubility of less soluble salts KCIO3 in water is suppressed by the addition of a more soluble salt Kcl by common ion effect. Kt is a common ion. The ionization of KCIO3 is suppressed and it settles down as precipitate. КСЮ 3(8) = KC (3) = (i) Solubility of H2S The dissociation of a weak acid H2S in water can be suppressed by the addition of stronger acid Hcl. Ht is a common ion. H¿S becomes less dissociated in acidic solution. In this way, low concentration of S2- ion is produced. H,S (g) = 2H (aq) + Sag) This low concentration of S? ions helps to do the precipitation of radicals of second group basic radicals during salt analysis. HC (aq) [Nacl] - K (a9) + CO 3(ag) K+ -(aq) + C(ag) → H (ag) + CLaq) (iii) Solubility of NH+Cl An addition of NH4Cl in NH3 solution suppresses the concentration of OH due to the presence from NH4Cl. Actually, NHaCl is a strong electrolyte. The of a large excess of NH4 combination of these two substances is used as a group reagent in third group basic radicals for salt analysis. NH, + H, (aq) NH 4(ag) + OH (ag) Common ion effect finds extensive applications in the qualitative analysis and the preparation of buffers.

Applications of common ion effect Purification of NaCl (i) NaCl is purified by passing hydrogen chloride gas through the saturated brine. Sodium chloride is fully ionized in the solution. H (ag) + Cl(ag) 1C (g) = On passing HCl gas, concentration of Cl ions is increased. Therefore, NaCl crystallizes out of the solution to maintain the constant value of equilibrium constant. (ii) Precipitation of KClO3 The solubility of a less soluble salt KClO3 in water is suppressed by the addition of more soluble salt KCl by common ion effect. K' is a common ion. The ionization of KClO3 is suppressed and it settles down as precipitate. KCIO 3(s) KC (8) F (iii) Precipitation of 2"° group basic radicals The dissociation of a weak acid H2S in water can be suppressed by the addition of stronger acid HCl. Ht is a common ion. H¿S becomes less dissociated in acidic solution. In this way, low concentration of S-2 is developed This low concentration of S2- ions helps to do the precipitation of radicals of second group basic radicals during salt analysis. (iv) Preparation of buffers Common ion effect has its extensive applications in the qualitative analysis and preparation of buffer solutions. BUFFER SOLUTION

Buffer Solutions Definition: The solutions which resist the change in their pH when a small amount of an acid or a base is added to them, are called buffer solutions. They have a specific constant value of pH and their pH values do not change on dilution and on keeping for a long time. Types of Buffer Buffer solutions are mostly prepared by mixing two substances. (i) Acidic Buffers • By mixing a weak acid and its salt with a strong base. Such solutions give acidic buffers with pH less than 7 Mixture of acetic acid and sodium acetate is one of the best examples of acidic butters. (ii) Basic Buffers • By mixing a weak base and its salt with a strong acid. Such solutions will give basic buffers with pH more than 7 • Mixture of NH4OH and NH4Cl is one of the best examples of basic buffers. Mechanism of action of Buffers • The example of an acidic buffer consisting of CH:COOH and CHCOONa. CH:COOH being a weak electrolyte undergoes very little dissociation. • When CH;COONa, which is a strong electrolyte, is added to CHCOOH solution, then the dissociation of CH COOH is suppressed, due to common ion effect of CHCOO CHCOOH (aq) + H20(0) CH, COO (ag) + H, (aq) CH, COON (ag) CH, COO(ag) + Na (ag) • If one goes on adding CH:COONa in CHCOOH solution, then the added concentrations of CH COO decrease the dissociation of CHCOOH and the pH of solution increases. The table 9.4 tells us how the pH value of a mixture of two compounds is maintained. Greater the concentration of acetic acid as compared to CHCOONa, lesser is the pH of solution. Table 9.4 Effect of addition of acetate ions on the pH of acetic acid solution ICH:C001 ICH: COOHI (mol dm3) (mol dm3) 0.10 0.00 0.10 0.05 0.10 0.10 0.10 0.15 • A buffer mentioned above is a large reservoir of CHCOOH and CHCOO components. • When an acid or H3O* ions are added to this buffer, they will react with CH COO to give back acetic acid and hence the pH of the solution will almost remain unchanged. The reason is that CH COOH being a weak acid will prefer to remain un-dissociated. • The buffer solution consisting of NH4Cl and NH4OH, can resist the change of pH and pOH, when acid or base is added from outside. When a base or OH ions are added in it, they will react with H3O* to give back H20 and the pH of the solution again will remain almost unchanged

Calculating the pH of a Buffer Consider a weak acid HA and its salt NaA with a strong base say NaOH. The reversible reactions for dissociation of HA and NaA are as follows: HA- =H* + A NaA- % Dissociation pH 2.89 1.3 4.44 0.036 0.018 4.74 0.012 4.92 = Na* + A pH of a buffer solution can be calculated by using Henderson equation as given below: [Acid] pH=pK, - log [Salt] and the numerator Interchanging denominator, the sign of log changes. [Salt] pH=pK -a + 1og [Acid] is called Henderson's The relationship equation. The equation shows that two factors evidently govern the pH of a buffer solution. • First the pKa of the acid used and the is the ratio of the second concentrations of the salt and the acid. • The best buffer is prepared by taking equal concentration of salt and acid. pH is controlled by pKa of the acid Example: For the buffer solution of acetic acid and sodium acetate, the pH can be calculated as follows. [CH, COOH|- (CH, COONa) e 111016 [CH, COONa] Then pH=pK. + log [CH,COOH] pH=pK, + log(L) So, pH=pK, + 0= pK pH=4.74 It means that the pH of this buffer is just equal to the pKa of the acid. Quick Check 9.4 (a) Explain the impact of common ion effect on solubility. Ans. The solubility of sparingly soluble salts decreases in the presence of common ions in a mixture. Common ion does not affect the solubility of completely soluble salt. Solubility of KCO3 The solubility of less soluble salts KCIO3 in water is suppressed by the addition of a more soluble salt KCl by common ion effect. K* is a common ion. The ionization of KCO3 is suppressed and it settles down as precipitate. KCIO3(s) KC (s)* (b) To a saturated solution of AgCl, some of Nacl solution was added. (i) State the effect of the concentration of Ag+ on the equilibrium. Interesting Information! S.Q. How buffer maintain pH of blood? Ans. The body's blood buffering system, involving bicarbonates helps maintain a stable pH the composition of injections are buffered to ensure that no change in blood's pH occur. - K (aq) + CO 3(a9) →K+ (ii) Explain your answer with respect to the common ion effect. Ans. (i) When NaCl is added to the saturated solution of AgCl, the concentration of Ag+ decreased or become low at equilibrium. - Na" Ag (s) F -Ag(aq) + C (aq) (ii) NaCl solution is strong electrolyte and it produces strong electrolyte Clions more rapidly. Due to common ion (Cl ion) equilibrium shift to the left. As a result, the concentration of Ag ions become lowered and the solubility of AgCl decreases. (c) How does a buffer maintain pH stability? Ans. A Butter maintain pH stability due to common ions. It maintains the pH and poh when acid or base is added from outside. For Example, buffer solution consisting of NH4C and NH4OH. When a base or OH ions are added, they will react with H3O* to give back H2O. The pH of solution will remain almost unchanged (d) Calculate the pH of a buffer consisting of 0.50 M HF and 0.45 M of a fluoride (F) salt before and after addition of 0.40 g NaOH to 1.0 dm? of the buffer (Ka of HF = 6.8 × 101). Ans. [Acid] = [HF] = 0:50 M [Salt] = [F] = 0.45 M Ka = 6.8 × 104 Mass of NaOH = 0.49 Molar Mass of NaOH = 40g mole 0.4 . = 0.01M Concentration of NaOH = 40 pH =? Solution: pKa = - log 6.8 × 104 = 3.17 [Salt] pH = pKa + log [Acid] 0.45 pH = 3.17 + log 0.50 pH = 3.17 - 0.046 = 3.12 When 0.01m of NaOH are added in Buffer, acid is consumed or concentration of acid is decreased while concentration of salt (F) is increased Now [HF] = 0.50 - 0.01 = 0.49M [F] = 0.45 + 0.01 = 0.46M New pH is [Salt] pH = pha + log [Acid] 0.46 pH =3.17 + 10g 0.49 pH = 3.17 - 0.028 pH = 3.14 (e) Calculate the pH of a buffer solution in which 0.11 molar CHCOONa and 0.09 molar acetic acid solutions are present. Ka for CH:COOH is 1.85 × 10°. Ans. [CH:COONa] = [Salt] = 0.11 M Ka = 1.85 × 10-5 pH=? Solution: Pka = -log (1.85 × 10-5) Pka = 4.74 [Salt] pH = pka + log [Acid] (0.11) pH = 4.74 + log (0.09) pH = 4.74 + 0.087 pH = 4.83 SOLUBILITY PRODUCT

Solubility Product Definition: The solubility product is the product of the concentrations of ions raised to an exponent equal to the co-efficient of the balanced equation. • The value of Ksp is a measure of the dissociation of sparingly soluble salt. • The dissociation is completed at equilibrium stage for Nacl when it is mixed in water. Because it is a completely soluble salt. • The dissociation is not complete at equilibrium stage in slightly soluble salts like PbCl2, PbSO4 etc. • Ksp is only for slightly or sparingly soluble salts. Slightly soluble ionic compounds (i) Lead sulfate (PbSO4): When it is shaken with water the solution contains Pb2t and undissociated PbSO4, it means that equilibrium exists between solid solute, PbSO and the dissolved ions, SOt and Pb?t PbSO 4(s) Being a sparingly soluble salt, the concentration of lead sulphate (PbSO4) almost remains constant. Bring [PbSO4] on left hand side (L.H.S) with Kc. If Ksp = [P][SO]=1.6x10- at 25°C Ksp is called the solubility product of PbSO4. It is the product of molar solubilities of two ions at equilibrium stage: +, SO4 (ag) + SO 3(ag) [POSOA] (ii) Lead sulphate (PbCl2): It is a well-known sparingly soluble compound and it dissociates to a very small extent like POCIe Effect of temperature Ksp is usually a very small quantity at room temperature. The value of Ksp is temperature dependent. It increases by increase in temperature. The following table 9.5 shows the Ksp values of some slightly soluble ionic compounds. Table 9.5 Ksp Values of some ionic compounds Salt Ion Product Ksp 5.0x10-13 [Ag+][Br] AgBr 1.8x10-10 AgCl [Ag ]CH] 3x10-34 AI(OH)3 [AHOH] 1.1×10-10 BаCO3 [Ba][ CO: ] 3.3×10-9 [Ca*][C03?] CaCO3 3.2×10-11 CaF2 Applications of solubility product (a) Determination of Solubility from Ksp • To calculate solubility from Ksp, the formula of the compound and Ksp Value is required. • The unknown molar solubility S is calculated and the concentration of the ions is determined Table 9.6 shows the relationship between the Ksp values and the solubility of some sparingly soluble compounds Table 9.6 The relation between solubility and solubility products of some salts Formula No. of ions 1.69 × 104 PbSO4 3 Ca(OH)2 6.5 × 10-6 CaF2 3.2 × 10-11 3 2.6 × 10-12 AgzCrO4 (b) Common Ion Effect • The presence of a common ion decreases the solubility of a slightly soluble ionic compound. • In order to explain it, consider a saturated solution of PbCrO4, which is a sparingly soluble ionic salt. PbCO 4(aq) Pb + Now add NazCrO4 which is a soluble salt. CrO4 is the common ion. It combines with Pb2+ to form more insoluble PbCrO4. So equilibrium is shifted to the left to keep Ksp constant. (c) Predicting Precipitation The solubility product can also help in predicting whether the precipitation of a salt will occur or not. Ion Product Salt Kop 8×10-34 CuS 1.4x10-85 Fe2S3 3.5x10-8 MgCO3 3×10-11 MnS [Mn2+][S2] 2.3×10-13 PbCr04 1.6x10-8 [Pb2+][SO4] PbSO4 Ksp Solubility gdm? 1.3 × 104 1.175 × 10-2 2.0 × 104 8.7 × 10-5 (aq) + Cr 4(aq) Example: The solubility product of CaSO4 is 2 x 10-5. If we add 10-2 mol dm solution Ca?+ to 10ª mol dm solution of SO4 ions at 25°C. The concentrations of each ionic species can be calculated as 10-2 - = 5.0×10- moldm3 [Ca?]= [SO,]=" 2 [Ca?][SO?] = 5.0×10-moldm × 5.0×103 moldm3 = 2.5x10-mol dm =2.5x105 mol dm > K. of CaSOs As the ionic product of concentrations is greater thanKsp, therefore CaSO4 will precipitate out. ionic Product Type of Solution Supersaturated > Ksp Saturated = Ksp Unsaturated < Ksp Quick Check 9.5 (a) The solubility product constant (Ksp) of silver chloride (AgCI) is 1.77x10-10 at 25°C. A solution already contains M of 0.10 M AgCl. Find the solubility of AgCl in a Ans. Ksp of AgCl = 1.77 × 10-10 solution in 010m darde achi ded 1m. Concentration of Nacl = 0.10 M Solubility of AgCI=S = ? Solution: Consider the solubility of Ag+' as 'S" AgCl == (S) (S) 0.10M Ksp = [Ag ][CI] As NaCl is strong electrolyte it ionizes, it Give 0.10M CI [CI] = 0.1 M Total [CI] from AgCl and NaCl [CH=S + 0.1 Due to Common ions, the solubility of AgCl is suppressed or 'S' is very small [CI] = S +0.1 ~ 0.1 [CI] = 0.1 M Now Ksp = [Ag ][CH] 1.77 × 10-10 = [S] [001] 1.77x10-10 0.1 S = 1.77 × 10-9 M So, solubility of AgCl is 1.77.x 10-9 M. Precipitation Yes No No (b) Predict whether CaSO4 will be precipitated or not when 10-3 mol dm3 of each of Cart and SO is mixed together. Ans. [Cat] = 10-3 mol dm-3 [SO: ] = 10-3 mol dm3 ionic product is [Ca] [SO:] = 10-mol dm ×10 mol dm = 10% moldm " Asp for CaSO4 = 2 × 10-5 mol dm ° of CaSO4 Hence, Kop nor dm < Rap of CasOn So, CaSO4 will not be precipitated out. (c) Solution of potassium carbonate is basic. (i) Explain why the solution is alkaline? (ii) Also give the equation for the hydrolysis of the above salt. Ans. (i) Solution of potassium carbonate in derived basic salt because it is from weak acid (H2CO3) and strong base (KOH). So, it is alkaline in nature. (ii) Hydrolysis of salt when it is dissolved in water it dissociates. CO, is conjugate base of H2O3 (weak acid and reacts with water to produce OH. → HCO, + OH Kt is Conjugate acid of KOH (strong base) and does not affect the ph. Result: The solution is basic due to CO% ion as it produces OH. Interesting Information! S.Q. What is the role of calcium carbonate in shell of Nautilus? Ans. The shell of this nautilus is composed mainly of calcium carbonate. The nautilus adjusts conditions so shell material is formed when the concentration of calcium ions and carbonate ions in seawater are high enough to precipitate calcium carbonate. SALT HYDROLYSIS

Salt hydrolysis Definition: The breakdown of a salt (made up of weak acid or base) by reacting with water, resulting in the bond breaking in that salt which changes the pH of water is called hydrolysis. Explanation When a salt dissolves in water, it dissociates into its constituent ions. These ions can interact with water, affecting the solution's pH depending on the nature of the acid and base from which the salt is derived. It is called salt hydrolysis. (i) Salts of strong acids and strong bases In salts of strong acids and strong bases like sodium chloride (NaCl). • The conjugate base of a strong acid (Ct from HCl) is very weak and does not significantly react with water. • The conjugate acid of a strong base (Na" from NaOH) is also very weak and does not significantly react with water. 1,0 Nat (ag) NaCC (s) F • Nat is the conjugate acid of NaOH (a strong base) and does not affect the ph. • Cl is the conjugate base of Hcl (a strong acid) and does not affect the ph The solution remains neutral. (ii) Salts of strong acids and weak bases In salts of strong acids and weak bases like ammonium chloride (NH4Cl). • The conjugate base of the strong acid (Cl from HCl) does not react with water. • The conjugate acid of the weak base (NH4* from NH3) reacts with water to produce H3O+ ions, making the solution acidic. NH,Cl (s) = NH4(aq) + CC • NH4* is the conjugate acid of NH3 (a weak base) and reacts with water: NH* (ag) + H2O,, NH 3(g) + H2O+ • Cl is the conjugate base of HCl (a strong acid) and does not affect the pH. The solution is acidic due to the ammonium (NH4*) ions. (iii) Salts of weak acids and strong bases In salts of strong of weak acid and strong base like sodium acetate (CHCOOH). CH:COONa(s) + H20(e) = CHCOO (aq) + Na* (aq) • CHCOO is the conjugate base of CHCOOH (a weak acid) and reacts with water: CHCOO (aq) + H20(c) -CHCOOH(aq) + OH (aq) • Na' is the conjugate acid of NaOH (a strong base) and does not affect the pH. The solution is basic due to the acetate (CHCOO) ions. (iv) Salts of weak acids and weak bases: In salts of weak acids and weak bases like ammonium acetate (CH,COO NH,). • The conjugate acid (NH4*) and the conjugate base (CH:COO) both affect the pH. CHCOO NH 4(ag) NH (ag) + CH3 COO (aq) • NH4* hydrolyses to produce H3O+ NH; + H2O==NH, + H2O* • CHCOO* hydrolysis to produce OH. The resultant pH of the solution depends on the relative strengths of the conjugate acid and conjugate base. The solution may be acidic, basic, or nearly neutral, depending on which reaction is more dominant. ACID-BASE INDICATORS

Acid base indicators Definition: An indicator is a substance that changes colour to mark a titration's endpoint. • Acid-base indicators exhibit one colour in acid and another in base. • Most indicators used in acid-base titration are weak organic acids or weak organic bases. Acid solution: In solution, a weak-acid indicator (HIn) can be represented by the equation below. = H + In HIn F In is the symbol of the anion part of the indicator. Because the reaction is reversible, both HIn and In are present. The colours displayed result from the fact that HIn and In have different colours. Acidic solutions: In acidic solutions, any In ions that are present act as Bronsted bases and accept protons from the acid. The indicator is then present in largely unionized form, HIn. Basic Solutions: In basic solutions, the OH ions form the base combine with the H* ions produced by the indicator. • The indicator molecules further ionize to equalize the loss of H* ions. • The indicator is thus present largely in the form of its anion, In. The solution now show the base-indicating colour, which for litmus is blue. In basic solution In In acidic solution Fig: Unionized and ionized acid-base indicators and their ionised forms exist in an equilibrium Colour of indicators Different indicator change colour at different pH values. The colour depends on the relative amount of HIn and In at a give pH Example: Methyl red changes from red to yellow between pH 4.4 and 6.2. At pH 4.4, the indicator exists mostly as HIn molecules, which appear red in the solution. The indicator ranges are given below for some of the indicators commonly used. Orange Methyl orange Colourless Phenalphthalein 4 3 1 2 Fig: Range and colour changes of some common Acid-Base indicators Selecting a suitable indicator The two general criteria for an indicator to be used in a titration are: • The pH at the end of the titration should be close to the indicator's neutral point. • The indicator should indicate a sharp colour change near the equivalence point of the titration. Each pH indicator changes colour over a defined range of pH, known as the indicator range. An indicator changes colour over a range of about 2 pH units.

Titration curve and Equivalence point Definition of titration curve: A pH curve is a graph of the pH of the solution verses the volume of titrant added Definition of equivalence point: The equivalence point is the point at which the amount of titrant added is just enough to neutralize the analyte solution completely. Explanation of titration curve • Titration curves show how the pH of an acidic or basic solution changes as a basic or acidic solution is added to it. • The titration curve to choose an indicator that will show when the titration is complete and we reach the equivalence point. In + Yellow 11 13 8 5 7 10 14 12 6 • The end point of the titration occurs when the indicator changes colour • We choose and indicator with an end point close to the equivalence point. (i) Strong Acid-strong base titration curve The titration of HCl with NaOH is an example of strong acid-strong base titration. • The solution contains only the strong acid at the start. • As the acid is strong, it completely a to dissociates leading high concentration dissociates, leading to a high concentration of Hions and the pHE is very low • By the addition of NaOH, OH ions from NaOH begin to neutralize the Ht ions from HCl. and pH of the solution rises. • The equivalence point is reached when the amount of OH added is stoichiometrically equal to the amount of H* originally present in the acid • The pH at the equivalence point in a strong acid-strong base titration is 7 • The titration curve at this stage shows a steep rise in pH, changing quickly form acidic to neutral • The pH value at the endpoint changes about from 4.0 to 10.0. Conclusion A pH of 7.0 is at equivalence, so, such phenolphthalein as indicator can be used as they show different colours in this range. (ii) Strong acid-weak base titration curve The titration of NH3 with HCl is an example of strong acid-weak base titration. (a)Initial stage • The solution contains only the weak base at start, dissociates partially, with a lower concentration of OH ions compared to a strong base. • The initial pH will be greater than 7 but lower than the pH of a strong base. • By the addition of HCl, the concentration of OH decreases, results in further decrease in 14- 12- • phenolphthalein 10- 8 - pH = 7.00 at equivalence point 6 5 methyl red 2 0 10 20 30 70 40 50 60 80 Volume of strong base added (mL) Fig: Strong acid-strong base titration curve 14- 12- 10• phenolphthalein buffer region 8 - 6 - methyl red 5 pH = 5.27 at equivalence point 2 0 30 20 40 10 60 70 0- Volume of strong base added (mL) Fig: Strong acid-weak base titration curve (b) Change in pH • Before reaching the equivalence point the solution is in a buffer-like region where the pH changes more gradually. • The presence of the weak base and its conjugate acid (from the salt formed) creates a buffering effect, which helps to moderate the pH change as the base is added. (c) Equivalence point The pH at the equivalence point will be less than 7 (pH = 5.27) because the conjugate acid slightly dissociates, releasing H* ions making the solution acidic (NH4* → NH3 + H*). (d) Titration curve • After the equivalence point, the pH of the solution decreases rapidly. Because the strong acid dissociates completely in water, providing a high concentration of hydrogen ions (HT), which significantly lowers the pH • Methyl orange has its colour change in this range, therefore, it can be used as an indicator for strong acid-weak base titrations. Table 9.7 pH ranges of common indicators Acid Indicators Base colour pH Range color Yellow Orange vi Methyl orange red colorless Phenolpthalein Quick Check 9.6 (a) Differentiate end point and equivalent point. Ans. End point End point is the point at which titration is complete and indicator changes color. End point is usually determined experimentally. (b) Explain how an indicator changes its colour in acidic and basic solution. Ans. Acid base indicator change colour due to shift in equilibrium their acidic and basic form which have different colour. When indicator is added to a solution it can exist in either acidic or basic form, causing the indicator display its colour. (c) Suggest a suitable indicator for weak acid and strong base titration. Ans. A suitable indicator for weak acid and strong base titration is phenolphthalein. (d) Suggest a suitable indicator for weak acid and weak base titration. Ans. A suitable indicator for weak acid and weak base titration is Bromothymol Blue. Type of Titration Strong acid-strong base 3.2-4.5 Strong acid-weak base 8.2-10.0 Weak acid-strong base Equivalence point Equivalence point is the point at which the amount of titrant is added is just enough to neutralize the analyte solution completely. Equivalent point is usually determined theoretically SAMPLE PROBLEMS Sample Problem 9.1 In a solution, the pH is 9.2. Determine the ionic product of water Kw at 25°C. Solution Calculate the pOH from the pH: = 14 pH+pOH = 14 - 9.2 pOH = 4.8 Find the concentration of Ht and OH, Taking antilog of pH ad pOH Antilog pH = antilog (9.2 H+ = 10-pH [Ht] = 10-92 [Ht] = 66.3 ×10-10 M Similarly, antilog pOH = antilog (4.8) [OH-] = 10-POH, [0] = 104.8 [0]=6.3×10-10 M Kw = (6.3 × 10- Kv = 1.01 × 1012(1.6 × 10-5) Sample Problem 9.2 The ionic product of water at a certain temperature is Kw = 1.0 x 10-14 at 25°C. If the concentration of Ht ions in a solution is 1.0 × 10- M. Calculate the concentration of OH ions, the pH and pOH of the solution. Solution Kw = [H][0H] 1.0 × 10-14 = (1.0 × 10-7) [OH ] 1.0x10-14 [OH ]= 1.0x10-=10x10- pH=- 10g[H*] pH = - log(1.0 × 10-7) = 7 pOH = - log(0H ] pOH = - log(1.0×10-7) pOH = 7 Calculate Kw Kw = (6.3 × 10) x (1.6 × 10-5) Kw = 1.008 × 10-14 So the ionic product of water at given temperature is = 1.01 × 10-14 Sample Problem 9.3 Calculate the concentration of [HOt] in a 0.1 M (mol dm ) solution of acetic acid [CHCOOH], given that the acid dissociation constant Ka is 1.8 × 10-5 Solution [H2O+]=/K.×C [H2O*] = v1.8x 10-x 0.1 [HOt] = V0.18×10-6 [HO*]=1.34×103M (moldm3) The concentration [H3O*] in the solution is approximately 1.34 × 10-3 mol dm3 Sample Problem 9.4 Ca(OH)2 is a sparingly soluble compound. Its solubility product is 6.5 × 106 solubility of Ca(OH)2. Solution Let the solubility be represented by S in terms of mol dm3 The balanced equation is: Ca(OH)2(aq) 0+.0 Initial stage Ca(OH)2(aq) = S+ S Equilibrium stage The concentration of OH is double than the concentration of Cat K$=[CaIOH] = Sx(25)' 453 = 6.5×10° S'= 6.5x10° 4 . S= (1.625)13x-102 S=1.18×102 moldm3 Hence, at equilibrium stage 1.18 × 10-2 mol dm3 of Cat and 2 × 1.18 × 10-2 mol dm3 = 2.36 × 10-2 mol dm3 OH are present in the solution. In this way, we have calculated the individual concentrations of Ca2+ and OH ions from the solubility product of Cal CalOH)2. •. Calculate the (aq) (ag) + 20H