MASTERING IONIC EQUILLIBRIA. By : S.K.Sinha A CENTRE FOR CONCEPTUAL & SYSTEMATIC STUDY

MASTERING IONIC EQUILLIBRIA By : S.K.Sinha A CENTRE FOR CONCEPTUAL & SYSTEMATIC STUDY

MASTERING PHYSICAL CHEMISTRY ACIDS AND BASES The Nature of Acids and Bases Arrhenius concept: Acid a substance that produces hydrogen ion in aqueous solution. Base a substance that produces hydroxide ion in aqueous solution. Examples of Acids: 1. HCl (aq) H (aq) Cl (aq); (a strong acid) 2. HNO 3(aq) H (aq) NO 3 (aq) ; (a strong acid) 3. CH 3 COOH (aq) H (aq) CH 3 COO (aq); (a weak acid) Examples of bases: 1. NaOH (aq) Na (aq) OH (aq); (a strong base) 2. Ba(OH) 2(aq) Ba 2 (aq) 2OH (aq); (a moderately strong base) 8. NH 3(aq) H 2 O NH 4 (aq) OH (aq); (a weak base) By S.K.SINHA 1 * All hydroxides and oxides of metal are bases. (Oxides of metals in veryhigh O.N. is acidic. Ex CrO 3, Mn 2 O 7. The BrønstedLowry concept: Acid a substance that acts as a proton donor in chemical reaction; Base a substance that acts as a proton acceptor in chemical reaction; * Reactions that involve the transfer of protons (H ) are acidbase reactions. The BronstedLowry acidbase reaction can be represented as follows: HA B BH A acid 1 base 2 conjugate conjugate acid 2 base 1 For examples: 1. HCl H 2 O H 3 O (aq) Cl (aq) acid 1 base 2 conjugate conjugate acid 2 base 1 2. HC 2 H 3 O 2 H 2 O H 3 O (aq) C 2 H 3 O 2 (aq) ; acid 1 base 2 conjugate conjugate acid 2 base 1 3. NH 3 H 2 O NH 4 (aq) OH (aq); base 1 acid 2 conjugate conjugate acid 1 base 2 In each reaction, a conjugate base is what remains of the acid molecule after it loses a proton (H ), and a conjugate acid is what becomes of the base after it gains a proton. acid 1 conjugate base 1 and acid 2 conjugate base 2 are called conjugate acidbase pairs; they are a pair of species that are related to each other by the loss or gain of a single

2 MASTERING PHYSICAL CHEMISTRY By S.K.SINHA proton (H ). Thus, H 2 O and H 3 O are conjugate acidbase pair; H 3 O is called hydronium ion. A BrØnstedLowry acidbase reaction involves a competition between two bases for a proton, in which the stronger base ends up being the most protonated at equilibrium. In the reaction: HCl H 2 O H 3 O (aq) Cl (aq), H 2 O is a stronger base than Cl ; at equilibrium, a solution of hydrochloric acid contains mostly H 3 O and Cl ions. In reaction: HC 2 H 3 O 2 H 2 O H 3 O (aq) C 2 H 3 O 2 (aq) ; C 2 H 3 O 2 is the stronger base; at equilibrium, a solution of acetic acid contains mostly HC 2 H 3 O 2 and a small amount of H 3 O and C 2 H 3 O 2 ions. In the reaction: NH 3 H 2 O NH 4 (aq) OH (aq), water is an acid. The competition for protons is between NH 3 and OH, in which OH is the strong base; the above equilibrium shifts to the left. That is, an aqueous ammonia solution contains mostly NH 3 molecule and a small amount of NH 4 OH, NH 4, and OH. Exercises: 1. Write the conjugate base for each of the following acids: (a) H 2 PO 4 ; (b) H 2 C 2 O 4 ; (c) [Al(H 2 O) 6 3 ; (d) NH 3 ; 2. Write the conjugate acid for each of the following bases: (a) NH 3 (b) [Al(H 2 O) 2 (OH ) 4 (c) SO 3 2 The Relative Strength of Acids (d) (CH 3 ) 3 N: The fects of Structure of AcidBase Properties The strength of acids is determined by a combination of various factors, such as the strength and polarity of XH bond in the molecule, and the hydration energy of the ionic species in aqueous solution. For binary acids such as HF, HCl, HBr, and HI, HX bond strength decreases down the group. The weaker the bond, the easier the molecule ionizes in aqueous solution. Hence, the stronger will be the acid. Thus, for this group of acids, their strength increases down the group: HF < HCl < HBr < HI; H 2 O < H 2 S < H 2 Se < H 2 Te Among the hydrohalic acids, HF is the only weak acid; the others are strong acids. The relative strength of HCl, HBr, and HI cannot be differentiated in aqueous solution, because each of them dissociates almost completely. Less polar solvents are used to determine their relative strength. For example, HCl, HBr and HI ionize only partially in acetone or methanol, which have a weaker ionizing strength than water. The ionization of HCl in acetone can be represented by the following equilibrium: (CH 3 ) 2 CO (l) HCl (acetone) (CH 3 ) 2 COH (acetone) Cl (acetone) The degree of ionization in acetone increases in the order of HCl < HBr < HI. Based on this observation, it was determined that the acidity of hydrogen halides increases down the

MASTERING PHYSICAL CHEMISTRY By S.K.SINHA group: HF < HCl < HBr < HI. A similar trend of relative acidity is also observed for the hydrides of Group VIA elements, such that, H 2 O < H 2 S < H 2 Se < H 2 Te. For the same period hydrides, the relative acidity increases from left to right, such that: 3 CH 4 < NH 3 < H 2 O < HF; PH 3 < H 2 S << HCl Water is a stronger acid than ammonia, and in acidbase reaction, H 2 O acts as a BrØnsted Lowry acid, which donates a proton to NH 3 : H 2 O (l) NH 3(aq) NH 4 (aq) OH (aq) In reaction with HF, water acts as a BrØnstedLowry base, which accept a proton: HF (aq) H 2 O (l) H 3 O (aq) F (aq) The Relative Strength of Oxoacids Oxoacids are acids that contain one or more OH groups covalently bonded to a central atom, which can be a metal or a nonmetal. The OH group ionizes completely or partially in aqueous solution producing hydrogen ions. Some common examples of oxoacids are H 2 CO 3, HNO 3, H 3 PO 4, H 2 SO 4, HClO 4, and HC 2 H 3 O 2 : O O O HOSOH HOPOH OClOH O O O H For these type of acids, their relative strengths depend on the electronegativity of the central atoms. The more electronegative the central atom, the more polarized the OH bond and the more readily it ionizes in aqueous solution to release the H ion. For example, N, S, and Cl are more electronegative than P, and HNO 3, H 2 SO 4, and HClO 4 are stronger acids compared to H 3 PO 4. The relative acid strengths are: HNO 3 > H 2 SO 4 > H 3 PO 4. For oxoacids with central atoms from the same group, their relative strength decreases from top to bottom as the electronegativity of the central atom decreases: HOCl > HOBr > HOI; HOClO > HOBrO > HOIO; HOClO 2 > HOBrO 2 > HOIO 2 For oxoacids containing identical central atom, acidity increases as more oxygen atoms are bonded to it. For example, acidity increases in the following order: HOCl < HOClO < HOClO 2 < HOClO 3 ; H 2 SO 4 > H 2 SO 3 ; HNO 3 > HNO 2 Oxygen is a very electronegative atom; when more oxygen atoms are bonded to the central atom, the OH group in the molecule becomes more polarized (due to inductive effect), and the more readily it ionizes to release H ion. Acetic acid is an organic acid, which contains the acidic carboxyl group, COOH. When acetic acid (CH 3 COOH) ionizes, only the OH bond of the carboxyl group breaks, but not the CH bonds. However, if one or more of the hydrogen atoms in the methyl group ( CH 3 ) is substituted with a more electronegative atom, the OH of COOH group becomes more polarized and ionizes more readily, and increasing the acidity. The following K a values illustrate the effect on the acidity of COOH when one or more of the methyl hydrogen is substituted with more electronegative atoms: CH 3 COOH (aq) CH 3 COO (aq) H (aq); K a = 1.8 x 10 5

4 MASTERING PHYSICAL CHEMISTRY By S.K.SINHA ClCH 2 COOH (aq) ClCH 2 COO (aq) H (aq); K a = 1.4 x 10 3 chloroacetic acid FCH 2 COOH (aq) FCH 2 COO (aq) H (aq); K a = 2.6 x 10 3 fluoroacetic acid CCl 3 COOH (aq) CCl 3 COO (aq) H (aq); K a = 3.0 x 10 1 trichloroacetic acid Exercise: 1. Rank the following acids in order of increasing strength: (a) CH 3 COOH, CH 3 CH 2 COOH, HCOOH; (b) CH 3 COOH, ClCH 2 COOH, CHCl 2 COOH, CCl 3 COOH, CF 3 COOH 2. Rank the following bases in order of increasing strength: (a) NH 3, CH 3 NH 2, (CH 3 ) 2 NH 2, (CH 3 ) 3 N, (CH 3 CH 2 ) 3 N AcidBase Properties of Oxides Metal oxides are either basic or have amphoteric properties. The oxides the Group IA metals are strongly basic because they are generally very soluble in water. Other metal oxides are less soluble, but they react with strong acids: Na 2 O (s) H 2 O (l) 2NaOH (aq) ; BaO (s) H 2 O (l) Ba(OH) 2(aq) MgO (s) 2HCl (aq) MgCl 2(aq) H 2 O (l) The oxide ion, O 2, has a very strong affinity for protons and reacts with water to produce hydroxide ions: O 2 (aq) H 2 O (l) 2OH (aq) Some metal oxides are amphoteric; for example, Al 2 O 3, Cr 2 O 3, PbO, SnO, and ZnO. They react with both strong acids and strong bases: Al 2 O 3(s) 6HCl (aq) 2AlCl 3(aq) 3H 2 O (l) ; Al 2 O 3(s) 2NaOH (aq) 3H 2 O (l) 2Na[Al(OH) 4 (aq) ; Oxides of nonmetals are acidic. They form acidic solution when dissolved in water. CO 2(g) H 2 O (l) H 2 CO 3(aq) H (aq) HCO 3 (aq) ; SO 2(g) H 2 O (l) H 2 SO 3(aq) H (aq) HSO 3 (aq) ; P 4 O 10(g) 6H 2 O (l) 4H 3 PO 4(aq) ; H 3 PO 4(aq) H (aq) H 2 PO 4 (aq) ; Metal Hydroxides and Hydrides Hydroxides of Group IA metals are strongly basic. For example, LiOH, NaOH, and KOH, are strong bases. The basicity of other metal hydroxides is limited by their low solubility. The hydroxides of some metals exhibit amphoteric properties. For example, aluminum hydroxide, Al(OH) 3 is an amphoteric hydroxides. Al(OH) 3(s) OH (aq) Al(OH) 4 (aq) ; Al(OH) 3(s) 3H 3 O (aq) [Al(H 2 O) 6 3 (aq) Cr(OH) 3, Zn(OH) 2, Sn(OH) 2, and Pb(OH) 2 are also amphoteric hydroxides.

MASTERING PHYSICAL CHEMISTRY By S.K.SINHA The hydrides of reactive metals, such as NaH and CaH 2, are strong bases. The hydride ion has a strong affinity for protons and reacts with water to produce hydroxide ions and hydrogen gas: H (aq) H 2 O (l) H 2(g) OH (aq) The Lewis Acids and Bases According to G.N. Lewis, an acid is the reactant that is capable of sharing a pair of electrons from another reactant to form a covalent bond; a base is the reactant that provides the pair of electrons to be shared to form a covalent bond. Based on the Lewis s definition, hydrogen ion (H ) is considered a Lewis acid and water and ammonia are the Lewis bases in the following reactions: H H 2 O H 3 O ; H NH 3 NH 4 ; Lewis Lewis Lewis Lewis acid base acid base Species with incomplete octet (or electron deficient species) may act as Lewis acid and those with lone pair of electrons may act as Lewis bases. For example, in the following reactions, BF 3, AlCl 3, and FeBr 3 are Lewis acids; while NH 3, Cl, and Br are Lewis bases. BF 3 NH 3 F 3 B:NH 3 ; AlCl 3 Cl AlCl 4 ; FeBr 3 Br FeBr 4 ; In the formation of complex ions, the positive ions act as Lewis acids and the ligands (anions or small molecules) are Lewis bases: Cu 2 (aq) 4NH 3(aq) Cu(NH 3 ) 2 4 (aq); Lewis acid Lewis base Al 3 (aq) 6H 2 O (l) [Al(H 2 O) 6 3 (aq); Exercise 1. Determine the Lewis acids and Lewis bases in the following reactions: (a) CO 2(g) H 2 O (l) H 2 CO 3(aq) (b) SO 3(g) H 2 O (l) H 2 SO 4(aq) (c) AlCl 3 (CH 3 ) 3 N: (CH 3 ) 3 N:AlCl 3 5 (d) Zn(OH) 2(s) 2OH (aq) Zn(OH) 4 2 (aq) ; (e) AuCl 3 (s) Cl (aq) AuCl 4 (aq) Acids Strength The strength of an acid is defined by the extent of its dissociation (ionization) in aqueous solution. HA (aq) H 2 O H 3 O (aq) A (aq) Strong acids, such as HClO 4, HCl, HNO 3, and H 2 SO 4, are those that are considered to completely ionize at 1 M concentration. The above equilibrium would lie far to the right. Acids that are only partially ionized at this concentration, such as HC 2 H 3 O 2, HF, HNO 2, H 2 SO 3, H 3 PO 4, and HClO, are considered weak acids. Their dissociation equilibria would lie far to the left. The equilibrium constant, K a, for the acid ionization is given by the following expression: [H 3 O [A [H [A K a = = [HA [HA

6 MASTERING PHYSICAL CHEMISTRY By S.K.SINHA For strong acids, K a would have a very large value; for weak acids, K a < 10 1. The following are the K a values of some weak acids: 1. HF (aq) H (aq) F (aq); K a = [H [F = 7.2 x 10 4 [HF 2. HNO 2(aq) H (aq) NO 2 (aq) ; K a = [H [NO 2 = 4.0 x 10 4 [HNO 2 3. HC 2 H 3 O 2(aq) H (aq) C 2 H 3 O 2 (aq) ; K a = [H [ C 2 H 3 O 2 = 1.8 x 10 5 [HC 2 H 3 O 2 4. HOCl (aq) H (aq) ClO (aq); K a = [H [ClO = 3.5 x 10 8 [HOCl 5. HCN (aq) H (aq) CN (aq); K a = [H [CN = 6.2 x 10 10 [HCN The value of K a is a measure of the extent of acid dissociation; hence, the strength of the acid. A stronger acids have larger K a values. For strong acids such as HClO 4, HCl, H 2 SO 4, and HNO 3, their K a s are very large (not given in any tables of K a values). The Conjugate AcidBase Relationships Strong acids have weak conjugate bases, and weak acids have strong conjugate bases; the weaker the acid, the stronger will be its conjugate base. Likewise, weak bases have strong conjugate acids; the weaker the base, the stronger will be its conjugate acid. Examples: HCl is a strong acid and Cl is a very weak base; HF is a weak acid, and F is a stronger conjugate base than Cl. According to BrØnstedLowry, acidbase reactions favor the direction from strong acidstrong base combinations to weak acidweak base combinations: The following acidbase reactions go completely in the forward direction: 1. HClO 4(aq) H 2 O (l) H 3 O (aq) ClO 4 (aq) ; (HClO 4 is a very strong acid) 2. HCl (aq) NH 3(aq) NH 4 (aq) Cl (aq); (HCl is a strong acid) 3. HSO 4 (aq) CN (aq) HCN (aq) SO 4 2 (aq) ; (HSO 4 is a stronger than HCN) Many acidbase reactions reach a state of equilibrium. The following acidbase reactions do not go to completion; their equilibrium positions may shift to the right or to the left, depending on the relative strength of the acid: 1. H 2 PO 4 (aq) C 2 H 3 O 2 (aq) HC 2 H 3 O 2(aq) HPO 4 2 (aq) ; Equilibrium shifts left; HC 2 H 3 O 2 is the stronger acid and HPO 4 2 is the stronger base. 2. HNO 2(aq) C 2 H 3 O 2 (aq) HC 2 H 3 O 2(aq) NO 2 (aq) ; Equilibrium shifts right; HNO 2 is the stronger acid, C 2 H 3 O 2 is the stronger base. Water as an Acid and a Base Water is amphoteric because it can behave as an acid and a base. The following equilibrium occurs in pure water: H 2 O H 2 O H 3 O (aq) OH (aq); Acid 1 base 1 conjugate conjugate acid 2 base 1

MASTERING PHYSICAL CHEMISTRY By S.K.SINHA The above equilibrium can be simplified as follows, which is also called the autoionization of water: H 2 O H (aq) OH (aq); K w = [H [OH = 1.0 x 10 14 at 25 o C. K w is called the ion product constant for water. An aqueous solution contains both H (in the form of H 3 O ) and OH, such that the product of their concentrations is constant equal to 1.0 x 10 14 at 25 o C. In pure water (or in a neutral solution, [H = [OH = 1.0 x 10 7 M. If [H increases (> 1.0 x 10 7 M), [OH will decreases (< 1.0 x 10 7 M), and vice versa. Thus, a solution with: [H > 1.0 x 10 7 M, or [H > [OH, the solution is acidic; [H < 1.0 x 10 7 M, or [H < [OH, the solution is basic; [H = [OH = 1.0 x 10 7 M, the solution is neutral; Exercise2: 1. What is [OH if [H = 0.0050 M? Is the solution acidic, basic or neutral? 2. What is the [H if [OH = 6.0 x 10 4 M? Is the solution acidic, basic or neutral? The ph Scale ph is a scale that measures the acidity of an aqueous solution where [H is very small. ph = log[h, If [H = 1.0 x 10 7 M, ph = log(1.0 x 10 7 ) = (7.00) = 7.00 We can also express basicity using the log scale for [OH, such that poh = log[oh That is, if a solution has [OH = 1.0 x 10 7 M, poh = log(1.0 x 10 7 ) = (7.00) = 7.00 Since, at 25 o C, K w = [H [OH = 1.0 x 10 14 pk w = log K w = log[h (log[oh ) = log(1.0 x 10 14 ) = (14.00) = 14.00 pk w = ph poh = 14.00; and poh = 14.00 ph Thus, in aqueous solutions, if ph > 7.00 poh < 7.00, and vice versa; In neutral solution, [H = [OH = 1.0 x 10 7 M, and ph = poh = 7.00; [H > 1.0 x 10 7 M, ph < 7.00, the solution is acidic. [H < 1.0 x 10 7 M, ph > 7.00, the solution is basic.. Calculating the ph of Strong Acid and Strong Base Solutions Strong acids are assumed to ionize completely in aqueous solution. For monoprotic acids (that is, those acids with a single ionizable hydrogen atom), the concentration of hydrogen ion in solution is the same as the molar concentration of the acid. That is [H = [HA For example, in 0.10 M HCl(aq), [H = 0.10 M, and ph = log(0.10) = 1.00. A strong base such as NaOH, contains OH which concentration is equal to that of dissolved NaOH. That is, a solution of 0.10 M NaOH(aq) contains [OH = 0.10 M poh = log[oh = log(0.10) = 1.00; ph = 14.00 1.00 = 13.00 7

8 MASTERING PHYSICAL CHEMISTRY By S.K.SINHA A strong base such as Ba(OH) 2 produces twice the concentration of OH as the concentration of Ba(OH) 2 in solution. Ba(OH) 2 dissociates as follows: Ba(OH) 2(aq) Ba 2 (aq) 2OH (aq); [OH = 2 x [Ba(OH) 2 In a solution of 0.010 M Ba(OH) 2, [OH = 0.020 M, poh = 1.70, and ph = 12.30 Exercise3: 1. Calculate the ph of the following solutions: (a) 0.0050 M HCl; (b) 0.00060 M NaOH. 2. What is [H and [OH, respectively, in a solution where, (a) ph = 4.50; (b) poh = 5.40 3. The ph of an HCl solution is found to be 3.00. To what final volume must a 100.mL sample of this acid be diluted so that the ph of the solution becomes 3.50? (316 ml) 4. What is the ph of a saturated aqueous solution of Ca(OH) 2 in which 0.165 g of Ca(OH) 2 is dissolved in 100. ml solution at 25 o C? (Answer: ph = 11.65). Calculating the ph of Weak Acid Solutions Weak acids do not ionize completely. At equilibrium, [H is much less than the concentration of the acid. The concentration of H in a weak acid solution depends on the initial acid concentration and the K a of the acid. To determine [H of a weak acid we can set up the ICE table as follows: Consider a solution of 0.10 M acetic acid and its ionization products: Concentration: HC 3 H 3 O 2(aq) H (aq) C 2 H 3 O 2 (aq) Initial [, M: 0.10 0.00 0.00 Change, [, M: x x x Equilibrium [, M : (0.10 x) x x The acid ionization constant, K a, is given by the expression: K a = [H [C 2 H 3 O 2 = x 2 = 1.8 x 10 5 [HC 2 H 3 O 2 (0.10 x) Since K a << 0.10, we can use the approximation that x << 0.10, and (0.10 x) ~ 0.10 Then, K a = x 2 ~ x 2 /0.10 = 1.8 x 10 5 ; (0.10 x) x 2 = (0.10)(1.8 x 10 5 ); x = (1.8 x 10 6 ) = 1.3 x 10 3 Note that, x = [H = 1.3 x 10 3 M; ph = log(1.3 x 10 3 ) = 2.89 Percent Dissociation follows: The percent dissociation (or degree of ionization) of a weak acid is defined as Percent dissociation = Amount of acid that dissociate (mol/l) Initial concentration of acid (mol/l)

MASTERING PHYSICAL CHEMISTRY By S.K.SINHA For strong acids, the percent dissociation at equilibrium is almost 100%. For weak acids, the percent dissociation depends on K a of the acid and their initial concentration. For example, the percent dissociation of acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 5 ) at 0.10 M concentration is, (1.3 x 10 3 M/0.10 M) x 100% = 1.3 % K a or pk a is a measure of the acid strength. The stronger the acid, the more of it ionizes, and the greater its K a (or the lower its pk a ). The percent dissociation of a weak acid then depends on its K a and also on the degree of dilution. In the more dilute acid, a higher percentage of the molecules will ionize. Now consider a solution of 0.010 M acetic acid and its ionization products: Concentration: HC 3 H 3 O 2(aq) H (aq) C 2 H 3 O 2 (aq) Initial [, M: 0.010 0.00 0.00 Change, [, M: x x x Equilibrium [, M : (0.010 x) x x The acid ionization constant, K a, is given by the expression: K a = [H [C 2 H 3 O 2 = x 2 = 1.8 x 10 5 [HC 2 H 3 O 2 (0.010 x) K a << 0.010, we can still use the approximation that x << 0.010, and (0.010 x) ~ 0.010 Then, K a = x 2 ~ x 2 /0.010 = 1.8 x 10 5 ; (0.010 x) x 2 = (0.010)(1.8 x 10 5 ); x = (1.8 x 10 7 ) = 4.2 x 10 4 Note that, x = [H = 4.2 x 10 4 M; ph = log(4.2 x 10 4 ) = 3.37 The degree of ionization of acetic acid at this concentration is, 4.2 x 10 4 M x 100% = 4.2 % 0.010 M Note that, the degree of ionization of the acid is higher in the more dilute acid solution. In fact, in 1.0 M acetic acid, the percent dissociation is only 0.42%, which is a 10fold lower than in 0.010 M acid solution. Exercise: 1. What is [H in 0.10 M HNO 2 solution and the ph of the solution? K a = 4.0 x 10 4? (Answer: [H = = 6.3 x 10 3 M; ph = 2.20) 2. A 0.10 M aqueous solution of lactic acid, HC 3 H 5 O 3, has a ph = 2.43. Determine [H in this solution and the K a of lactic acid. (Answer: [H = 3.7 x 10 3 M; K a = 1.4 x 10 4 ) 3. If a 0.10 M benzoic acid, HC 7 H 5 O 2 is 2.5% ionized, what is [H and ph of 0.10 M benzoic acid? Calculate the K a of benzoic acid. (Answer: [H = 2.5 x 10 3 M; ph = 2.60; K a = 6.4 x 10 5 ) Bases Strong Bases: Hydroxides of Group IA metals (LiOH, NaOH, KOH, RbOH, and CsOH) are strong bases, but only NaOH and KOH are commercially important and are common 9

10 MASTERING PHYSICAL CHEMISTRY By S.K.SINHA laboratory bases, since the others are quite expensive. These bases are quite soluble in water and they completely dissociate in aqueous solution, producing a high concentration of hydroxide ions. A moderately dilute solution of NaOH contains [OH = Initial [NaOH. NaOH (s) Na (aq) OH (aq) A 0.10 M solution of NaOH contains [OH = 0.10 M. Among the hydroxides of the alkaline earth metals, Ca(OH) 2, Sr(OH) 2, and Ba(OH) 2 are also relatively strong bases. When completely dissolved, these bases produce two moles of OH per mole of the base: Ba(OH) 2(aq) Ba 2 (aq) 2 OH (aq) A 0.10 M Ba(OH) 2 solution contains 0.20 M of OH ions. The hydroxides of other metals are only slightly soluble. Even in a saturated solution, the concentration of OH is still quite low. For example, a saturated solution of Mg(OH) 2 has [OH ~ 3.1 x 10 4 M. Weak Bases Ammonia is the only weak base that is of commercial importance. Ammonia is very soluble in water and ionizes as follows: NH 3(aq) H 2 O (l) NH 4 (aq) OH (aq) The base dissociation constant, K b, is given by the expression: K b = [OH 4 [ NH 3 [NH = 1.8 x 10 5 The concentration of OH of a weak base solution, such as NH 3 (aq), depends on its K b and the initial concentration of the base. For example, to determine [OH, poh and ph of 0.10 M NH 3 (aq), we can set up the following ICE table: Concentration: NH 3(aq) H 2 O (l) NH 4 (aq) OH (aq) Initial [, M: 0.10 0.00 0.00 Change, [, M: x x x Equilibrium [, M : (0.10 x) x x K b = [NH 4 [OH = x 2 = 1.8 x 10 5 [NH 3 (0.10 x) Using approximation, K b = x 2 ~ x 2 /0.10 = 1.8 x 10 5 ; (0.10 x) x 2 = (0.10)(1.8 x 10 5 ); x = (1.8 x 10 6 ) = 1.3 x 10 3 where, x = [OH = 1.3 x 10 3 M; poh = log(1.3 x 10 3 ) = 2.87; ph = 11.13 For ammonia solution at 0.10 M concentration, the percent dissociation is: Percent dissociation = 1.3 x 10 3 M x 100% = 1.3% 0.10 M

MASTERING PHYSICAL CHEMISTRY By S.K.SINHA Like weak acids, the percent dissociation of weak bases depends on the K b of the base and the degree of dilution larger K b and higher dilution higher percent dissociation. Exercise: 1. What is [OH in 0.0055 M KOH and what is the ph of the solution? 2. What is [OH in a 0.020 M solution of Ba(OH) 2? What is the ph of this solution? 11 3. A 0.10 M solution of methylamine, CH 3 NH 2 is 6.5% ionized at 25 o C. Determine the K b for methylamine? What is the ph of this solution? (Answer: K b = 4.5 x 10 4 ; ph = 11.81) CH 3 NH 2(aq) H 2 O (l) CH 3 NH 3 (aq) OH (aq) Polyprotic Acids Acids such as HCl, HF, HOCl, HNO 2, HC 2 H 3 O 2, and other similar type of acids are called monoprotic acids each contains a single ionizable hydrogen ion. Some acids contain more than one ionizable hydrogen, and they are called polyprotic acids. Examples of polyprotic acids are H 2 SO 4, H 2 SO 3, H 2 C 2 O 4, and H 3 PO 4. In each case, the hydrogen ionizes in stages and with different ionization constant, such as the following example with H 3 PO 4 : H 3 PO 4(aq) H (aq) H 2 PO 4 (aq) ; K a1 = 7.5 x 10 3 ; H 2 PO 4 (aq) H (aq) HPO 4 2 (aq) ; K a2 = 6.2 x 10 8 ; HPO 4 2 (aq) H (aq) PO 4 3 (aq) ; K a3 = 4.8 x 10 13 ; Acid strength decreases in the order: H 3 PO 4 >> H 2 PO 4 >> HPO 4 2 ; Sulfuric acid is a strong acid because the first hydrogen ionizes completely: H 2 SO 4 (aq) H (aq) HSO 4 (aq); K a1 = very large However, the second hydrogen does not dissociate complete. Thus, HSO 4 is a weak acid, which dissociates as follows: HSO 4 (aq) H (aq) SO 4 2 (aq) ; K a2 = 1.2 x 10 2 Exercise: 1. What is [H, [HSO 4, and [SO 4 2 in 0.10 M H 2 SO 4 (aq)? What is the ph of the solution? (K a1 is very large; K a2 = 1.2 x 10 2 ) (Answer: ph = 0.96) 2. What is the ph of 0.10 M oxalic acid, H 2 C 2 O 4, and what is [C 2 O 4 2 in this solution? (K a1 = 6.5 x 10 2 ; K a2 = 6.1 x 10 5 ) (Answer: [H = 0.054 M; ph = 1.26; [C 2 O 4 2 = 6.1 x 10 5 M) AcidBase Properties of Salts When salts (ionic compounds) dissolve in water, we assume that they completely dissociate into separate ions. Some of these ions can react with water and behave as acids or bases. Salts are also products of acidbase reactions. The acidic or basic nature of a salt solution depends on whether it is a product of: a strong acidstrong base reaction; a weak acidstrong base reaction; a strong acidweak base reaction, or a weak acidweak base reaction.

12 MASTERING PHYSICAL CHEMISTRY By S.K.SINHA Salts of Strong AcidStrong Base Reactions: such as NaCl, NaNO 3, KBr, etc. This type of salt forms a solution that is neutral, because neither the cation ion nor the anion reacts with water and offset the equilibrium between the H 3 O and OH concentrations in the aqueous solution. Salts of Weak AcidStrong Base Reactions: such as NaF, NaNO 2, NaC 2 H 3 O 2, etc. A salt that is the product of a reaction between a weak acid and a strong base will form an aqueous solution that is basic in nature. the anion of such salt reacts with water to increase [OH. For example, Sodium acetate (NaC 2 H 3 O 2 ) is a product of reaction between a weak acid (HC 2 H 3 O 2 ) and a strong base (NaOH). HC 2 H 3 O 2(aq) NaOH (aq) NaC 2 H 3 O 2(aq) H 2 O (l) When sodium acetate is dissolved in water, it dissociates into sodium and acetate ions: NaC 2 H 3 O 2(aq) Na (aq) C 2 H 3 O 2 (aq) The acetate ion undergoes hydrolysis in aqueous solution as follows: C 2 H 3 O 2 (aq) H 2 O (l) HC 2 H 3 O 2(aq) OH (aq); K b = [HC 2 H 3 O 2 [OH [C 2 H 3 O 2 Note the for the dissociation of acetic acid: HC 2 H 3 O 2(aq) H 2 O (l) H 3 O (aq) C 2 H 3 O 2 (aq) ; K a = [H 3 O [C 2 H 3 O 2 [HC 2 H 3 O 2 K a x K b = [H 3 O [C 2 H 3 O 2 x [HC 2 H 3 O 2 [OH = [H 3 O [OH = K w = 1.0 x 10 14 [HC 2 H 3 O 2 [C 2 H 3 O 2 Thus, for C 2 H 3 O 2, K b (for = K w = 1.0 x 10 14 = 5.6 x 10 10 K a (for HC2H3O2) 1.8 x 10 5 Thus, the hydrolysis of acetate ion has K b = 5.6 x 10 10 (> K w ), which implies that the solution of sodium acetate will be basic (ph > 7). Salts of Strong AcidWeak Base Reactions: NH 4 Cl, NH 4 NO 3, (CH 3 ) 2 NH 2 Cl, C 5 H 5 NHCl. Aqueous solutions of this type of salts are acidic, because the cation (such as NH 4 ion) is the conjugate acid of a weak base (NH 3 ) the cations react with water, which increases [H 3 O in solutions. For example, ammonium chloride, NH 4 Cl, is produced by the reaction of a strong acid HCl and a weak base NH 3. HCl (aq) NH 3(aq) NH 4 Cl (aq) NH 4 (aq) Cl (aq) In an aqueous solution, ammonium ion (NH 4 ) establishes the following equilibrium that increases [H 3 O and makes the solution acidic: NH 4 (aq) H 2 O (l) H 3 O (aq) NH 3(aq) ; K a = [H 3 O [NH 3 [NH 4 For the reaction of NH 3 with water, the following equilibrium occurs: NH 3(aq) H 2 O (l) NH 4 (aq) OH (aq); K b = [NH 4 [OH [NH 3

MASTERING PHYSICAL CHEMISTRY By S.K.SINHA K a x K b = [H 3 O [NH 3 x [NH 4 [OH = [H 3 O [OH = K w = 1.0 x 10 14 [NH 4 [NH 3 13 Thus, for NH 4, K a = K w = 1.0 x 10 14 = 5.6 x 10 10 K b (for NH 3 ) 1.8 x 10 5 Thus, the hydrolysis of ammonium ion has K a = 5.6 x 10 10 aqueous solution of ammonium chloride will be acidic (ph < 7). (> K w ), which implies that Salts of Weak AcidWeak Base Reactions: such as NH 4 C 2 H 3 O 2, NH 4 CN, NH 4 NO 2, etc.. Aqueous solutions of this type of salts can either be neutral, acidic, or basic, depending on the relative magnitude of the K a of the weak acid and the K b of the weak base. If K a ~ K b, the salt solution will be approximately neutral; If K a > K b, the salt solution will be acidic, and If K a < K b, the salt solution will be basic. For example, K a of HC 2 H 3 O 2 = 1.8 x 10 5, K b of NH 3 = 1.8 x 10 5, and aqueous solution of ammonium acetate, NH 4 C 2 H 3 O 2, is approximately neutral. NH 4 C 2 H 3 O 2(aq) NH 4 (aq) C 2 H 3 CO 2 (aq) NH 4 (aq) H 2 O (l) H 3 O (aq) NH 3(aq) ; K a = 5.6 x 10 10 ; C 2 H 3 O 2 (aq) H 2 O (l) HC 2 H 3 O 2(aq) OH (aq); K b = 5.6 x 10 10 ; K a =K b a neutral solution, because the hydrolyses result in a solution with [H 3 O = [OH. For NH 4 CN, K a (HCN) = 6.2 x 10 10, K b (NH 3 ) = 1.8 x 10 5, and aqueous solution of NH 4 CN will be basic. NH 4 CN (aq) NH 4 (aq) CN (aq) NH 4 (aq) H 2 O (l) H 3 O (aq) NH 3(aq) ; K a = 5.6 x 10 10 ; CN (aq) H 2 O (l) HCN (aq) OH (aq); K b = 1.6 x 10 5 ; K a < K b basic solution, because the hydrolyses result in a solution in which [H 3 O < [OH. For NH 4 NO 2, K a (HNO 2 ) = 4.0 x 10 4, K b (NH 3 ) = 1.8 x 10 5, and aqueous solution of NH 4 NO 2 will be acidic. NH 4 NO 2(aq) NH 4 (aq) NO 2 (aq) NH 4 (aq) H 2 O (l) H 3 O (aq) NH 3(aq) ; K a = 5.6 x 10 10 ; NO 2 (aq) H 2 O (l) HNO 2(aq) OH (aq); K b = 2.5 x 10 11 ; K a > K b acidic solution, because the hydrolyses result in a solution with [H 3 O > [OH. Calculating the ph of a Basic or an Acidic salt Solution 1. Consider a solution of 0.050 M sodium acetate, which dissociates completely as follows: NaC 2 H 3 O 2(aq) Na (aq) C 2 H 3 O 2 (aq) The acetate ion undergoes hydrolysis in aqueous solution as follows: C 2 H 3 O 2 (aq) H 2 O (l) HC 2 H 3 O 2(aq) OH (aq); K b = [HC 2 H 3 O 2 [OH = 5.6 x 10 10 [C 2 H 3 O 2 By approximation, [OH = (K b [C 2 H 3 O 2 ) = {(5.6 x 10 10 )(0.050)} = 5.3 x 10 6 M poh = log(5.3 x 10 6 ) = 5.28, and ph = 8.72, ( a basic solution)

14 MASTERING PHYSICAL CHEMISTRY By S.K.SINHA 2. Consider a solution of 0.050 M NH 4 Cl, which dissociates as follows: NH 4 Cl (aq) NH 4 (aq) Cl (aq) In an aqueous solution, the following equilibrium occurs: NH 4 (aq) H 2 O (l) H 3 O (aq) NH 3(aq) ; K a = [H 3 O [NH 3 = 5.6 x 10 10 [NH 4 By approximation, [H 3 O = (K a [NH 4 ) = {(5.6 x 10 10 )(0.050)} = 5.3 x 10 6 M ph = log(5.3 x 10 6 ) = 5.28, ( an acidic solution) Exercise: 1. Predict whether aqueous solution of each of the following salts is acidic, basic, or neutral? (a) KNO 3 (b) CH 3 NH 3 Cl (c) Na 2 SO 3 (d) (NH 4 ) 2 HPO 4 K a (for HSO 3 ) = 1.0 x 10 7 ; K a (for H 2 PO 4 ) = 6.2 x 10 8 ; K b (for CH 3 NH 2 ) = 4.4 x 10 4 2. Calculate the ph of a solution that is 0.10 M in sodium cyanide, NaCN(aq). (K a for HCN = 6.2 x 10 10 ) (Answer: ph = 11.10) 3. What is the ph of a solution containing 1.5 g of pyridinium chloride, C 5 H 5 NHCl, in 100.0 ml solution? (K b of C 5 H 5 N = 1.7 x 10 9 ) In aqueous solution C 5 H 5 NHCl dissociates as follows: C 5 H 5 NHCl(aq) C 5 H 5 NH (aq) Cl (aq) (Answer: ph = 3.56) The Common Ion Effect in AcidBase Equilibria Common ions are ions produced by more than one solute in the same solution. For example, in a solution containing sodium acetate and acetic acid, the acetate ion (C 2 H 3 O 2 ) is the common ion. According to the Le Chatelier's principle, the following equilibrium for acetic acid: HC 2 H 3 O 2(aq) H 2 O (l) H 3 O (aq) C 2 H 3 O 2 (aq) will shift to the left if C 2 H 3 O 2 (from another source) is added to the solution. This will reduce the acid dissociation, which lowers [H 3 O and increases the ph of the solution. Addition of ammonium chloride, NH 4 Cl, to ammonia solution will cause the following equilibrium to shift to the left: NH 3(aq) H 2 O (l) NH 4 (aq) OH (aq) NH 4 Cl dissociates to produce NH 4 and Cl ions, where NH 4 is a common ion in the equilibrium of ammonia solution. The introduction of NH 4 ion reduces the ionization of NH 3, which decreases [OH and lowers the ph of the solution. Calculating [H and ph of a solution containing weak acid and the salt of its conjugate base. Consider a solution containing 0.10 M HNO 2 and 0.050 M NaNO 2, where the latter dissociates completely as follows: NaNO 2(aq) Na (aq) NO 2 (aq) The concentration of H can be calculated using the following ICE table:

MASTERING PHYSICAL CHEMISTRY By S.K.SINHA 15 Concentration: HNO 2(aq) H (aq) NO 2 (aq) Initial [, M 0.10 0.00 0.050 (from salt) Change, [, M x x x Equilibrium [, M (0.10 x) x (0.050 x) K a = [H [NO2 [HNO 2 = ( x)(0.050 x) (0.10 x) = 4.0 x 10 4 Since K a << 0.10, x is assumed to very small, and by approximation, ( x)(0.050 x) (0.10 x) ~ (x)(0.050) (0.10) = 4.0 x 10 4 ; x = 8.0 x 10 4 where x = [H = 8.0 x 10 4 M, and ph = log(8.0 x 10 4 ) = 3.10 In general, for a solution containing a weak acid HX, with an dissociation constant K a, and the salt NaX, [H can be calculated by approximation, such that, [ HX [H = K a x [ X Exercises: 1. What are [H and ph of a solution containing 4.0 g of sodium acetate (NaC 2 H 3 O 2 ) in 1.0 L of 0.050 M HC 2 H 3 O 2(aq)? (K a = 1.8 x 10 5 for acetic acid) 2. What are [OH, [H, and ph of a solution containing 2.5 g of NH 4 Cl dissolved in 1.0 L of 0.050 M NH 3(aq)? (K b = 1.8 x 10 5 for NH 3 ) Buffered Solutions A buffered solution is one that has the ability to maintain its ph relatively constant even when a small amount of strong acid or strong base is added. Buffered solutions contain a comparable amount of weak acid and its conjugate base (or a weak base and its conjugate acid). Consider a buffer made up of acetic acid and sodium acetate, in which the major species present in solution are HC 2 H 3 O 2 and C 2 H 3 O 2. If a little HCl is added to this solution, most of H (from HCl) is absorbed by the conjugate base, C 2 H 3 O 2, in the following reaction: H (aq) C 2 H 3 O 2 (aq) HC 2 H 3 O 2 (aq) Since C 2 H 3 O 2 is present in a much larger quantity than the added H, the above reaction is considered to go almost completely to the right. This buffering reaction prevents a significant increase in [H and minimizes the change in its ph. If a strong base such as NaOH is added, most of the OH ions (from NaOH) are reacted by the acid component of the buffer as follows: OH (aq) HC 2 H 3 O 2(aq) H 2 O C 2 H 3 O 2 (aq). Again, because of the larger concentration of HC 2 H 3 O 2 compared to OH, this reaction also goes almost to completion. This buffering reaction prevents a large increase in the [OH and minimizes any change in the ph of the solution.

16 MASTERING PHYSICAL CHEMISTRY By S.K.SINHA Buffer solution is generally prepared by dissolving a mixture of either a weak acid and a "salt" that contains its conjugate base, or a weak base and a "salt" that contains its conjugate acid, in the desired molar ratio to obtain the correct ph. A given buffer is effective within a range of ph that is typically within approximately 1 of the pk a of its acid component. Examples of some common buffer systems: Buffer pk a ph Range HCHO 2 NaCHO 2 3.74 2.75 4.75 HC 2 H 3 O 2 NaC 2 H 3 O 2 4.74 3.75 5.75 KH 2 PO 4 K 2 HPO 4 7.21 6.20 8.20 (one of the buffer systems in blood); CO 2 /H 2 O NaHCO 3 6.37 5.40 7.40 (one of the buffer systems in blood) NH 3 NH 4 Cl 9.25 8.25 10.25 Calculating The [H and ph of Buffer Solutions: For a buffer containing the weak acid HB and the salt NaB, such that B is the conjugate base to the acid, the concentration [H and ph of the buffer depend on the dissociation constant, K a, of the acid component and the concentration ratio [B /[HB in the buffer solution. Consider the equilibrium: HX (aq) H (aq) X (aq); K a = Rearranging, we obtain, [H = K a x [ H [ X [ HX [ HX [ HX ; ph = pka log( ) [ X [ X The last expression is called the HendersonHasselbalch equation, which is useful for calculation the ph of solutions when the ratio [X /[HX is known. For a given buffer system, [H and ph varies with the ratio: [Base/[Acid. Exercise: 1. A 100.0 ml buffer solution contains 5.0 g of 0.038 mol acetic acid and 0.061 mol of sodium acetate. Calculate the ph of the buffer? (K a = 1.8 x 10 5 for acetic acid) If 0.0050 mol of HCl is added to the buffer, what is the final ph of the solution? (Assume that the volume of buffer does not change when HCl(aq) is added.) 2. A buffer solution is prepared by mixing 25 ml of 0.20 M H 2 SO 3 and 75 ml of 0.20 M NaHSO 3. Calculate the ph of this buffer. (K a = 1.5 x 10 2 for H 2 SO 3 ) 3. (a) Suppose you want to prepare a 100mL buffer solution with ph = 7.30, which buffer system would you consider? [Hint: look for acids that have pk a close to the buffer ph. (b) Calculate the ratio [Base/[Acid that the buffer should have. (c) If the concentration of the acid is 0.50 M, what should be the base concentration? (d) Assuming your starting materials are solids, how many grams of each must be dissolved to make the 100. ml buffer solution? (Both acid and its conjugate base may be in the form of sodium or potassium salts) Buffering Capacity The buffering capacity of a solution represents the amount of H and/or OH the buffer solution can absorb without significantly altering its ph. A buffer that contains large concentrations of buffering components and able to absorb significant quantities of strong acid or strong base with little change in its ph is said to have a large buffering capacity.

MASTERING PHYSICAL CHEMISTRY By S.K.SINHA Whereas the ph of a buffered solution is determined by the ratio [X /[HX, buffer capacity is determined by the sizes of [HX and [X in the buffer. Calculating the change in ph of buffered solution after a strong acid is added: 1. Calculate the change in ph that occurs when 0.010 mol of HCl is added to 1.0 L of each of the following buffers: BufferA: 1.0 M HC 2 H 3 O 2 1.0 M NaC 2 H 3 O 2 ; BufferB: 0.020 M HC 2 H 3 O 2 0.020 M NaC 2 H 3 O 2 Solutions For these buffers the ph can be calculated using the HendersonHasselbalch equation: ph = pk a log([c 2 H 3 O 2 /[HC 2 H 3 O 2 ) = log(1.8 x 10 5 ) log(1) = 4.74 When 0.010 mol HCl is added to BufferA, the following reaction takes place: H (aq) C 2 H 3 O 2 (aq) ( HC2H3O2(aq) [ before reaction: 0.010 M 1.0 M 1.0 M [ after reaction: 0 0.99 M 1.01 M 17 The new ph = 4.74 log(0.99/1.01) = 4.74 0.010 = 4.73 (ph is hardly changed ~ 0.21%) When 0.010 mol HCl is added to BufferB, the following reaction takes place: H(aq) C2H3O2(aq) ( HC2H3O2(aq) [ before reaction: 0.010 M 0.020 M 0.020 M [ after reaction: 0 0.010 M 0.030 M New ph = 4.74 log(0.010/0.030) = 4.74 0.48 = 4.26 (ph decreases by 0.48 unit or 10%) This shows that BufferA, which contains larger quantities of buffering components, has a much higher buffering capacity that BufferB. For the BufferA to decrease its ph by 0.48 unit (or 10%), it would have to absorb the equivalent of 0.50 mol of HCl. Exercise 1. An acetate buffer solution is prepared by mixing 35.0 ml of 1.0 M HC2H3O2 and 65.0 ml of 1.0 M NaC2H3O2. Calculate the ph of this buffer. If 1.0 ml of 6.0 M HCl is added to the buffer solution, what will be the new ph of the buffer. 2. A 100mL buffer solution contains 0.20 M KH 2 PO 4 and 0.32 M K 2 HPO 4. What is the ph of the buffer? If 0.0050 mol of HCl is added to the buffer without changing its volume, what is the final ph of the buffer? (K a = 6.2 x 10 8 for H 2 PO 4 ) Titration and ph Curves A ph curve is a graph of ph versus the volume of titrant in an acidbase titration. The data used to plot a ph curve may be obtained either by computation or by measuring the ph directly with a ph meter during titration. There are four types of ph curves for the different types of acidbase titration: 1. strong acid strong base titration. 2. strong acid weak base titration.

18 MASTERING PHYSICAL CHEMISTRY By S.K.SINHA 3. weak acid strong base titration. 4. weak acid weak base titration. (MEMORIZE AND UNDERSTAND THE GENERAL SHAPES AND FEATURES OF EACH ph CURVES!) Strong AcidStrong Base Titration Net reaction: H (aq) OH (aq) H 2 O (l) Calculating The ph of Solutions During Titration: Consider the titration of 20.0 ml of 0.100 M HCl with 0.100 M NaOH solution. Calculate the ph of the solution: (a) before any of the NaOH is added; (b) after 15.0 ml of NaOH is added; (c) after 19.5 ml of NaOH is added; (d) after 20.0 ml of NaOH is added; (e) after 21.0 ml of NaOH is added. (a) Before titration, [H = 0.100 M, ph = 1.000 (b) After 15.0 ml of NaOH is added, calculation of [H will be as follows: H (aq) OH (aq) H 2 O [ before mixing: 0.100 M 0.100 M [ after mixing, but before reaction: 0.057 M 0.043 M [ after reaction: 0.014 M 0.000 [H = 0.014 M, ph = log(0.014) = 1.85 (c) After 19.0 ml of NaOH is added, calculation of [H will be as follows: H (aq) OH (aq) H 2 O [ before mixing: 0.100 M 0.100 M [ after mixing, but before reaction: 0.051 M 0.049 M [ after reaction: 0.002 M 0.000 [H = 0.002 M, ph = log(0.002) = 2.70 (d) After 20.0 ml of NaOH is added, calculation of [H will be as follows: H (aq) OH (aq) ( H2O [ before mixing: 0.100 M 0.100 M [ after mixing, but before reaction: 0.050 M 0.050 M [ after reaction: 0.000 M 0.000 At this point, the solution contains only water and salt (NaCl) and [H = [OH = 1.0 x 107 M, which is due to water ionization. The ph of the solution will be 7.00 (ph of neutral solution) (e) When 21.0 ml of 0.100 M NaOH has been added, there will be excess OH: H(aq) OH(aq) ( H2O [ after mixing, but before reaction: 0.049 M 0.051 M [ after reaction: 0.000 0.002 M [OH = 0.002 M, ( poh = 2.70, and ph = 11.30

MASTERING PHYSICAL CHEMISTRY By S.K.SINHA Note that in strong acidstrong base titration, an abrupt change from about ph 3 to 11 occurs within (0.5 ml of NaOH added near the equivalent point. All strong acidstrong base titration have the same ph curves. Weak AcidStrong Base Titration Net reaction: HX(aq) OH(aq) ( H2O X(aq) For example, when acetic acid is titrated with sodium hydroxide, the net reaction is HC2H3O2(aq) OH(aq) ( C2H3O2(aq) H2O 19 1. Consider the titration of 20.0 ml of 0.100 M HC2H 3 O 2 with 0.100 M NaOH solution. Calculate the ph of the solution: (a) before any of the NaOH is added; (b) after 10.0 ml of NaOH is added; (c) after 15.0 ml of NaOH is added; (d) after 20.0 ml of NaOH is added; (e) after 25.0 ml of NaOH is added. Solution: (a) Before titration, [H = {(0.100 M)(1.8 x 10 5 )} = 1.34 x 10 3 M, ph = log(1.34 x 10 3 ) = 2.873 (b) After adding 10.0 ml of 0.100 M NaOH, [H is calculated as follows: Concentration, M: HC 2 H 3 O 2(aq) OH (aq) C 2 H 3 O 2 (aq) H 2 O Before mixing: 0.100 0.100 0.000 After mixing, before rxn: 0.0667 0.0333 0.000 After reaction: 0.0333 0.000 0.0333 Using HendersonHasselbalch equation, [H = K a x [HC [C H O 2 3 2 2H 3O 2 = 1.8 x 10 5 x (0.033 M) (0.033 M) = 1.8 x 10 5 M ph = pk a log( [C2H 3O 2 [C H O 2 3 2 = 4.74 log(1) = 4.74 When a weak acid is halfneutralized, 50% of the acid is converted to its conjugate base. That is, at halfway to the equivalent point of the titration, [C 2 H 3 O 2 = [HC 2 H 3 O 2. Under this condition, [H = K a, and ph = pk a. (c) At the equivalent point (when 20.0 ml of 0.100 M NaOH has been added), all of the acid is converted to its conjugate base. The latter undergoes hydrolysis as follows: C 2 H 3 O 2 (aq) H 2 O (l) HC 2 H 3 O 2(aq) OH (aq) Initial [, M: 0.0500 0.000 0.000 Change, [, M: x x x Equilibrium [, M: (0.0500 x) x x By approximation, [OH 10 = x = K b x [C2 H 3O 2 0 = (5.6 x 10 )(0.050) = 5.3 x 10 6 M poh = log(5.3 x10 6 ) = 5.28; ph = 8.72

20 MASTERING PHYSICAL CHEMISTRY By S.K.SINHA The hydrolysi of the conjugate base of a weak acid produces OH, which leads to ph > 7 at the equivalent point. In the case of aceticnaoh titration, the ph at the equivalent point is about 8.72. The larger the K a of the weak acid, the closer the ph to 7 at equivalent point. The smaller K a, the higher the ph than 7 at equivalent point. Exercise 1. In a titration of 20.00 ml of 0.100 M formic acid, HCHO 2, with 0.100 M NaOH, calculate the ph: (a) before NaOH was added; (b) after 15.00 ml of NaOH is added; (b) after 20.00 ml of NaOH is added. (K a = 1.8 x 10 4 for HCHO 2 ) 2. In a titration of 20.00 ml of 0.100 M NH 3 (aq) with 0.100 M HCl(aq), calculate the ph: (a) before titration begins; (b) after 10.0 ml of HCl(aq) is added; (c) at equivalent point, that is after 20.0 ml of HCl(aq) is added. (K b = 1.8 x 10 5 for NH 3 ) The Use of Indicators in AcidBase Titration An indicator a substance that changes color to mark the endpoint of a titration. Most indicators used in acidbase titration are weak organic acids. Indicators exhibit one color in the acid or protonated form (HIn) and another in the base or deprotonated form (In ). Each indicator has a range of ph = pk a 1, where the change of colors occurs; A suitable indicator gives the endpoint that corresponds to the equivalent point of the titration; this would be one with a range of ph that falls within the sharp increase (or decrease) of ph in the titration curves. Like a weak acid, an indicator exhibits the following equilibrium in aqueous solution: HIn (aq) K a = H (aq) In (aq); [H [In [HIn ; [In Rearranging, we obtain [H = K a x ([HIn/[In ); ph = pk a log( [HIn ) When [HIn = 10 x [In, [In ph = pk a log( [HIn ) = pk a 1; indicator assumes the color of acid form When [In = 10 x [HIn, [In ph = pk a log( [HIn ) = pka 1; indicator assumes the color of base. Phenolphthalein, which is the most common acidbase indicator, has K a ~ 10 9. Its acid form (HIn) is colorless and the conjugate base form (In ) is pink. It is colorless when the solution ph < 8, that is when 90% or more of the species are in the acid form (HIn), and pink at ph > 10, when 90% or more of the species are in the conjugate base form (In ). The ph range at which an indicator changes depends on the K a. For phenolphthalein with K a ~ 10 9, its color changes at ph = 8 10. It is suitable for strong acidstrong base titration and weak

MASTERING PHYSICAL CHEMISTRY By S.K.SINHA acidstrong base titration. The range of ph for other common acidbase indicators are listed below: Indicators: Acid Color Base Color ph Range Type of Titration Methyl orange orange yellow 3.2 4.5 strong acidstrong base strong acidweak base Bromocresol green yellow blue 3.8 5.4 strong acidstrong base strong acidweak base Methyl red Red Yellow 4.5 6.0 strong acidstrong base strong acidweak base Bromothymol blue Yellow blue 6.0 7.6 strong acidstrong base Phenol Red orange red 6.8 8.2 strong acidstrong base weak acidstrong base Solubility Equilibria Solubility Equilibria and the Solubility Product Constants When a slightly soluble salt such as silver chloride, AgCl, is dissolved in water, a saturated is quickly obtained, because only a very small amount of the solid dissolves, while the rest remains undissolved. The following equilibrium between solid AgCl and the free ions occurs in solution: AgCl(s) Ag(aq) Cl(aq); K sp = [Ag [Cl K sp is called the solubility product constant or just solubility product. 21 General Expressions of Solubility Product Constant, K sp For solubility equilibrium: M a X b(s) am b (aq) bx a (aq), K sp = [M b a [X a b A. For ionic equilibria of the type: MX (s) M n (aq) X n (aq); K sp = [M n [X n If the solubility of compounds is S mol/l, K sp = S 2 ; S = (K sp ) For example, the solubility equilibrium for BaSO 4 is BaSO 4(s) Ba 2 (aq) SO 4 2 (aq) ; K sp = [Ba 2 [SO 4 2 = 1.5 x 10 9 If the solubility of BaSO 4 is S mol/l, a saturated solution of BaSO 4 has [Ba 2 = [SO 4 2 = S mol/l, K sp = S 2 ; and S = (K sp ) = (1.5 x 10 9 ) = 3.9 x 10 5 mol/l B. For ionic equilibria of the type: MX 2(s) M 2 (aq) 2X (aq), K sp = [M 2 [X 2 ; and for the type: M 2 X (s) 2 M (aq) X 2 (aq); K sp = [M 2 [X 2 For both types, if the solubility is S mol/l, K sp = 4S 3 ; S = (K sp /4) 1/3 For example, CaF 2(s) Ca 2 (aq) 2F (aq); K sp = [Ca 2 [F 2 = 4.0 x 10 11 The solubility of calcium fluoride is S = (K sp /4) 1/3 = (4.0 x 10 11 /4) 1/3 = 2.2 x 10 4 mol/l

22 MASTERING PHYSICAL CHEMISTRY By S.K.SINHA C. For solubility equilibria of the type: MX 3(s) M 3 (aq) 3X (aq), K sp = [M 3 [X 3 Or one of the type: M 3 X (s) 3 M (aq) X 3 (aq), K sp = [M 3 [X 3 If the solubility of the compound (MX 3 or M 3 X) is S mol/l, K sp = 27S 4 ; S = 4 K / 27) ( sp For example, Ag 3 PO 4(s) 3Ag 3 (aq) PO 4 (aq) ; K sp = [Ag 3 [PO 3 4 = 1.8 x 10 18 The solubility of silver phosphate is S = 4 K / 27) = 4 (1.8 x 10 18 )/27 = 1.6 x 10 5 mol/l ( sp D. For solubility equilibria: M 2 X 3(s) 2M 3 (aq) 3X 2 (aq), K sp = [M 3 2 [X 2 3 OR, one of the type: M 3 X 2(s) 3M 2 (aq) 2X 3 (aq), K sp = [M 2 3 [X 3 2 If the solubility of the compound M 3 X 2 is S mol/l, then [M 2 = 3S, and [X 3 = 2S; K sp = (3S) 3 (2S) 2 = 108S 5 ; S = 5 K sp / 108 For example, Ca 3 (PO 4 ) 2(s) 3Ca 2 3 (aq) 2PO 4 (aq), K sp = [Ca 2 3 [PO 3 4 2 = 1.3 x 10 32 ; the solubility of Ca 3 (PO 4 ) 2 is S = 5 (1.3 x 10 32 )/108 = 1.6 x 10 7 mol/l Calculating K sp from Solubility 1. Suppose the solubility of PbSO 4 in water is 4.3 x 10 3 g/100 ml solution at 25 o C. What is the K sp of PbSO 4 at 25 o C? Solution: Solubility of PbSO 4 in mol/l = 4.3 x 10 3 g x 1000 ml/l = 1.40 x 10 4 mol/l 100. ml 303.26 g/mol That is, a saturated solution of PbSO 4, contains [Pb 2 = [SO 4 2 = 1.4 x 10 4 mol/l For the equilibrium: PbSO 4(s) Pb 2 (aq) SO 4 2 (aq) ; K sp = [Pb 2 [SO 4 2 = S 2 = (1.4 x 10 4 mol/l) 2 = 2.0 x 10 8 Exercise 1. If the solubility of MgF 2 is 7.3 x 10 3 g/100 ml solution at 25 o C, what is the K sp of MgF 2? 2. A saturated solution of calcium hydroxide has a ph = 12.17. Calculate the solubility of Ca(OH) 2 and its K sp at 25 o C. Determining Solubility from K sp

MASTERING PHYSICAL CHEMISTRY By S.K.SINHA 1. If the K sp of Mg(OH) 2 is 6.3 x 10 10 at 25 o C, what is its solubility at 25 o C (a) in mol/l, and (b) in g/100 ml solution at 25 o C? 23 (a) Solubility equilibrium for Mg(OH) 2 is: Mg(OH) 2(s) Mg 2 (aq) 2 OH (aq) K sp = [Mg 2 [OH 2 = 4S 3 = 6.3 x 10 10 Solubility of Mg(OH) 2 : S = 3 K / 4 = 3 ( 6.3 x 10 10 )/4 = 5.4 x 10 4 mol/l sp (b) Solubility in g/100 ml solution = (5.4 x 10 4 mol/l)(58.32 g/mol)(0.1l/100 ml) = 3.1 x 10 3 g/100 ml solution. Exercise: 1. The K sp of PbI 2 is 1.4 x 10 8. What is its solubility in grams per 100. ml of solution at 25 o C? 2. Calcium hydroxide, Ca(OH) 2, has K sp = 1.3 x 10 6. How many grams of Ca(OH) 2 can be dissolved in 250mL solution to make a saturated solution at 25 o C? Relative Solubility from K sp For compounds whose formula yields the same number of ions, their K sp value can be used to determine relative solubility. That is, larger K sp implies greater solubility. Common Ion Effect of Solubility The presence of common ion decreases the solubility of a slightly soluble ionic compound. For example, in the following equilibrium PbCl 2(s) Pb 2 (aq) 2 Cl (aq) If some NaCl is added to a saturated solution of PbCl 2, the [Cl will increase, and according to Le Chateliler s principle, the equilibrium will shift in the direction that tends to reduce [Cl. In this case, the equilibrium will shift left, to form more PbCl 2 solid, hence decreasing the amount of PbCl 2 that dissolves into solution. Sample Problem The K sp of BaSO 4 is 1.5 x 10 9 at 25 o C. (a) What is its solubility in water at 25 o C? (b) What is the solubility in 0.10 M Na 2 SO 4 at 25 o C? Solubility equilibrium for BaSO 4 : BaSO 4(s) Ba 2 (aq) SO 4 2 (aq) ; K sp = [Ba 2 [SO 4 2 = S 2 = 1.5 x 10 9 Solubility of BaSO 4 in water is S = (K sp ) = (1.1 x 10 10 ) = 3.9 x 10 5 mol/l Let the solubility of BaSO 4 in 0.10 M Na 2 SO 4 be x mol/l Then, the solution contains [Ba 2 = x mol/l and [SO 4 2 = (0.10 x) mol/l K sp = [Ba 2 [SO 4 2 = (x)(0.10 x) = 1.5 x 10 9 By approximation, since x << 0.10 M, (0.10 x) ~ 0.10, and K sp = (x)(0.10 x) ~ (x)(0.10) = 1.5 x 10 9 ; x = 1.5 x 10 8 mol/l The Effect of ph on Solubility ph affect the solubility of slightly soluble compounds containing anions that are conjugate bases of weak acids, such as F, NO 2, OH, SO 3 2, and PO 4 3, but not those containing anions which are conjugate bases of strong acids, such as SO 4 2, Cl, Br. For example, in a saturated solution of calcium fluoride, CaF 2, the following equilibrium exists: