Chapter 19: Ionic Equilibria in Aqueous Systems

Transcription

1 CHEM 1B: GENERAL CHEMISTRY Chapter 19: Ionic Equilibria in Aqueous Systems 19-1 Instructor: Dr. Orlando E. Raola Santa Rosa Junior College

2 Chapter 19 Ionic Equilibria in Aqueous Systems 19-2

3 Ionic Equilibria in Aqueous Systems 19.1 Equilibria of Acid-Base Buffers 19.2 Acid-Base Titration Curves 19.3 Equilibria of Slightly Soluble Ionic Compounds 19.4 Equilibria Involving Complex Ions 19-3

4 Acid-Base Buffers An acid-base buffer is a solution that lessens the impact of ph from the addition of acid or base. An acid-base buffer usually consists of a conjugate acidbase pair where both species are present in appreciable quantities in solution. An acid-base buffer is therefore a solution of a weak acid and its conjugate base, or a weak base and its conjugate acid. 19-4

5 Figure 19.1 The effect of adding acid or base to an unbuffered solution. A 100-mL sample of dilute HCl is adjusted to ph The addition of 1 ml of strong acid (left) or strong base (right) changes the ph by several units. 19-5

6 Figure 19.2 The effect of adding acid or base to a buffered solution. A 100-mL sample of an acetate buffer is adjusted to ph The addition of 1 ml of strong acid (left) or strong base (right) changes the ph very little. The acetate buffer is made by mixing 1 mol/l CH 3 COOH ( a weak acid) with 1 mol/l CH 3 COONa (which provides the conjugate base, CH 3 COO - ). 19-6

7 Buffers and the Common-ion Effect A buffer works through the common-ion effect. Acetic acid in water dissociates slightly to produce some acetate ion: CH 3 COOH(aq) + H 2 O(l) acetic acid CH 3 COO - (aq) + H 3 O + (aq) acetate ion 19-7 If NaCH 3 COO is added, it provides a source of CH 3 COO - ion, and the equilibrium shifts to the left. CH 3 COO - is common to both solutions. The addition of CH 3 COO - reduces the % dissociation of the acid.

8 Table 19.1 The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH 3 COOH] init [CH 3 COO - ] added % Dissociation * [H 3 O + ] ph x x x x * % Dissociation = [CH 3COOH] dissoc [CH 3 COOH] init x

9 How a Buffer Works The buffer components (HA and A - ) are able to consume small amounts of added OH - or H 3 O + by a shift in equilibrium position. CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) Added OH - reacts with CH 3 COOH, causing a shift to the right. Added H 3 O + reacts with CH 3 COO -, causing a shift to the left. The shift in equilibrium position absorbs the change in [H 3 O + ] or [OH - ], and the ph changes only slightly. 19-9

10 Figure 19.3 Buffer has more HA after addition of H 3 O +. How a buffer works. Buffer has equal concentrations of A - and HA. Buffer has more A - after addition of OH -. H 3 O + OH - H 2 O + CH 3 COOH H 3 O + + CH 3 COO - CH 3 COOH + OH - CH 3 COO - + H 2 O 19-10

11 Relative Concentrations of Buffer Components CH 3 COOH(aq) + H 2 O(l) K a = [CH 3COO - ][H 3 O + ] [CH 3 COOH] CH 3 COO - (aq) + H 3 O + (aq) [H 3 O + ] = K a x [CH 3COOH] [CH 3 COO - ] Since K a is constant, the [H 3 O + ] of the solution depends on the ratio of buffer component concentrations. If the ratio If the ratio [HA] [A - ] [HA] [A - ] increases, [H 3 O + ] decreases. decreases, [H 3 O + ] increases

12 Sample Problem 19.1 Calculating the Effect of Added H 3 O + or OH - on Buffer ph PROBLEM: Calculate the ph: (a) Of a buffer solution consisting of 0.50 mol/l CH 3 COOH and 0.50 mol/l CH 3 COONa (b) After adding mol of solid NaOH to 1.0 L of the buffer solution (c) After adding mol of HCl to 1.0 L of the buffer solution in (a). K a of CH 3 COOH = 1.8 x (Assume the additions cause a negligible change in volume.) PLAN: We can calculate [CH 3 COOH] init and [CH 3 COO - ] init from the given information. From this we can find the starting ph. For (b) and (c) we assume that the added OH - or H 3 O + reacts completely with the buffer components. We write a balanced equation in each case, set up a reaction table, and calculate the new [H 3 O + ]

13 Sample Problem 19.1 SOLUTION: (a) Concentration CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) (mol/l) Initial Change x - +x +x Equilibrium x x x Since K a is small, x is small, so we assume [CH 3 COOH] = 0.50 x 0.50 mol/l and [CH 3 COO - ] = x 0.50 mol/l x = [H 3 O + ] = K a x 1.8x10-5 mol/l x 100 = 3.6x10-3 % (< 5%; assumption is justified.) 0.50 mol/l [CH 3 COOH] [CH 3 COO - ] 1.8x10-5 x = 1.8x10-5 mol/l Checking the assumption: ph = -log(1.8x10-5 ) = 4.74

14 Sample Problem 19.1 (b) [OH mol ] added = = mol/l 1.0 L soln OH Setting up a reaction table for the stoichiometry: Concentration CH 3 COOH(aq) + OH - (aq) CH 3 COO - (aq) + H 2 O(l) (mol/l) Initial Change Equilibrium Setting up a reaction table for the acid dissociation, using new initial [ ]: Concentration CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) (mol/l) Initial Change x - +x +x Equilibrium x x x 19-14

15 Sample Problem 19.1 Since K a is small, x is small, so we assume [CH 3 COOH] = 0.48 x 0.48 mol/l and [CH 3 COO - ] = x 0.52 mol/l x = [H 3 O + ] = K a x [CH 3 COOH] [CH 3 COO - ] 1.8x10-5 x = 1.7x10-5 mol/l ph = -log(1.7x10-5 ) =

16 Sample Problem 19.1 (c) [H 3 O mol ] added = = mol/l 1.0 L soln H 3 O + Setting up a reaction table for the stoichiometry: Concentration CH 3 COO - (aq) + H 3 O + (aq) CH 3 COOH(aq) + H 2 O(l) (mol/l) Initial Change Equilibrium Setting up a reaction table for the acid dissociation, using new initial [ ]: Concentration CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) (mol/l) Initial Change x - +x +x Equilibrium x x x 19-16

17 Sample Problem 19.1 Since K a is small, x is small, so we assume [CH 3 COOH] = 0.52 x 0.52 mol/l and [CH 3 COO - ] = x 0.48 mol/l x = [H 3 O + ] = K a x [CH 3 COOH] [CH 3 COO - ] 1.8x10-5 x = 2.0x10-5 mol/l ph = -log(2.0x10-5 ) =

18 The Henderson-Hasselbalch Equation HA(aq) + H 2 O(l) K a = [H 3O + ][A - ] [HA] A - (aq) + H 3 O + (aq) [H 3 O + ] = K a x [HA] [A - ] -log[h 3 O + ] = -logk a log [HA] [A - ] ph = pk a + log [base] [acid] 19-18

19 Buffer Capacity The buffer capacity is a measure of the strength of the buffer, its ability to maintain the ph following addition of strong acid or base. The greater the concentrations of the buffer components, the greater its capacity to resist ph changes. The closer the component concentrations are to each other, the greater the buffer capacity

20 Figure 19.4 The relation between buffer capacity and ph change. When strong base is added, the ph increases least for the most concentrated buffer This graph shows the final ph values for four different buffer solutions after the addition of strong base.

21 Buffer Range The buffer range is the ph range over which the buffer is effective. Buffer range is related to the ratio of buffer component concentrations. [HA] The closer is to 1, the more effective the buffer. [A - ] If one component is more than 10 times the other, buffering action is poor. Since log10 = 1, buffers have a usable range within ± 1 ph unit of the pk a of the acid component

22 Sample Problem 19.2 Using Molecular Scenes to Examine Buffers PROBLEM: The molecular scenes below represent samples of four HA/A - buffers. (HA is blue and green, A - is green, and other ions and water are not shown.) (a) Which buffer has the highest ph? (b) Which buffer has the greatest capacity? (c) Should we add a small amount of concentrated strong acid or strong base to convert sample 1 to sample 2 (assuming no volume changes)? 19-22

23 Sample Problem 19.2 PLAN: Since the volumes of the solutions are equal, the scenes represent molarities as well as numbers. We count the particles of each species present in each scene and calculate the ratio of the buffer components. SOLUTION: [A - ]/[HA] ratios: sample 1, 3/3 = 1; sample 2, 2/4 = 0.5; sample 3, 4/4 = 1; and sample 4, 4/2 =

24 Sample Problem 19.2 (a) As the ph rises, more HA will be converted to A -. The scene with the highest [A - ]/[HA] ratio is at the highest ph. Sample 4 has the highest ph because it has the highest ratio. (b) The buffer with the greatest capacity is the one with the [A - ]/[HA] closest to 1. Sample 3 has the greatest buffer capacity. (c) Sample 2 has a lower [A - ]/[HA] ratio than sample 1, so we need to increase the [A - ] and decrease the [HA]. This is achieved by adding strong acid to sample

25 Preparing a Buffer Choose the conjugate acid-base pair. The pk a of the weak acid component should be close to the desired ph. Calculate the ratio of buffer component concentrations. ph = pk a + log [base] [acid] Determine the buffer concentration, and calculate the required volume of stock solutions and/or masses of components. Mix the solution and correct the ph

26 Sample Problem 19.3 Preparing a Buffer PROBLEM: An environmental chemist needs a carbonate buffer of ph to study the effects of the acid rain on limsetone-rich soils. How many grams of Na 2 CO 3 must she add to 1.5 L of freshly prepared 0.20 mol/l NaHCO 3 to make the buffer? K a of HCO 3 - is 4.7x PLAN: The conjugate pair is HCO 3 - (acid) and CO 3 2- (base), and we know both the buffer volume and the concentration of HCO 3-. We can calculate the ratio of components that gives a ph of 10.00, and hence the mass of Na 2 CO 3 that must be added to make 1.5 L of solution. SOLUTION: [H 3 O + ] = 10 -ph = = 1.0x10-10 mol/l HCO 3- (aq) + H 2 O(l) H 3 O + (aq) + CO 2-3 (aq) K a = [CO 3 2- ][H 3 O + ] [HCO 3- ] 19-26

27 Sample Problem 19.3 Preparing a Buffer [CO 3 2- ] = K a[hco 3- ] [H 3 O + ] = (4.7x10-11 )(0.20) 1.0x10-10 = mol/l Amount (mol) of CO 3 2- needed = 1.5 L soln x mol CO L soln = 0.14 mol CO mol Na g Na 2 CO 2 CO 3 x 3 = 15 g Na 2 CO 3 1 mol Na 2 CO 3 The chemist should dissolve 15 g Na 2 CO 3 in about 1.3 L of 0.20 mol/l NaHCO 3 and add more 0.20 mol/l NaHCO 3 to make 1.5 L. Using a ph meter, she can then adjust the ph to by dropwise addition of concentrated strong acid or base

28 Acid-Base Indicators An acid-base indicator is a weak organic acid (HIn) whose color differs from that of its conjugate base (In - ). The ratio [HIn]/[In - ] is governed by the [H 3 O + ] of the solution. Indicators can therefore be used to monitor the ph change during an acid-base reaction. The color of an indicator changes over a specific, narrow ph range, a range of about 2 ph units

29 Figure 19.5 Colors and approximate ph range of some common acid-base indicators. ph 19-29

30 Figure 19.6 The color change of the indicator bromthymol blue. ph < 6.0 ph = ph >

31 Acid-Base Titrations In an acid-base titration, the concentration of an acid (or a base) is determined by neutralizing the acid (or base) with a solution of base (or acid) of known concentration. The equivalence point of the reaction occurs when the number of moles of OH - added equals the number of moles of H 3 O + originally present, or vice versa. The end point occurs when the indicator changes color. - The indicator should be selected so that its color change occurs at a ph close to that of the equivalence point

32 Figure 19.7 Curve for a strong acid strong base titration. The ph increases gradually when excess base has been added. The ph rises very rapidly at the equivalence point, which occurs at ph = The initial ph is low

33 Calculating the ph during a strong acid strong base titration Initial ph [H 3 O + ] = [HA] init ph = -log[h 3 O + ] ph before equivalence point initial mol H 3 O + = V acid x M acid mol OH - added = V base x M base mol H 3 O + remaining = (mol H 3 O + init) (mol OH - added) [H 3 O + ] = mol H 3 O + remaining ph = -log[h 3 O + ] V acid + V base 19-33

34 Calculating the ph during a strong acid strong base titration ph at the equivalence point ph = 7.00 for a strong acid-strong base titration. ph beyond the equivalence point initial mol H 3 O + = V acid x M acid mol OH - added = V base x M base mol OH - excess = (mol OH - added) (mol H 3 O + init) [OH - ] = mol OH - excess V acid + V base poh = -log[oh - ] and ph = poh 19-34

35 Example: ml of mol/l HCl is titrated with mol/l NaOH. The initial ph is simply the ph of the HCl solution: [H 3 O + ] = [HCl] init = mol/l and ph = -log(0.1000) = 1.00 To calculate the ph after ml of NaOH solution has been added: Initial mol of H 3 O + = L HCl x mol = 4.000x10-3 mol H 3 O + 1 L OH - added = L NaOH x mol = 2.000x10-3 mol OH - 1 L The OH - ions react with an equal amount of H 3 O + ions, so H 3 O + remaining = 4.000x x10-3 = 2.000x10-3 mol H 3 O

36 [H 3 O x10 ] = -3 mol L L ph = -log( ) = 1.48 = mol/l The equivalence point occurs when mol of OH - added = initial mol of HCl, so when ml of NaOH has been added. To calculate the ph after ml of NaOH solution has been added: OH - added = L NaOH x mol 1 L = 5.000x10-3 mol OH - OH - in excess = 5.000x x10-3 = 1.000x10-3 mol OH - [OH - ] = 1.000x10-3 mol L L = mol/l poh = -log( ) = 1.95 ph = =

37 Figure 19.8 Curve for a weak acid strong base titration. The ph increases slowly beyond the equivalence point. The curve rises gradually in the buffer region. The weak acid and its conjugate base are both present in solution. The ph at the equivalent point is > 7.00 due to the reaction of the conjugate base with H 2 O. The initial ph is higher than for the strong acid solution

38 Calculating the ph during a weak acid strong base titration [H 3 O + ][A - ] K a = [HA] ph = -log[h 3 O + ] Initial ph [H 3 O + ] = ph before equivalence point [H 3 O + ] = K a x [HA] or [A - ] ph = pk a + log [base] [acid] 19-38

39 Calculating the ph during a weak acid strong base titration A - (aq) + H 2 O(l) [OH - ] = ph at the equivalence point HA(aq) + OH - (aq) where [A - ] = mol HA init and K b = K w V acid + V base K a K [H 3 O + ] w and ph = -log[h 3 O + ] ph beyond the equivalence point [OH - ] = mol OH - excess V acid + V base [H 3 O + ] = K w [OH - ] ph = -log[h 3 O + ] 19-39

40 Sample Problem 19.4 Finding the ph During a Weak Acid Strong Base Titration PROBLEM: Calculate the ph during the titration of ml of mol/l propanoic acid (HPr; K a = 1.3x10-5 ) after adding the following volumes of mol/l NaOH: (a) 0.00 ml (b) ml (c) ml (d) ml PLAN: The initial ph must be calculated using the K a value for the weak acid. We then calculate the number of moles of HPr present initially and the number of moles of OH - added. Once we know the volume of base required to reach the equivalence point we can calculate the ph based on the species present in solution. SOLUTION: (a) [H 3 O + ] = = 1.1x10-3 mol/l ph = -log(1.1x10-3 ) =

41 Sample Problem 19.4 (b) ml of mol/l NaOH has been added. Initial amount of HPr = L x mol/l = 4.000x10-3 mol HPr Amount of NaOH added = L x mol/l = 3.000x10-3 mol OH - Each mol of OH - reacts to form 1 mol of Pr -, so Concentration HPr(aq) + OH - (aq) Pr - (aq) + H 2 O(l) (mol/l) Initial Change Equilibrium [H 3 O + ] = K a x [HPr] [Pr - ] ph = -log(4.3x10-6 ) = 5.37 = (1.3x10-5 ) x = 4.3x10-6 mol/l

42 Sample Problem 19.4 (c) ml of mol/l NaOH has been added. This is the equivalence point because mol of OH - added = = mol of HA init. All the OH - added has reacted with HA to form mol of Pr -. [Pr - ] = mol L L = mol/l Pr - is a weak base, so we calculate K b = [H 3 O + ] K w = 1.0x10-14 K w K a = 1.0x x10-5 = 7.7x10-10 = 1.6x10-9 mol/l ph = -log(1.6x10-9 ) =

43 Sample Problem 19.4 (d) ml of mol/l NaOH has been added. Amount of OH - added = L x mol/l = mol Excess OH - = OH - added HA init = = mol [OH - ] = mol OH- excess total volume = mol L = mol/l [H 3 O + ] = K w [OH - ] = 1x = 9.0x10-13 mol/l ph = -log(9.0x10-13 ) =

44 Figure 19.9 Curve for a weak base strong acid titration. The ph decreases gradually in the buffer region. The weak base and its conjugate acid are both present in solution. The ph at the equivalence point is < 7.00 due to the reaction of the conjugate acid with H 2 O

45 Figure Curve for the titration of a weak polyprotic acid. Titration of ml of mol/l H 2 SO 3 with mol/l NaOH pk a2 = 7.19 pk a1 =

46 Amino Acids as Polyprotic Acids An amino acid contains a weak base (-NH 2 ) and a weak acid (-COOH) in the same molecule. Both groups are protonated at low ph and the amino acid behaves like a polyprotic acid

47 Figure Abnormal shape of red blood cells in sickle cell anemia. Several amino acids have charged R groups in addition to the NH 2 and COOH group. These are essential to the normal structure of many proteins. In sickle cell anemia, the hemoglobin has two amino acids with neutral R groups instead of charged groups. The abnormal hemoglobin causes the red blood cells to have a sickle shape, as seen here

48 Equilibria of Slightly Soluble Ionic Compounds Any insoluble ionic compound is actually slightly soluble in aqueous solution. We assume that the very small amount of such a compound that dissolves will dissociate completely. For a slightly soluble ionic compound in water, equilibrium exists between solid solute and aqueous ions. PbF 2 (s) Q c = [Pb2+ ][F - ] 2 [PbF 2 ] Pb 2+ (aq) + 2F - (aq) Q sp = Q c [PbF 2 ] = [Pb 2+ ][F - ]

49 Q sp and K sp Q sp is called the ion-product expression for a slightly soluble ionic compound. For any slightly soluble compound M p X q, which consists of ions M n+ and X z-, Q sp = [M n+ ] p [X z- ] q When the solution is saturated, the system is at equilibrium, and Q sp = K sp, the solubility product constant. The K sp value of a salt indicates how far the dissolution proceeds at equilibrium (saturation)

50 Metal Sulfides Metal sulfides behave differently from most other slightly soluble ionic compounds, since the S 2- ion is strongly basic. We can think of the dissolution of a metal sulfide as a two-step process: MnS(s) Mn 2+ (aq) + S 2- (aq) S 2- (aq) + H 2 O(l) HS - (aq) + OH - (aq) MnS(s) + H 2 O(l) Mn 2+ (aq) + HS - (aq) + OH - (aq) K sp = [Mn 2+ ][HS - ][OH - ] 19-50

51 Sample Problem 19.5 Writing Ion-Product Expressions PROBLEM: Write the ion-product expression at equilibrium for each compound: (a) magnesium carbonate (b) iron(ii) hydroxide (c) calcium phosphate (d) silver sulfide PLAN: We write an equation for a saturated solution of each compound, and then write the ion-product expression at equilibrium, K sp. Note the sulfide in part (d). SOLUTION: (a) MgCO 3 (s) Mg 2+ (aq) + CO 3 2- (aq) K sp = [Mg 2+ ][CO 3 2- ] (b) Fe(OH) 2 (s) Fe 2+ (aq) + 2OH - (aq) K sp = [Fe 2+ ][OH - ] 2 (c) Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3- (aq) K sp = [Ca 2+ ] 3 [PO 4 3- ]

52 Sample Problem 19.5 (d) Ag 2 S(s) 2Ag + (aq) + S 2- (aq) S 2- (aq) + H 2 O(l) HS - (aq) + OH - (aq) Ag 2 S(s) + H 2 O(l) 2Ag + (aq) + HS - (aq) + OH - (aq) K sp = [Ag + ] 2 [HS - ][OH - ] 19-52

53 Table 19.2 Solubility-Product Constants (K sp ) of Selected Ionic Compounds at 25 C Name, Formula Aluminum hydroxide, Al(OH) 3 Cobalt(II) carbonate, CoCO 3 Iron(II) hydroxide, Fe(OH) 2 Lead(II) fluoride, PbF 2 Lead(II) sulfate, PbSO 4 Mercury(I) iodide, Hg 2 I 2 Silver sulfide, Ag 2 S Zinc iodate, Zn(IO 3 ) 2 K sp 3x x x x x x x x

54 Sample Problem 19.6 Determining K sp from Solubility PROBLEM: (a) Lead(II) sulfate (PbSO 4 ) is a key component in leadacid car batteries. Its solubility in water at 25 C is 4.25x10-3 g/100 ml solution. What is the K sp of PbSO 4? (b) When lead(ii) fluoride (PbF 2 ) is shaken with pure water at 25 C, the solubility is found to be 0.64 g/l. Calculate the K sp of PbF 2. PLAN: We write the dissolution equation and the ion-product expression for each compound. This tells us the number of moles of each ion formed. We use the molar mass to convert the solubility of the compound to molar solubility (molarity), then use it to find the molarity of each ion, which we can substitute into the K sp expression

55 Sample Problem 19.6 SOLUTION: (a) PbSO 4 (s) Pb 2+ (aq) + SO 4 2- (aq) K sp = [Pb 2+ ][SO 4 2- ] Converting from g/ml to mol/l: 4.25x10-3 g PbSO ml soln x 1000 ml 1 L x 1 mol PbSO g PbSO 4 = 1.40x10-4 mol/l PbSO 4 Each mol of PbSO 4 produces 1 mol of Pb 2+ and 1 mol of SO 4 2-, so [Pb 2+ ] = [SO 4 2- ] = 1.40x10-4 mol/l K sp = [Pb 2+ ][SO 4 2- ] = (1.40x10-4 ) 2 = 1.96x

56 Sample Problem 19.6 (b) PbF 2 (s) Pb 2+ (aq) + F - (aq) K sp = [Pb 2+ ][F - ] 2 Converting from g/l to mol/l: 0.64 g PbF 2 1 L soln x 1 mol PbF g PbF 2 = 2.6x10-3 mol/l PbF 2 Each mol of PbF 2 produces 1 mol of Pb 2+ and 2 mol of F -, so [Pb 2+ ] = 2.6x10-3 mol/l and [F - ] = 2(2.6x10-3 ) = 5.2x10-3 mol/l K sp = [Pb 2+ ][F - ] 2 = (2.6x10-3 )(5.2x10-3 ) 2 = 7.0x

57 Sample Problem 19.7 Determining Solubility from K sp PROBLEM: Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH) 2 are used in industry as a strong, inexpensive base. Calculate the molar solubility of Ca(OH) 2 in water if the K sp is 6.5x10-6. PLAN: We write the dissolution equation and the expression for K sp. We know the value of K sp, so we set up a reaction table that expresses [Ca 2+ ] and [OH - ] in terms of S, the molar solubility. We then substitute these expressions into the K sp expression and solve for S. SOLUTION: Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) K sp = [Ca 2+ ][OH - ] 2 = 6.5x

58 Sample Problem 19.7 Concentration Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) (mol/l) Initial Change - +S + 2S Equilibrium - S 2S K sp = [Ca 2+ ][OH - ] 2 = (S)(2S) 2 = 4S 3 = 6.5x10-6 S = = = 1.2x10-2 mol/l 19-58

59 Table 19.3 Relationship Between K sp and Solubility at 25 C No. of Ions Formula Cation/Anion K sp Solubility (mol/l) 2 MgCO 3 1/1 3.5x x PbSO 4 1/1 1.6x x BaCrO 4 1/1 2.1x x Ca(OH) 2 1/2 6.5x x BaF 2 1/2 1.5x x CaF 2 1/2 3.2x x Ag 2 CrO 4 2/1 2.6x x10-5 The higher the K sp value, the greater the solubility, as long as we compare compounds that have the same total number of ions in their formulas

60 Figure The effect of a common ion on solubility. PbCrO 4 (s) Pb 2+ (aq) + CrO 4 2- (aq) If Na 2 CrO 4 solution is added to a saturated solution of PbCrO 4, it provides the common ion CrO 4 2-, causing the equilibrium to shift to the left. Solubility decreases and solid PbCrO 4 precipitates

61 Sample Problem 19.8 Calculating the Effect of a Common Ion on Solubility PROBLEM: In Sample Problem 19.7, we calculated the solubility of Ca(OH) 2 in water. What is its solubility in 0.10 mol/l Ca(NO 3 ) 2? K sp of Ca(OH) 2 is 6.5x10-6. PLAN: The addition of Ca 2+, an ion common to both solutions, should lower the solubility of Ca(OH) 2. We write the equation and K sp expression for the dissolution and set up a reaction table in terms of S, the molar solubility of Ca(OH) 2. We make the assumption that S is small relative to [Ca 2+ ] init because K sp is low. We can then solve for S and check the assumption. SOLUTION: Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) K sp = [Ca 2+ ][OH - ]

62 Sample Problem 19.8 [Ca 2+ ] init = 0.10 mol/l because Ca(NO 3 ) 2 is a soluble salt, and dissociates completely in solution. Concentration Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) (mol/l) Initial Change - +S + 2S Equilibrium S 2S K sp = [Ca 2+ ][OH - ] 2 = 6.5x10-6 (0.10)(2S) 2 = (0.10)(4S 2 ) 4S 2 6.5x so S = 4.0x10-3 mol/l Checking the assumption: 4.0x10-3 mol/l x 100 = 4.0% < 5% 0.10 mol/l 19-62

63 Effect of ph on Solubility Changes in ph affects the solubility of many slightly soluble ionic compounds. The addition of H 3 O + will increase the solubility of a salt that contains the anion of a weak acid. CaCO 3 (s) Ca 2+ (aq) + CO 3 2- (aq) CO 3 2- (aq) + H 3 O + (aq) HCO 3- (aq) + H 2 O(l) HCO 3- (aq) + H 3 O + (aq) [H 2 CO 3 (aq)] + H 2 O(l) CO 2 (g) + 2H 2 O(l) The net effect of adding H 3 O + to CaCO 3 is the removal of CO 3 2- ions, which causes an equilibrium shift to the right. More CaCO 3 will dissolve

64 Figure Test for the presence of a carbonate. When a carbonate mineral is treated with HCl, bubbles of CO 2 form

65 Sample Problem 19.9 Predicting the Effect on Solubility of Adding Strong Acid PROBLEM: Write balanced equations to explain whether addition of H 3 O + from a strong acid affects the solubility of each ionic compound: (a) lead(ii) bromide (b) copper(ii) hydroxide (c) iron(ii) sulfide PLAN: We write the balanced dissolution equation for each compound and note the anion. The anion of a weak acid reacts with H 3 O +, causing an increase in solubility. SOLUTION: (a) PbBr 2 (s) Pb 2+ (aq) + 2Br - (aq) Br - is the anion of HBr, a strong acid, so it does not react with H 3 O +. The addition of strong acid has no effect on its solubility

66 Sample Problem 19.9 (b) Cu(OH) 2 (s) Cu 2+ (aq) + 2OH - (aq) OH - is the anion of H 2 O, a very weak acid, and is in fact a strong base. It will react with H 3 O + : The addition of strong acid will cause an increase in solubility. (c) FeS(s) Fe 2+ (aq) + S 2- (aq) OH - (aq) + H 3 O + (aq) 2H 2 O(l) S 2- is the anion of HS -, a weak acid, and is a strong base. It will react completely with water to form HS - and OH -. Both these ions will react with added H 3 O + : HS - (aq) + H 3 O + (aq) H 2 S(aq) + H 2 O(l) OH - (aq) + H 3 O + (aq) 2H 2 O(l) The addition of strong acid will cause an increase in solubility

67 Figure Limestone cave in Nerja, Málaga, Spain. Limestone is mostly CaCO 3 (K sp = 3.3x10-9 ) CO 2 (g) CO 2 (aq) CO 2 (aq) + 2H 2 O(l) CaCO 3 (s) + CO 2 (aq) + H 2 O(l) Ground water rich in CO 2 trickles over CaCO 3, causing it to dissolve. This gradually carves out a cave. Water containing HCO - 3 and Ca 2+ ions drips from the cave ceiling. The air has a lower P CO than the soil, causing CO 2 2 to come out of solution. A shift in equilibrium results in the precipitation of CaCO 3 to form stalagmites and stalactites. H 3 O + (aq) + HCO 3- (aq) Ca 2+ (aq) + 2HCO 3- (aq)

68 Predicting the Formation of a Precipitate For a saturated solution of a slightly soluble ionic salt, Q sp = K sp. When two solutions containing the ions of slightly soluble salts are mixed, If Q sp = K sp, the solution is saturated and no change will occur. If Q sp > K sp, a precipitate will form until the remaining solution is saturated. If Q sp =< K sp, no precipitate will form because the solution is unsaturated

69 Sample Problem Predicting Whether a Precipitate Will Form PROBLEM: A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate form when L of 0.30 mol/l Ca(NO 3 ) 2 is mixed with L of mol/l NaF? PLAN: First we need to decide which slightly soluble salt could form, look up its K sp value in Appendix C, and write the dissolution equation and K sp expression. We find the initial ion concentrations from the given volumes and molarities of the two solutions, calculate the value for Q sp and compare it to K sp. SOLUTION: The ions present are Ca 2+, NO 3-, Na +, and F -. All Na + and NO 3 - salts are soluble, so the only possible precipitate is CaF 2 (K sp = 3.2x10-11 ) CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = [Ca 2+ ][F - ] 2

70 Sample Problem Ca(NO 3 ) 2 and NaF are soluble, and dissociate completely in solution. We need to calculate [Ca 2+ ] and [F - ] in the final solution. Amount (mol) of Ca 2+ = mol/l Ca 2+ x L = mol Ca 2+. [Ca 2+ ] init = mol Ca L L = 0.10 mol/l Ca 2+ Amount (mol) of F - = mol/l F - x L = mol F -. [F - ] init = mol F L L = mol/l F - Q sp = [Ca 2+ ] init [F - ] 2 init = (0.10)(0.040) 2 = 1.6x10-4 Since Q sp > K sp, CaF 2 will precipitate until Q sp = 3.2x

71 Sample Problem Using Molecular Scenes to Predict Whether a Precipitate Will Form PROBLEM: These four scenes represent solutions of silver (gray) and carbonate (black and red) ions above solid silver carbonate. (The solid, other ions, and water are not shown.) (a) Which scene best represents the solution in equilibrium with the solid? (b) In which, if any, other scene(s) will additional solid silver carbonate form? (c) Explain how, if at all, addition of a small volume of concentrated strong acid affects the [Ag + ] in scene 4 and the mass of solid present

72 Sample Problem PLAN: We need to determine the ratio of the different types of ion in each solution. A saturated solution of Ag 2 CO 3 should have 2Ag + ions for every 1 CO 3 2- ion. For (b) we need to compare Q sp to K sp. For (c) we recall that CO 3 2- reacts with H 3 O +. SOLUTION: First we determine the Ag + /CO 2-3 ratios for each scene. Scene 1: 2/4 or 1/2 Scene 2: 3/3 or 1/1 Scene 3: 4/2 or 2/1 Scene 4: 3/4 (a) Scene 3 is the only one that has an Ag + /CO 3 2- ratio of 2/1, so this scene represents the solution in equilibrium with the solid

73 Sample Problem (b) We use the ion count for each solution to determine Q sp for each one. Since Scene 3 is at equilibrium, its Q sp value = K sp. Scene 1: Q sp = (2) 2 (4) = 16 Scene 3: Q sp = (4) 2 (2) = 32 Scene 2: Q sp = (3) 2 (3) = 27 Scene 4: Q sp = (3) 2 (4) = 36 Scene 4 is the only one that has Q sp > K sp, so a precipitate forms in this solution. (c) Ag 2 CO 3 (s) 2Ag + (aq) + CO 3 2- (aq) CO 3 2- (aq) + 2H 3 O + (aq) [H 2 CO 3 (aq)] + 2H 2 O(l) 3H 2 O(l) + CO 2 (g) The CO 2 leaves as a gas, so adding H 3 O + decreases the [CO 3 2- ] in solution, causing more Ag 2 CO 3 to dissolve. [Ag + ] increases and the mass of Ag 2 CO 3 decreases

74 Selective Precipitation Selective precipitation is used to separate a solution containing a mixture of ions. A precipitating ion is added to the solution until the Q sp of the more soluble compound is almost equal to its K sp. The less soluble compound will precipitate in as large a quantity as possible, leaving behind the ion of the more soluble compound

75 Sample Problem Separating Ions by Selective Precipitation PROBLEM: A solution consists of 0.20 mol/l MgCl 2 and 0.10 mol/l CuCl 2. Calculate the [OH - ] that would separate the metal ions as their hydroxides. K sp of Mg(OH) 2 = is 6.3x10-10 ; K sp of Cu(OH) 2 is 2.2x PLAN: Both compounds have 1/2 ratios of cation/anion, so we can compare their solubilities by comparing their K sp values. Mg(OH) 2 is times more soluble than Cu(OH) 2, so Cu(OH) 2 will precipitate first. We write the dissolution equations and K sp expressions. Using the given cation concentrations, we solve for the [OH - ] that gives a saturated solution of Mg(OH) 2. Then we calculate the [Cu 2+ ] remaining to see if the separation was successful

76 Sample Problem SOLUTION: Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) K sp = [Mg 2+ ][OH - ] 2 = 6.3x10-10 Cu(OH) 2 (s) Cu 2+ (aq) + 2OH - (aq) K sp = [Cu 2+ ][OH - ] 2 = 2.2x10-20 [OH - ] = = = 5.6x10-5 mol/l This is the maximum [OH - ] that will not precipitate Mg 2+ ion. Calculating the [Cu 2+ ] remaining in solution with this [OH - ] K [Cu 2+ ] = sp = [OH - ] 2 2.2x10-20 (5.6x10-5 ) 2 = 7.0x10-12 mol/l Since the initial [Cu 2+ ] is 0.10 mol/l, virtually all the Cu 2+ ion is precipitated.

77 Chemical Connections Figure B19.1 Formation of acidic precipitation. Since ph affects the solubility of many slightly soluble ionic compounds, acid rain has far-reaching effects on many aspects of our environment

78 Figure Cr(NH 3 ) 6 3+, a typical complex ion. A complex ion consists of a central metal ion covalently bonded to two or more anions or molecules, called ligands

79 Figure The stepwise exchange of NH 3 for H 2 O in M(H 2 O) The overall formation constant is given by K f = [M(NH 3 ) 4 2+ ] [M(H 2 O) 4 2+ ][NH 3 ]

80 19-80 Table 19.4 Formation Constants (K f ) of Some Complex Ions at 25 C

81 Sample Problem Calculating the Concentration of a Complex Ion PROBLEM: An industrial chemist converts Zn(H 2 O) 2+ 4 to the more stable Zn(NH 3 ) 2+ 4 by mixing 50.0 L of mol/l Zn(H 2 O) 2+ 4 and 25.0 L of 0.15 mol/l NH 3. What is the final [Zn(H 2 O) 2+ 4 ] at equilibrium? K f of Zn(NH 3 ) 2+ 4 is 7.8x10 8. PLAN: We write the reaction equation and the K f expression, and use a reaction table to calculate equilibrium concentrations. To set up the table, we must first find [Zn(H 2 O) 2+ 4 ] init and [NH 3 ] init using the given volumes and molarities. With a large excess of NH 3 and a high K f, we assume that almost all the Zn(H 2 O) 2+ 4 is converted to Zn(NH 3 ) SOLUTION: Zn(H 2 O) 4 2+ (aq) + 4NH 3 (aq) Zn(NH 3 ) 4 2+ (aq) + 4H 2 O(l) K f = [Zn(NH 3 ) 4 2+ ] [Zn(H 2 O) 4 2+ ][NH 3 ]

82 Sample Problem [Zn(H 2 O) 2+ 4 ] initial = 50.0 L x mol/l = 1.3x L L mol/l 25.0 L x 0.15 mol/l [NH 3 ] initial = = 5.0x L L mol/l 4 mol of NH 3 is needed per mol of Zn(H 2 O 4 ) 2+, so [NH 3 ] reacted = 4(1.3x10-3 mol/l) = 5.2x10-3 mol/l and [Zn(NH 3 ) 2+ 4 ] 1.3x10-3 mol/l Concentration Zn(H 2 O) 2+ 4 (aq) + 4NH 3 (aq) Zn(NH 3 ) 2+ 4 (aq) + 4H 2 O(l) (mol/l) Initial 1.3x x Change ~(-1.3x10-3 ) ~(-5.2x10-3 ) ~(+1.3x10-3 ) - Equilibrium x 4.5x x

83 Sample Problem K f = [Zn(NH 3 ) 4 2+ ] [Zn(H 2 O) 4 2+ ][NH 3 ] 4 = 7.8x108 = (1.3x10-3 ) x(4.5x10-2 ) 4 x = [Zn(H 2 O) 4 2+ = 4.1x10-7 mol/l 19-83

84 Sample Problem Calculating the Effect of Complex-Ion Formation on Solubility PROBLEM: In black-and-white film developing, excess AgBr is removed from the film negative by hypo, an aqueous solution of sodium thiosulfate (Na 2 S 2 O 3 ), which forms the complex ion Ag(S 2 O 3 ) Calculate the solubility of AgBr in (a) H 2 O; (b) 1.0 mol/l hypo. K f of Ag(S 2 O 3 ) 2 3- is 4.7x10 13 and K sp AgBr is 5.0x PLAN: After writing the equation and the K sp expression, we use the given K sp value to solve for S, the molar solubility of AgBr. For (b) we note that AgBr forms a complex ion with S 2 O 3 2-, which shifts the equilibrium and dissolves more AgBr. We write an overall equation for the process and set up a reaction table to solve for S

85 Sample Problem SOLUTION: (a) AgBr(s) Ag + (aq) + Br - (aq) K sp = [Ag + ][Br - ] = 5.0x10-13 S = [AgBr] dissolved = [Ag + ] = [Br - ] K sp = [Ag + ][Br - ] = S 2 = 5.0x10-13 S = 7.1x10-7 mol/l (b) Write the overall equation: AgBr(s) Ag + (aq) + 2S 2 O 3 2- (aq) AgBr(s) + 2S 2 O 3 2- (aq) Ag + (aq) + Br - (aq) Ag(S 2 O 3 ) 2 3- (aq) Br - (aq) + Ag(S 2 O 3 ) 2 3- (aq) K overall = K sp x K f = [Br- ][Ag(S 2 O 3 ) 2 3- ] [S 2 O 3 2- ] 2 = (5.0x10-13 )(4.7x10 13 ) =

86 Sample Problem Concentration (mol/l) AgBr(s) + 2S 2 O 2-3 (aq) Br - (aq) + Ag(S 2 O 3 ) 3-2 (aq) Initial Change - -2S +S +S Equilibrium S S S K overall = S 2 (1.0-2S) 2 = 24 so S 1.0-2S = = 4.9 S = 4.9 mol/l 0.9S and 10.9S = 4.9 mol/l S = [Ag(S 2 O 3 ) 2 3- ] = 0.45 mol/l 19-86

87 Figure The amphoteric behavior of aluminum hydroxide. When solid Al(OH) 3 is treated with H 3 O + (left) or with OH - (right), it dissolves as a result of the formation of soluble complex ions