Copyright The McGraw-Hill Companies, Inc. Permission required for

Copyright The McGraw-Hill Companies, Inc. Permission required for

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 19 Ionic Equilibria in Aqueous Systems 19-1 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Ionic Equilibria in Aqueous Systems 19.1 Equilibria of Acid-Base Buffer Systems 19.2 Acid-Base Titration Curves 19.3 Equilibria of Slightly Soluble Ionic Compounds 19.4 Equilibria Involving Complex Ions 19.5 Application of Ionic Equilibria to Chemical Analysis

19-2 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Buffered solutions A solution that resists a change in its pH when either OH- or H+ ions are added. Ex: Blood, absorbs acid and bases without change in pH. The components of a buffer are the conjugate acid-base pair of a weak acid. WA and its salt-HF and NaF WB and its salt- NH3 and NH4Cl. 19-3 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.1 The effect of addition of acid or base to

acid added Figure 19.2 an unbuffered solution acid added 19-4 base added or a buffered solution base added Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Common ion effect The shift in equilibrium position that occurs

because of the addition of an ion already involved in equilibrium reaction is called common ion effect. Consider a weak acid HF and its salt NaF. NaF (s) _______ Na+ (aq)+ F- (aq) , Major species: Na+ F- HF H2O HF (aq) H+ (aq) + F- (aq) 19-5 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Common ion effect F- is the common ion. F- from the added NaF moves the equ position to the left according to LC principle. If we dissolve HF in a NaF solution, the Fion and H+ ion enter the solution. The Fion already present combines with the H+, which lowers the H+. 19-6

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Buffered Solution Characteristics Buffers contain relatively large amounts of weak acid and corresponding base. Added H+ is consumed by A Added OH is consumed by HA The pH is determined by the ratio of the concentrations of the weak acid and its conjugate base. 19-7 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 19.1 The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH3COOH]initial [CH3COO-]added

pH 0.10 0.00 1.3 2.89 0.10 0.050 0.036 4.44 0.10

0.10 0.018 4.74 0.10 0.15 0.012 4.92 * % Dissociation = [CH3COOH]dissoc [CH3COOH]initial 19-8

% Dissociation* x 100 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.3 How a buffer works. Buffer after addition of H3O+ Buffer with equal concentrations of conjugate base and acid H3O+ H2O + CH3COOH 19-9

H3O+ + CH3COO- Buffer after addition of OH- OH- CH3COOH + OH- H2O + CH3COO- Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Problem P.1. A buffered solution contains 0.50 M acetic acid (Ka= 1.8 x 10-5) and 0.50 M sodium acetate. a) Calculate pH of this solution. b) Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of buffered solution described in the above problem.Compare this pH change with the one that occurs when

0.010 mol solid NaOH is added to 1.0 L of water. c) Calculate the change in pH that occurs when 0.010 mol solid HCl is added to 1.0 L of buffered solution. 19-10 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Henderson-Hasselbalch Equation HA + H2O Ka = H3O+ + A[H3O+] [A-] [HA] [H3O+] = Ka [HA] [A-] - log[H3O+] = - log Ka + log

pH = pKa + log 19-11 [A-] [HA] [base] [acid] Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Problems P. 2. Calculate the pH of a solution 0.75 M lactic acid (Ka= 1.4 x10-4 ) and 0.25 M sodium lactate. P.3. A buffer solution contains 0.25 M NH3 (Kb= 1.8 x 10-5) and 0.40 M NH4Cl. Calculate pH of this solution.

19-12 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Buffering Capacity It represents the amount of H+ or OH the buffer can absorb without a significant change in pH. More concentrated the components of a buffer , the greater the buffer capacity. pH of a buffered solution is determined by the ratio [A-]/[HA] . If an acid or a base is added , the concentration ratio changes less when the buffer component concentrations are similar than they are different. 19-13 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Buffering Capacity: Buffer has highest capacity when

component concentrations are equal. pH=pKa highest buffer capacity. The Buffer Range is the pH range over which the buffer works effectively. The further the buffering component concentration ratio is from 1 less effective is the buffering action. 19-14 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Buffer Capacity and Buffer Range Buffer capacity is the ability to resist pH change. The more concentrated the components of a buffer, the greater the buffer capacity. The pH of a buffer is distinct from its buffer capacity. A buffer has the highest capacity when the component concentrations are equal. Buffer range is the pH range over which the buffer acts effectively. Buffers have a usable range within 1 pH unit of the pKa of

its acid component. 19-15 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.4 19-16 The relation between buffer capacity and pH change. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Preparing a Buffer

19-17 Choose the conjugate acid-base pair. Ratio of concentrations should be close to1. pH=pKa Calculate ratio of buffer component concentrations. Use the Henderson Hasselbalch equation. Determine the buffer concentration. Higher the buffer concentrations greater the buffer capacity. Mix the solutions and adjust the pH by adding strong acid or strong base with the help of a pH probe. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Preparing a Buffer 1. Choose the conjugate acid-base pair.

2. Calculate the ratio of buffer component concentrations. 3. Determine the buffer concentration. 4. Mix the solution and adjust the pH. 19-18 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Problem P.4. A chemist needs a solution buffered at pH= 4.30 and can choose from the following acids and their sodium salts: 1. chloroacetic acid Ka= 1.35 x 10-3 2. 2. propanoic acid Ka= 1.3 x 10-5 3. benzoic acid Ka= 6.4 x 10-5 4. hypochlorous acid. Ka= 3.5 x 10-8 Calculate the ratio [HA]/[A-] required for each system to yield a pH of 4.30, which system will

work the best? 19-19 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Acid base Titration Curves A plot of pH of the solution being analyzed as a function of the amount of titrant added. 19-20 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Acid-Base Indicators: It marks the end point of a titration by changing color. Ex: phenolphthalein is colorless in its HIn

form and pink in In- form, or basic form. Indicator changes color over a range of about 2 pH units. 19-21 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.5 Colors and approximate pH range of some common acid-base indicators. pH 19-22 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.6 The color change of the indicator bromthymol blue. basic

acidic 19-23 change occurs over ~2pH units Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Strong Acid- Strong Base Titration curves: pH is low initially, as base is added the pH increases slowly. The pH rises steeply when the moles of OH- nearly equals the moles of H3O+ The additional drop of base neutralizes the tiny excess acid and introduces a tiny

excess of base. Then pH increases smoothly as more base is added. 19-24 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Equivalence (stoichiometric) point Enough titrant has been added to react exactly with the solution being analyzed. Equivalence point is nearly the vertical portion of the curve. The point at which the number of moles of added OH- =the number of moles of H3O+ For a SA vs SB titration pH =7(neutral ions) 19-25 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

End point It occurs when indicator changes color. The indicator chosen should be close to the equivalence point. 19-26 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.7 19-27 Curve for a strong acid-strong base titration Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Calculating pH: Initial pH is the pH of the acid.

Before the equ. Point find initial moles present, moles reacted, change in volume and then the pH. At the equ point, pH =7 After the equ point find the excess of OH- present and then calculate pH. 19-28 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Problem P.5. 50.0 mL of 0.200 m HNO3 is titrated with 0.100M NaOH. Calculate the pH of the solution at selected points during the course of titration, where 0 mL, 20.0 mL, 100.0 mL, 150.0 mL of 0.100 M NaOH has been added. 19-29

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Weak acid-Strong Base Titrations: Initial pH is high as weak acid dissociates slightly. Buffer Region: A gradual rising portion of the curve appearing before the equ.point. At midpoint of the buffer region pH=pKa. pH at equ point is > 7. Beyond equ point pH increases slowly as excess OH- is added. 19-30 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.8 Titration of 40.00mL of 0.1000M HPr with

0.1000M NaOH Curve for a weak acidstrong base titration pKa of HPr = 4.89 pH = 8.80 at equivalence point [HPr] = [Pr-] 19-31 methyl red Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Calculating pH 1. When only HA present, Use ICE table

calculate H+ and pH like a weak acid type calculation. 2. As we add base, find change in moles and solve for H+ 3. At equ. Point pH >7, Find Kb from Ka and then calculate OH- and pH. 4. Beyond the equ. point , calculate excess of OH- present and calculate pH. 19-32 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Problem P.6. 50.0 mL of 0.1 M acetic acid (Ka=1.8 x 10-5) with 0.10 M NaOH. Calculate the pH at various points representing volumes of 0mL, 10.0 mL ,

25.0 mL, 50.0 mL , 60.0 mL of added NaOH. 19-33 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Weak base strong acid Titration curves: Same shape curve as WA vs SB , but inverted. Initially pH above 7 as it is a weak base. pH decreases in buffer region. At midpoint pH=pKa

After buffer region curve drops vertically to equ.point . pH at equ.point is below 7 Beyond equ point , pH decreases slowly as more acid is added. 19-34 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.9 Titration of 40.00mL of 0.1000M NH3 with 0.1000M HCl pKa of NH4+ = 9.25 Curve for a weak basestrong acid titration

19-35 pH = 5.27 at equivalence point Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Problems P.7. 100.0 mL of 0.050 M NH3 is titrated with 0.10 M HCl. Calculate the pH at various points: 1) before adding HCl 2) Before equ.point 3) At equ point (setup) P.8 If 2.00 mmol of solid acid in 100.0 mL water is titrated with 0.0500 M NaOH. After 20.0 mL NaOH has been added the pH is 6.00. What is the Ka value for acid? 19-36 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Titration curves of polyprotic acids: Number of curves =Number of H+ ions. 19-37 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.10 Curve for the titration of a weak polyprotic acid. pKa = 7.19 pKa = 1.85 Titration of 40.00mL of 0.1000M H2SO3 with 0.1000M NaOH 19-38

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 19.3 Calculating the pH During a Weak AcidStrong Base Titration PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000M NaOH: (a) 0.00mL (b) 30.00mL (c) 40.00mL (d) 50.00mL PLAN: The amounts of HPr and Pr- will be changing during the titration. Remember to adjust the total volume of solution after each addition. SOLUTION: (a) Find the starting pH using the methods of Chapter 18. Ka = [Pr-][H3O+]/[HPr] [Pr-] = x = [H3O+] x (1.3x10 5)(0.10)

(b) 19-39 [Pr-] = x = [H3O+] x = 1.1x10-3 ; pH = 2.96 Amount (mol) HPr(aq) + OH-(aq) Before addition Addition 0.04000 0.03000 After addition 0.01000

0 Pr-(aq) + H2O (l) 0 0.03000 - Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 19.3 Calculating the pH During a Weak AcidStrong Base Titration continued [H3O+] = 1.3x10-5 0.001000 mol 0.003000 mol

= 4.3x10-6M pH = 5.37 (c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr -] will be (0.004000 mol) = 0.05000M (0.004000L) + (0.004000L) Ka x Kb = Kw [H3O+] = Kw / Kb = Kw/Ka = 1.0x10-14/1.3x10-5 = 7.7x10-10 K bx[Pr =] 1.6x10-9M pH = 8.80 (d) 50.00mL of NaOH will produce an excess of OH-. mol XS base = (0.1000M)(0.05000L - 0.04000L) = 0.00100mol [H3O+] = 1.0x10-14/0.01111 = 9.0x10-11M pH = 12.05

19-40 M = (0.00100) (0.0900L) M = 0.01111 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.11 19-41 Sickle shape of red blood cells in sickle cell anemia. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Equilibria of slightly soluble ionic compounds:

An ionic substance dissociates completely in water to form hydrate cations and anions. Both the forward and reverse reactions take place. A stage is reached when no more solid dissolves. This is called as dynamic equilibrium. CaF2 (s) Ca2+ (aq) + 2F- (aq) Ksp= [Ca2+ ] [F-]2 Ksp is called as the solubility product constant or solubility product for the equilibrium expression. 19-42 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Equilibria of slightly soluble

ionic compounds: For the hypothetical compound, MpXq At equilibrium Qsp = [Mn+]p [Xz-]q= Ksp Higher the Ksp greater the solubility for formulas containing same total number of ions. If salts being compared produce different number of ions. Compare their solubilities, which is opposite of Ksp 19-43 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Ion-Product Expression (Qsp)

and Solubility Product Constant (Ksp) For the hypothetical compound, MpXq At equilibrium 19-44 Qsp = [Mn+]p [Xz-]q = Ksp Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 19.4 Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds PROBLEM: Write the ion-product expression for each of the following: (a) Magnesium carbonate (c) Calcium phosphate (b) Iron (II) hydroxide

(d) Silver sulfide PLAN: Write an equation which describes a saturated solution. Take note of the sulfide ion produced in part (d). SOLUTION: (a) MgCO3(s) Mg2+(aq) + CO32-(aq) (b) Fe(OH)2(s) Fe2+(aq) + 2OH- (aq) (c) Ca3(PO4)2(s) (d) Ag2S(s) 3Ca2+(aq) + 2PO43-(aq) Ksp = [Fe2+][OH-] 2 Ksp = [Ca2+]3[PO43-]2

2Ag+(aq) + S2-(aq) S2-(aq) + H2O(l) Ag2S(s) + H2O(l) 19-45 Ksp = [Mg2+][CO32-] HS-(aq) + OH-(aq) Ksp = [Ag+]2[HS-][OH-] 2Ag+(aq) + HS-(aq) + OH-(aq) Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 19.2 Solubility-Product Constants (K sp) of Selected Ionic Compounds at 250C Name, Formula

19-46 Ksp Aluminum hydroxide, Al(OH)3 3 x 10-34 Cobalt (II) carbonate, CoCO3 1.0 x 10-10 Iron (II) hydroxide, Fe(OH)2 4.1 x 10-15 Lead (II) fluoride, PbF2 3.6 x 10-8 Lead (II) sulfate, PbSO4

1.6 x 10-8 Mercury (I) iodide, Hg2I2 4.7 x 10-29 Silver sulfide, Ag2S 8 x 10-48 Zinc iodate, Zn(IO3)2 3.9 x 10-6 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Problems P.10. CuBr has a measured solubility of 2.0 x 10-4 mol/L at 25 C. Calculate Ksp.

P.11. Calculate Ksp value for Bi2S3 which has a solubility of 1.0 x 10-15 mol/L at 25 C. P.12. Calculate the solubility of copper (II) iodate at 25 C, if Ksp = 1.4 x 10-7 19-47 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Table 19.3 Relationship Between Ksp and Solubility at 250C No. of Ions 19-48 Formula Cation:Anion Ksp

Solubility (M) 2 MgCO3 1:1 3.5 x 10-8 1.9 x 10-4 2 PbSO4 1:1 1.6 x 10-8 1.3 x 10-4

2 BaCrO4 1:1 2.1 x 10-10 1.4 x 10-5 3 Ca(OH)2 1:2 5.5 x 10-6 1.2 x 10-2

3 BaF2 1:2 1.5 x 10-6 7.2 x 10-3 3 CaF2 1:2 3.2 x 10-11 2.0 x 10-4 3

Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Common ion Effect: Presence of a common ion decreases solubility of a slightly soluble ionic compound. 19-49 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Figure 19.12 The effect of a common ion on solubility CrO42- added PbCrO4(s) 19-50 Pb2+(aq) + CrO42-(aq) PbCrO4(s) Pb2+(aq) + CrO42-(aq) Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Problem .13. Calculate the solubility of CaF2

(Ksp=4.0 x 10-11) in a 0.025 M NaF solution. 19-51 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.13 19-52 Test for the presence of a carbonate. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. pH and solubility: If the compound contains the anion of a weak acid, addition of H3O+ increases its solubility.

19-53 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 19.8 Predicting the Effect on Solubility of Adding Strong Acid PROBLEM: Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of these ionic compounds: (a) Lead (II) bromide (b) Copper (II) hydroxide (c) Iron (II) sulfide PLAN: Write dissolution equations and consider how strong acid would affect the anion component. SOLUTION: (a) PbBr2(s)

Pb2+(aq) + 2Br-(aq) Br- is the anion of a strong acid. No effect. (b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) OH- is the anion of water, which is a weak acid. Therefore it will shift the solubility equation to the right and increase solubility. (c) FeS(s) Fe2+(aq) + S2-(aq) FeS(s) + H2O(l) S2- is the anion of a weak acid and will react with water to produce OH-. Fe2+(aq) + HS-(aq) + OH-(aq) Both weak acids serve to increase the solubility of FeS.

19-54 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Predicting formation of precipitate: If Qsp > Ksp ppt occurs If Qsp < Ksp solution is unsaturated, no ppt occurs. If Qsp = Ksp solution is saturated and no change occurs. 19-55 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 19.9 PROBLEM: PLAN:

Predicting Whether a Precipitate Will Form A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO3)2 is mixed with 0.200L of 0.060M NaF? Write out a reaction equation to see which salt would be formed. Look up the Ksp valus in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < Ksp. Remember to consider the final diluted solution when calculating concentrations. SOLUTION: CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = 3.2x10-11

mol Ca2+ = 0.100L(0.30mol/L) = 0.030mol [Ca2+] = 0.030mol/0.300L = 0.10M mol F- = 0.200L(0.060mol/L) = 0.012mol [F-] = 0.012mol/0.300L = 0.040M Q = [Ca2+][F-]2 = (0.10)(0.040)2 = 1.6x10-4 Q is >> Ksp and the CaF2 WILL precipitate. 19-56 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Equilibria Involving Complex Ions: Complex ion: A central metal ion

covalently bonded to two or more anions, or molecules called ligands. Ligand: Lewis base that contains a lone pair of electron that can be donated to an empty orbital on the metal ion to form a covalent bond. Ex: H2O, NH3, CN-,Cl19-57 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Equilibria Involving Complex Ions: Coordination number: Number of ligands attached to a metal ion.ex; 6,4,2 Formation constants/Stability Constants : Metal ions add ligands one at a time in steps. A ligand increases the solubility of a

slightly soluble ionic compound if it forms a complex ion with the cation. 19-58 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.14 19-59 Cr(NH3)63+, a typical complex ion. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.15 The stepwise exchange of NH3 for H2O in M(H2O)42+. NH3 M(H2O)42+

3NH3 M(H2O)3(NH3)2+ M(NH3)42+ 19-60 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 19-61 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 19.10 Calculating the Concentration of a Complex Ion PROBLEM: An industrial chemist converts Zn(H2O)42+ to the more stable Zn(NH3)42+ by mixing 50.0L of 0.0020M Zn (H2O)42+ and 25.0L of

0.15M NH3. What is the final [Zn (H2O)42+]? Kf of Zn(NH3)42+ is 7.8x108. PLAN: Write the reaction equation and Kf expression. Use a reaction table to list various concentrations. Remember that components will be diluted when mixed as you calculate final concentrations. It is obvious that there is a huge excess of NH3 and therefore it will drive the reaction to completion. SOLUTION: Zn(H2O)42+(aq) + 4NH3(aq) Kf = [Zn(NH3)42+] Zn(NH3)42+(aq) + 4H2O(l) [Zn(H2O)42+]initial = (50.0L)(0.0020M) [Zn(H2O)42+][NH3]4 [NH3]initial =

75.0L (25.0L)(0.15M) 75.0L 19-62 = 1.3x10-3 M = 5.0x10-2 M Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 19.10 Calculating the Concentration of a Complex Ion continued Since we assume that all of the Zn(H2O)42+ has reacted, it would use 4 times its amount in NH3. [NH3]used = 4(1.3x10-3M) = 5.2x10-3M [Zn(H2O)42+]remaining = x(a very small amount)

Concentration(M) Zn(H2O)42+(aq) + 4NH3(aq) Initial Change 1.3x10-3 ~(-1.3x10-3) Equilibrium Kf = [Zn(NH3)42+] [Zn(H2O)42+][NH3]4 19-63 5.0x10-2 ~(-5.2x10-3)

x 4.5x10-2 = 7.8x10 = 8 (1.3x10-3) x(4.5x10-2) Zn(NH3)42+(aq) + 4H2O(l) 0 - ~(+1.3x10-3) - 1.3x10-3

- x = 4.1x10-7M Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 19.11 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 (Na2S2O3), through formation of the complex ion Ag(S2O3)23-. Calculate the solubility of AgBr in (a) H2O; (b) 1.0M hypo. Kf of Ag(S2O3)23- is 4.7x1013 and Ksp AgBr is 5.0x10-13. PLAN: Write equations for the reactions involved. Use Ksp to find S, the molar solubility. Consider the shifts in equilibria upon the addition of the complexing agent. SOLUTION:

AgBr(s) Ag+(aq) + Br-(aq) (a) S = [AgBr]dissolved = [Ag+] = [Br-] (b) 19-64 AgBr(s) Ksp = [Ag+][Br-] = 5.0x10-13 Ksp = S2 = 5.0x10-13 ; S = 7.1x10-7M Ag+(aq) + Br-(aq) Ag+(aq) + 2S2O32-(aq) Ag(S2O3)23-(aq) AgBr(s) + 2S2O32-(aq)

Br -(aq) + Ag(S2O3)23-(aq) Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 19.11 Calculating the Effect of Complex-Ion Formation on Solubility continued Koverall = Ksp x Kf = Concentration(M) [Br-][Ag(S2O3]23[AgBr][S2O32-]2 AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)23-(aq)

Initial - 1.0 0 0 Change - -2S +S +S

Equilibrium - 1.0-2S S S Koverall = S2 (1.0-2S)2 S = 24 1.0-2S

S = [Ag(S2O3)23-] = 0.45M 19-65 = (5.0x10-13)(4.7x1013) = 24 = (24)1/2 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.16 The amphoteric behavior of aluminum hydroxide. 3H2O(l) + Al(H2O)3(OH)3(s) 19-66 Al(H2O)3(OH)3(s) Al(H2O)3(OH)4-(s) + H2O(l)

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 19.12 Separating Ions by Selective Precipitation PROBLEM: A solution consists of 0.20M MgCl2 and 0.10M CuCl2. Calculate the [OH-] that would separate the metal ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of Cu(OH)2 is 2.2x10-20. PLAN: Both precipitates are of the same ion ratio, 1:2, so we can compare their Ksp values to determine which has the greater solubility. It is obvious that Cu(OH)2 will precipitate first so we calculate the [OH-] needed for a saturated solution of Mg(OH)2. This should ensure that we do not precipitate Mg(OH)2. Then we can check how much Cu2+ remains in solution. SOLUTION: Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = 6.3x10-10 Cu(OH)2(s)

Cu2+(aq) + 2OH-(aq) [OH-] needed for a saturated Mg(OH)2 solution = Ksp = 2.2x10-20 6.3x10 10 2 0.20 [Mg ] K sp = 5.6x10-5M 19-67 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 19.12 Separating Ions by Selective Precipitation

continued Use the Ksp for Cu(OH)2 to find the amount of Cu remaining. [Cu2+] = Ksp/[OH-]2 = 2.2x10-20/(5.6x10-5)2 = 7.0x10-12M Since the solution was 0.10M CuCl2, virtually none of the Cu2+ remains in solution. 19-68 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 19.17 The general procedure for separating ions in qualitative analysis. 19-69 Add precipitating ion

Centrifuge Centrifuge Add precipitating ion Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A qualitative analysis scheme for separating cations into five ion groups. 19-70 Add (NH4)2HPO4 Centrifuge Add

NH3/NH4+ buffer(pH 8) Centrifuge Centrifuge Add 6M HCl Acidify to pH 0.5; add H2S Centrifuge Figure 19.18 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A qualitative analysis scheme for Ag+,Al3+,Cu2+, and Fe3+

Centrifuge Step 1 Add NH3(aq) Step 2 Add HCl Centrifuge Extra: Step 3 Add NaOH Centrifuge Step 4 Add HCl, Na2HPO4

19-71 Step 5 Dissolve in HCl and add KSCN Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B19.1 19-72 A view inside Carlsbad Caverns, New Mexico Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B19.3 19-73

Formation of acidic precipitation. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B19.4 A forest damaged by acid rain 19-74 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B19.5 The effect of acid rain on marble statuary. 1944 1994 Location: New York City

19-75

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