Understanding Buffers and Acid-Base Equilibria: A Detailed Analysis, Study notes of Chemistry

Download Understanding Buffers and Acid-Base Equilibria: A Detailed Analysis and more Study notes Chemistry in PDF only on Docsity! Applications of Aqueous Equilibria Chapter 8 Chapter 8: Applications of Aqueous Equilibria 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions 8.4 Buffer Capacity 8.5 Titrations and pH Curves 8.6 Acid-Base Indicators 8.7 Titration of Polyprotic Acids 8.8 Solubility Equilibria and The Solubility Product 8.9 Precipitation and Qualitative Analysis 8.10 Complex Ion Equilibria Molecular model: F-, Na+, HF, H2O Like Example 8.1 (P 278-9) - I Nitrous acid, a very weak acid, is only 2.0% ionized in a 1.0 M solution. Calculate the [H+], the pH, and the percent dissociation of HNO2 in a 1.0 M solution that is also 1.0 M in NaNO2! HNO2(aq) H+(aq) + NO2-(aq) Ka = = 4.0 x 10-4 [H+] [NO2 -] [HNO2] Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [HNO2]0 = 1.0 M [HNO2] = 1.0 – x (from dissolved HNO2) [NO2-]0 = 1.0 M [NO2-] = 1.0 + x (from dissolved NaNO2) [H+]0 = 0 [H+] = x (neglect the contribution from water) Like Example 8.1 (P 274-5) - II Ka = = = 4.0 x 10-4 [H+] [NO2-] [HNO2] ( x ) ( 1.0 + x ) (1.0 – x ) Assume x is small as compared to 1.0: X (1.0) (1.0) = 4.0 x 10 -4 or x = 4.0 x 10-4 = [H+] Therefore pH = - log [H+] = - log ( 4.0 x 10-4 ) = 3.40 The percent dissociation is: 4.0 x 10-4 1.0 x 100 = 0.040 % Nitrous acid Nitrous acid alone + NaNO2 [H+] 2.0 x 10-2 4.0 x 10-4 pH 1.70 3.40 % Diss 2.0 0.040 Example 8.2 (P279-82) - III Initial Concentration (mol/L) Equilibrium concentration (mol/L) [HC2H3O2]0 = 0.49 [HC2H3O2] = 0.49 – x [C2H3O2-]0 = 0.51 [C2H3O2-] = 0.51 + x [H+]0 = 0 [H+] = x X mol/L of HC2H3O2 Dissociates to reach equilibrium Ka = 1.8 x 10-5 = = = [H+][C2H3O2-] [HC2H3O2] (x)(0.51+ x) 0.49 - x (x)(0.51) 0.49 x = 1.7 x 10-5 and pH = 4.76 If the base is added to pure water without the buffer being present we get an entirely different solution: If the 0.01 mol of NaOH is added to 1.0 L of pure water the Concentration of hydroxide ion is 0.01 M. [H+] = = = 1.0 x 1012 and the pH = 12.00Kw[OH-] 1.0 x 10-14 1.0 x 10-2 Acid lonization Equilibrium When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium calculations. This procedure can be represented as follows: How Does a Buffer Work Lets add a strong base to a weak acid and see what happens: OH- + HA A- + H2O Original buffer pH Final pH of buffer close to original- Added OH- ions Replaced by A- ions Ka = [H+] [A-] [HA] [H +] = Ka [HA] [A-] The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH3COOH] [CH3COO-]added % Dissociation* 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 = x 100 [CH3COOH]dissoc [CH3COOH]init Molecular model: HC,H,O,, C,H,0,- aa Bb 2. Bore VO g HC,H,0, Na* ~& ——— H,O Human blood is a buffered solution Source: Visuals Unlimited CO2 (g), H2CO3 (aq) and HCO3-(aq) are the buffering components in Blood that hold the pH to a range that will allow Hemoglobin to transport oxygen from the lungs to the cells of the body for metabolism. How a Buffer Works–I A buffer consists of a solution that contains “high” concentrations of the acidic and basic components. This is normally a weak acid and the anion of that weak acid, or a weak base and the corresponding cation of the weak base. When small quantities of H3O+ or OH- are added to the buffer, they cause a small amount of one buffer component to convert into the other. As long as the amounts of H3O+ and OH- are small as compared to the concentrations of the acid and base in the buffer, the added ions will have little effect on the pH since they are consumed by the buffer components. Consider a buffer made from acetic acid and sodium acetate: CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Ka = or [H3O+] = Ka x [CH3COO-] [H3O+] [CH3COOH] [CH3COOH] [CH3COO-] How a Buffer Works–II Let’s consider a buffer made by placing 0.25 mol of acetic acid and 0.25 mol of sodium acetate per liter of solution. What is the pH of the buffer? And what will be the pH of 100.00 mL of the buffer before and after 1.00 mL of concentrated HCl (12.0 M) is added to the buffer? What will be the pH of 100.00 mL of pure water if the same acid is added? [H3O+] = Ka x = 1.8 x 10-5 x = 1.8 x 10-5 [CH3COOH] [CH3COO-] (0.25) (0.25) pH = -log[H3O+] = -log(1.8 x 10-5) = pH = 4.74 Before acid added! 1.00 mL conc. HCl 1.00 mL x 12.0 mol/L = 0.012 mol H3O+ Added to 300.00 mL of water : 0.012 mol H3O+ 301.00 mL soln. = 0.0399 M H3O+ pH = -log(0.0399 M) pH = 1.40 Without buffer! How a Buffer Works–V By adding the 1.00mL base to 300.00 mL of pure water we would get a hydroxide ion concentration of: 301.00 mL 0.012 mol OH-[OH-] = = 3.99 x 10-5 M OH- This calculates out to give a pH of: The hydrogen ion concentration is: [H3O+] = = = 2.506 x 10-10 Kw [OH-] 1 x 10-14 3.99 x 10-5 M pH = -log(2.5 6 x 10-10) = 10.000 - 0.408 = 9.59 With 1.0 mL of the base in pure water! In summary: Buffer alone pH = 4.74 Buffer plus 1.0 mL base pH = 4.79 Base alone pH = 9.59 Buffer plus 1.0 mL acid pH = 4.70 Acid alone pH = 1.40 The Relation Between Buffer Capacity and pH Change Ea —_h, Oo 0.30 0.10 Concentration (M) of buffer components 0.030 (initial pH = 4.74) 4.8 pH | | T oT T | | | | T oT T 4.9 5.0 5.1 Molecular model: Cl-, NH4+ Like Example 8.3 (P 285-7) - II Substituting into the equation for Kb: Kb = = 1.8 x 10-5 = [NH4+] [OH-] [NH3] (0.040 + x) (x) (0.060 – x) Assume : 0.060 – x = 0.060 ; 0.040 + x = 0.040~~ Kb = 1.8 x 10-5 = x = 2.7 x 10-5 0.040 (x) 0.060 Check assumptions: 0.040 + 0.000027 = 0.040 or 0.068% 0.060 – 0.000027 = 0.060 or 0.045% [OH-] = 2.7 x 10-5 ; pOH = - log[OH-] = - log (2.7 x 10-5) = 5 – 0.43 pOH = 4.57 pH = 14.00 – pOH = 14.00 – 4.57 = 9.43 The Henderson-Hasselbalch Equation Take the equilibrium ionization of a weak acid: HA(aq) + H2O(aq) = H3O+(aq) + A-(aq) Ka = [H3O+] [A-] [HA] Solving for the hydronium ion concentration gives: [H3O+] = Ka x [HA] [A-] Taking the negative logarithm of both sides: -log[H3O +] = -log Ka - log( )[HA] [A-] pH = -log Ka - log( )[HA] [A-] Generalizing for any conjugate acid-base pair : pH = log Ka + log( )[base][acid] Henderson-Hasselbalchequation Summary: Characteristics of Buffered Solutions Buffered solutions contain relatively large concentrations of a weak acid and its corresponding weak base. They can involve a weak acid HA and the conjugate base A- or a weak base B and the conjugate acid BH+. When H+ is added to a buffered solution, it reacts essentially to completion with the weak base present: H+ + A- HA or H+ + B BH+ When OH- is added to a buffered solution, it reacts essentially to completion with the weak acid present. OH- + HA A- + H2O or OH- + BH+ B + H2O The pH of the buffered solution is determined by the ratio of the concentrations of the weak base and weak acid. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A- or G and BH+) are large compared with the amounts of H+ or OH- added. Example 8.4 (P 289-90) - I Calculate the pH of a buffered solution containing 3.0 x 10-4 M HOCl (Ka = 3.5 x 10-8) and 1.0 x 10-4 M NaOCl. Ka = = 3.5 x 10-8 [H+] [OCl-] [HOCl] Let x = [H+] then: [OCl-] = 1.0 x 10-4 + x [HOCl] = 3.0 x 10-4 - x 3.5 x 10-8 = =[H +] [OCl-] [HOCl] (x)(1.0 x 10-4 + x) (3.0 x 10-4 – x ) Assuming x is small compared to 1.0 x 10-4 and solving for x we have: [H+] = x = = 1.05 x 10-7 M = 1.1 x 10-7 M1.05 x 10 -11 1.0 x 10-4 Since this is close to that of water we must use the equation that uses water, and takes it’s ionization into account. Example 8.4 (P 289-90) - II Ka = 3.5 x 10-8 = [H+]{[OCl-]0 + } [H+]2 – 1.0 x 10-14 [H+] [H+]2 – 1.0 x 10-14 [HOCl]0 - [H+] Where: [OCl-]0 = 1.0 x 10-4 M [HOCl]0 = 3.0 x 10-4 M We expect [H+] to be close to 1.0 x 10-7, so [H+]2 to be about 1.0 x 10-14 [H+]2 – 1.0 x 10-14[OCl-]0 = 1.0 x 10-4 M >>> [H+] [H+]2 – 1.0 x 10-14[HOCl]0 = 1.0 x 10-4 M >>> [H+]The expression becomes: [H+][OCl-] [HOCl]3.5 x 10 -8 = = [H +](1.0 x 10-4) (3.0 x 10-4) [H+][OCl-] [HOCl]3.5 x 10 -8 = = [H +](1.0 x 10-4) (3.0 x 10-4) Example 8.4 (P 289-90) - III [H+] = 1.05 x 10-7 M = 1.1 x 10-7 M Using this result, we can check the magnitude of the neglected term: [H+]2 – 1.0 x 10-14 [H+] = (1.05 x 10-7)2 – 1.0 x 10-14 1.05 x 10-7 = 9.8 x 10-9 This result suggests that the approximation was fine! Original solution and new solution Original solution New solution H: [A-] 5.00 [A-] 4.99 —— = ——- = 1.00 = = 0.996 [HA] 5.00 added [HA] 5.01 Original solution New solution [A] _ .050_ 15 TA] _ 9.040 _ 947 [HA] .050 — [HA] 0.060 Example 8.5 (P 290-2) - II For Solution A: H+ + C2H3O2- H C2H3O2 Before reaction 0.010 M 5.00 M 5.00 M After reaction 0 4.99 M 5.01 M Calculate the new pH using the Henderson-Hasselbalch equation: pH = pKa + log ( ) = 4.74 + log ( ) = 4.74 – 0.0017 = 4.74 [C2H3O2-] [H C2H3O2] 4.99 5.01 For Solution B: H+ + C2H3O2- H C2H3O2 Before reaction 0.010 M 0.050 M 0.050 M After reaction 0 0.040 M 0.060 M The new pH is: pH = 4.74 + log( ) = 4.74 – 0.18 = 4.560.040 0.060 Figure 8.1: The pH curve for the titration of 50.0 ml of Nitric acid with 0.10M NaOH Weak acid Weak acid Strong acid Vol NaOH Figure 8.3: The pH curve for the titration of 50.0 ml of Acetic acid with 0.10 M NaOH Treat the stoichiometry Calculating the pH During a Weak Acid-Strong Base Titration–I Problem: Calculate the pH during the titration of 20.00 mL of 0.250 M nitrous acid (HNO2; Ka = 4.5 x 10-4) after adding different volumes of 0.150 M NaOH : (a) 0.00 mL (b) 15.00 mL (c) 20.00 mL (d) 35.00 mL. Plan: (a) We just calculate the pH of a weak acid. (b)-(d) We calculate the amounts of acid remaining after the reaction with the base, and the anion concentration, and plug these into the Henderson-Hasselbalch eq. Solution: HNO2 (aq) + NaOH(aq) H2O(l) + NaNO2 (aq) HNO2 (aq) + H2O(l) H3O+(aq) + NO2-(aq) (a) Ka = = = 4.5 x 10-4 [H3O+] [NO2-] [HNO2] x (x) 0.250 M x2 = 1.125 x 10-4 x = 1.061 x 10-2 pH = -log(1.061 x 10-2) = 2.000 - 0.0257 = 1.97 no base added pH = pKa + log = 3.35 + log(0.00225/0.00275) ( )[NO2-][HNO2] Calculating the pH During a Weak Acid-Strong Base Titration–II (b) 15.00 mL of 0.150 M NaOH is added to the 20.00 mL of 0.250 M HNO2 (20.00 mL x 0.250 mmol/mL = 5.00 mmol HNO2) which will neutralize (15.00 mL x 0.150 mmol/mL = 2.25 mmol of HNO2), leaving 2.75 mmol HNO2, and generating 2.25 mmol of nitrite anion. HNO2 (aq) + H2O(l) H3O+(aq) + NO2-(aq)Concentration (M) Initial 0.00275 ---- 0 0.00225 Change -x ---- +x +x Equilibrium 0.00275 - x ---- x 0.00225 + x pH = 3.35 -0.0872 = 3.26 with 15.0 mL of NaOH added Calculating the pH During a Weak Acid-Strong Base Titration–III (c) 20.00 mL of 0.150 M NaOH is added to the 20.00 mL of 0.250 M HNO2 (20.00 mL x 0.250 mmol/mL = 5.00 mmol HNO2) which will neutralize (20.00 mL x 0.150 mmol/mL = 3.00 mmol of HNO2), leaving 2.00 mmol HNO2, and generating 3.00 mmol of nitrite anion. HNO2 (aq) + H2O(l) H3O+(aq) + NO2-(aq)Concentration (M) Initial 0.00200 ---- x 0.00300 pH = pKa + log = 3.35 + log(0.00300/0.00200) ( )[NO2-][HNO2] pH = 3.35 + 0.176 = 3.53 with 20.00 mL of base added (d) 35.00 mL of 0.150 M NaOH is added to the 20.00 mL of 0.250 M HNO2 (20.00 mL x 0.250 mmol/mL = 5.00 mmol HNO2) which will neutralize (35.00 mL x 0.150 mmol/mL = 5.25 mmol of HNO2), leaving no HNO2, and generating 5.00 mmol of nitrite anion. There will be an excess of 0.25 mmol of NaOH which will control the pH. TABLE 8.2 Summary of Results for the Titration of 100.0 mL 0.050 M NH; with 0.10 M HCl Volume of 0.10 M HCl Added (mL) [NH3]o [NH4*]o ([H*] pH 0 0.050 M 0 11x10°''M 10.96 4.0 mmol 1.0 mmol TO i 4X ‘ ane (Won Comim i" “~ 7 : 2.5 mmol 2.5 mmol m0 .Q* 6% : 2s) (100 + 25) mL (100 + 25) mL St i a= 5.0 mmol - i SEE EESEREe x 6 50.0 0 (100 + 50) mL 43x10 °M 5.36 5.0 mmol 1.0 mmol + nae 60,0 ° (100 + 60) mL 160 mL 2. =62x103°M *Halfway point. tEquivalence point. '[H*] determined by the 1.0 mmol of excess H*. Figure 8.5: The pH curve for the titration of 100.0 ml of 0.050 M NH3 with ).10 M HCl weak Base: ‘a satretbeniot 20,00 mil ot 2-120 ENE rong Aci Titration 12} — INHy]=INH,*) Curve Ph = 5.27 at equivalence point 0 T T T T T T 1 10 20 30 40 50 #60 70 £80 Volume of HCI added (mL)  TABLE 8.3 Selected pH Values Near the Equivalence Point in the Titration of 100.0 mL of 0.100 M HCI with 0.100 M NaOH NaOH Added (mL) pH 0999 5.3 100.00 7.0 100.01 8.7 Colors and Approximate pH Range of Some Common Acid-Base Indicators Figure 8.9: pH curve of 0.10 M HCI being titrated with 0.10 M NaOH Figure 8.11: A summary of the important equilibria at various points in the titration of a triprotic acid TABLE 8.4 A Summary of Various Points in the Titration of a Triprotic Acid Major Species Equilibrium Expression Point in the Titration Present Used to Obtain the pH H*|[HoA No base added H;A, H,0 K = [H"}1H24 | Base added Before the first equivalence point At the first equivalence point Berween the first and second equivalence points At the second equivalence point Between the second and third equivalence points At the third equivalence point Bevond the third equivalence point H3A, H2A~, H20 H,A ,H,0 H»A~, HA?~, HO HA?-, H,0 HA?~, A?-, H,0 Ae”, H2O A®-, OH”, H30 “1 [HA] K= [H* JHA" | “1 [HA] See the following discussion x = 1 [HA*] [H)A"] See the following discussion ce |H* Ae : (HA*] y= Ku Kp Ka, _ [HA?|[OH- [Ay] pH determined by excess QH™ Precipitation of bismuth PbClo, a Slightly Soluble lonic Compound © MeGrew-Hill Higher Education’Stephen Frisch, photographer — Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds Problem: Write the ion-product expression for (a) silver bromide; (b) strontium phosphate; (c) aluminum carbonate; (d) nickel(III) sulfide. Plan: Write the equation for a saturated solution, then write the expression for the solubility product. Solution: (a) Silver bromide: AgBr(s) Ag+(aq) + Br -(aq) Ksp = [Ag+] [Br -] (b) Strontium phosphate: Sr3(PO4)(s) 3 Sr2+(aq) + 2 PO43-(aq) Ksp = [Sr2+]3[PO43-]2 (c) Aluminum carbonate: Al2(CO3)3 (s) 2 Al3+(aq) + 3 CO32-(aq) Ksp = [Al3+]2[CO32-]3 (d) Nickel(III) sulfide: Ni2S3 (s) + 3 H2O(l) 2 Ni3+(aq) + 3 HS -(aq) + 3 OH-(aq) Ksp =[Ni3+]2[HS-]3[OH-]3 Table 8.5: Ksp Values at 25 C for Common Ionic Solids Determining Ksp from Solubility Problem: Lead chromate is an insoluble compound that at one time was used as the pigment in the yellow stripes on highways. It’s solubility is 4.33 x 10 -6g/100mL water. What is the Ksp? Plan: We write an equation for the dissolution of the compound to see the number of ions formed, then write the ion-product expression. Solution: PbCrO4 (s) Pb2+(aq) + CrO42-(aq) Molar solubility of PbCrO4 = x x 4.33 x 10 -6g 100 mL 1000 ml 1 L 1mol PbCrO4 323.2 g = 1.34 x 10 -8 M PbCrO4 1 Mole PbCrO4 = 1 mole Pb2+ and 1 mole CrO42- Therefore [Pb2+] = [CrO42-] = 1.34 x 10-8 M Ksp = [Pb2+] [CrO42-] = (1.34 x 10 -8 M)2 = 1.54 x 10-16 Determining Solubility from Ksp Problem: Lead chromate used to be used as the pigment for the yellow lines on roads, and is a very insoluble compound. Calculate the solubility of PbCrO4 in water if the Ksp is equal to 2.00 x 10-16. Plan: We write the dissolution equation, and the ion-product expression. Solution: Writing the dissolution equation, and the ion-product expression: PbCrO4 (s) Pb2+(aq) + CrO42-(aq) Ksp = 2.00 x 10-16 =[Pb2+][CrO42] Concentration (M) PbCrO4 Pb2+ CrO42- Initial ---------- 0 0 Change ---------- +x +x Equilibrium ---------- x x Ksp = [Pb2+] [CrO42-] = (x)(x ) = 2.00 x 10-16 x = 1.41 x 10-8 Therefore the solubility of PbCrO4 in water is 1.41 x 10-8 M TABLE 8.6 Calculated Solubilities for CuS, Ag2S, and BizSz at 25°C Calculated Solubility Salt i (mol/L) @uSss 10) 9210 AgS 1.6x10°7 3.4x 1071’ BioS3; 1.110 7% 1.0x 10° The Effect of a Common Ion on Solubility PbCrO4(s) Pb2+(aq) + CrO42-(aq) PbCrO4(s) Pb2+(aq) + CrO42-(aq; added) Calculating the Effect of a Common Ion on Solubility Problem: What is the solubility of silver chromate in 0.0600 M silver nitrate solution? Ksp = 2.6 x 10-12 . Plan: From the equation and the ion-product expression for Ag2CrO4, we predict that the addition of silver ion will decrease the solubility. Solution: Writing the equation and ion-product expression: Ag2CrO4 (s) 2 Ag+(aq) + CrO42-(aq) Ksp = [Ag+]2[CrO42-] Concentration (M) Ag2CrO4 (s) 2 Ag+(aq) + CrO42-(aq) Initial --------- 0.0600 0 Change --------- +2x +x Equilibrium --------- 0.0600 + 2x x Assuming that Ksp is small, 0.0600 M + 2x = 0.600 M Ksp = 2.6 x 10-12 = (0.0600)2(x) x = 7.22 x 10-10 M Therefore, the solubility of silver chromate is 7.22 x 10-10 M 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: (a) Iron(II) cyanide (b) Potassium bromide (c) Aluminum hydroxide Plan: Write the balanced dissolution equation and note the anion. Anions of weak acids react with H3O+ and shift the equilibrium position toward more dissolution. Strong acid anions do not react, so added acid has no effect. Solution: (a) Fe(CN)2 (s) Fe2+(aq) + 2 CN-(a) Increases solubility We noted earlier that CN- ion reacts with water to form the weak acid HCN, so it would be removed from the solubility expression. (b) KBr(s) K+(aq) + Br -(aq) No effect This occurs since Br- is the anion of a strong acid, and K+ is the cation of a strong base. (c) Al(OH)3 (s) Al3+(aq) + 3 OH-(aq) Increases solubility The OH- is the anion of water, a very weak acid, so it reacts with the added acid to produce water in a simple acid-base reaction. Predicting the Formation of a Precipitate: Qsp vs. Ksp Qsp = Ksp : When a solution becomes saturated, no more solute will dissolve, and the solution is called “saturated.” There will be no changes that will occur. Qsp > Ksp : Precipitates will form until the solution becomes saturated. Qsp< Ksp : Solution is unsaturated, and no precipitate will form. The solubility produce constant, Ksp, can be compared to the ion-product constant, Qsp to understand the characteristics of a solution with respect to forming a precipitate. Predicting Whether a Precipitate Will Form–I Problem: Will a precipitate form when 0.100 L of a solution containing 0.055 M barium nitrate is added to 200.00 mL of a 0.100 M solution of sodium chromate? Plan: We first see if the solutions will yield soluble ions, then we calculate the concentrations, adding the two volumes together to get the total volume of the solution, then we calculate the product constant (Qsp), and compare it to the solubility product constant to see if a precipitate will form. Solution: Both Na2CrO4 and Ba(NO3) are soluble, so we will have Na+, CrO42-, Ba2+ and NO3- ions present in 0.300 L of solution. We change partners, look up solubilities, and we find that BaCrO4 would be insoluble, so we calculate it’s ion-product constant and compare it to the solubility product constant of 2.1 x 10-10: [Ba2+] = = 0.183 M in Ba2+ For Ba2+: [0.100 L Ba(NO3)2] [0.55 M] = 0.055mol Ba2+ 0.055 mol Ba2+ 0.300 L Predicting Whether a Precipitate Will Form–II Solution cont. For CrO42- : [0.100 M Na2CrO4] [0.200 L] = 0.0200 mol CrO42- [CrO42-] = = 0.667 M in CrO42- 0.0200 mol CrO42- 0.300 liters Qsp = [Ba2+] [CrO42-] =(0.183 M Ba2+)(0.667 M CrO42-) = 0.121 Since Ksp = 2.1 x 10-10 and Qsp = 0.121, Qsp >> Ksp and a precipitate will form. The Stepwise Exchange of NH, for H,O in M(H,O),2* 3NH, = M(H,0),’*(aq) + NH,(aq) == M(H, 0), (NH.)"*(aq) = M(NH,,), "(aq) + 4H, O(/) 3 more steps Formation Constants (Kf) of Some Complex Ions at 25oC–I Complex Ion Kf Ag(CN)2- 3.0 x 1020 Ag(NH3)2+ 1.7 x 107 Ag(S2O3)23- 4.7 x 1013 AlF63- 4 x 1019 Al(OH)4- 3 x 1033 Be(OH)42- 4 x 1018 CdI42- 1 x 106 Co(OH)42- 5 x 109 Cr(OH)4- 8.0 x 1029 Cu(NH3)42+ 5.6 x 1011 Fe(CN)64- 3 x 1035 Fe(CN)63- 4.0 x 1043 Formation Constants (Kf) of Some Complex Ions at 25oC–II Complex Ion Kf Hg(CN)42- 9.3 x 1038 Ni(OH)42- 2 x 1028 Pb(OH)3 - 8 x 1013 Sn(OH)3 - 3 x 1025 Zn(CN)42- 4.2 x 1019 Zn(NH3)42+ 7.8 x 108 Zn(OH)42- 3 x 1015 The Amphoteric Behavior of Aluminum Hydroxide 3H,07 3H,0(/) + AI(H,0}2" (aq) AKH30)3(OH)a(s) AW(H20)2(OH)r (ag) + H2O(/) Separating Ions by Selective Precipitation–I Problem: A solution consists of 0.10 M AgNO3 and 0.15 M Cu(NO3)2. Calculate the [I -] that would separate the metal ions as their iodides. Kspof AgI = 8.3 x 10-17; Kspof CuI = 1.0 x 10-12. Plan: Since the two iodides have the same formula type (1:1), we compare their Ksp values and we see that CuI is about 100,000 times more soluble than AgI. Therefore, AgI precipitates first, and we solve for [I -] that will give a saturated solution of AgI. Solution: Writing chemical equations and ion-product expressions: AgI(s) Ag+(aq) + I -(aq) Ksp = [Ag+][I -] CuI(s) Cu+(aq) + I -(aq) Ksp = [Cu+][I -] H2O H2O Calculating the quantity of iodide needed to give a saturated solution of CuI: [I -] = = = 6.7 x 10-11 M Ksp [Cu+] 1.0 x 10-12 0.15 M Separating Ions by Selective Precipitation–II Thus, the concentration of iodide ion that will give a saturated solution of copper(I) iodide is 1.0 x 10-11 M, and this will not precipitate the copper(I) ion, but should remove most of the silver ion. Calculating the quantity of silver ion remaining in solution we get: [Ag+] = = = 1.2 x 10-6 M Ksp [I -] 8.3 x 10-17 6.7 x 10-11 Since the initial silver ion was 0.10 M, most of it has been removed, and essentially none of the copper(I) was removed, so the separation was quite complete. If the iodide was added as sodium iodide, you would have to add only a few nanograms of NaI to remove nearly all of the silver from solution: 6.7 x 10-11 mol I - x x = 10.0 ng NaI 149.9 g NaI mol NaI 1 molNaI mol I -