1. CH. 17 ACID -- BASE EQUILIBRA & BUFFERS 17.1 Common ion effect 17.1 Calculation Henderson- Hasselbalch eqn Buffers how works ion effect pH range 17.3 Titration Curves indicators SA/SB - SA/WB WA/SB - Poly Calculation pH 17.4 Solubility Calculation Ksp (Qsp) precipation 17.5 Solubility; pH Complex Ions 17.6 &.7 Ion Groups Separation

2. COMMON ION EFFECT The shift in the position of an equilibrium on addition of a subst that provides an ion in common w / one of the ions already involved in the equilibrium process. Also, decr solubility of soluble aqueous portion of ionic cmpd Dissolve HNO 2 in NaNO 2 soln (H 2 O ) LeChatilier add NO 2 - ; from NaNO 2 H 3 O + + NO 2 - H 3 O + lowers pH; shifts away NO 2 - already formed

3. BUFFERS resist  es in pH composed of: WA & CB or WB & CA Add sm amt base to buffer soln Acidic component neutralize base added BIO SYS: blood pH 7.4 control buffer of H 2 O / HCO 3 - Add sm amt acid to buffer soln Basic component neutralize acid added

4. Fig. 16.7 pg 664

5. HENDERSON -- HASSELBALCH To prep buffer soln select WA w / pK a close to  1 pH unit of buffer soln How pH affects % dissoc WA pH of buffer soln not depend on soln vol but on pK a & molar amt WA-CB EX. Titrant 0.100 M NaOH and neutralize 40.0 ml of 0.100 M butanoic acid (HC 4 H 7 O 2 ) K a = 1.54*10 -4 Find pH add 20.00 ml of titrant Find pH at equivalence pt & determine best indicator to use

6. 1 st determine # mmol HC 4 H 7 O 2 present mol HC 4 H 7 O 2 = 40.0 mL * (0.100 mmol / 1 mL) = 4.00 mmol need 4.00 mmol NaOH to reach equiv pt means need 40.0 ml of 0.100 M NaOH b) Add 20.00 ml NaOH, also midpoint so pH = pKa moles WA: 4.00 mmol mol NaOH added: 20.00 ml*0.100 M= 2.00 mmol 2.0 mmol / 60 ml = 0.0033 M ratio of [HC 4 H 7 O 2 ] / [C 4 H 7 O 2 - ] = 0.033 / 0.033 = 1 a) No base, NaOH, added; butanoic WA: K a = x 2 / [HA] x 2 = (1.54*10 -4 )*(0.100) =.00392 pH = -Log(0.00392) = 2.41

7. c) Add 40.0 ml NaOH @ equiv pt All HC 4 H 7 O 2 neutralized, find [C 4 H 7 O 2 - ] [C 4 H 7 O 2 - ] = mmol / tot mL pH = -Log(5.56*10 -9 ) = 8.25 indicator: PHENOLPHTHALEIN mmol C 4 H 7 O 2 - = 40.00 ml * (0.100 mmol / ml) = 4.0 mmol [C 4 H 7 O 2 - ] = 4.0 mmol / 80.0 ml = 0.05 M

8. TITRATION CURVES Monoprotic Acids pH Vol SB 7 SA - SB

9. TITRATION CURVES Monoprotic Acids pH Vol SB 9 WA - SB

10. TITRATION CURVES Monoprotic Acids pH Vol SA 3.5 SA - WB

11. TITRATION CURVES Polyprotic Acids pH Vol Base H 2 SO 4 HSO 4 - pK a1 SO 4 -2 pK a2

12. Problem 25.0-mL of 0.145 M HCl soln is titrated by 0.200 M KOH. How many mL of base must be added to reach the end point. Want: mL KOH Data: 25.0-mL HCl @ 0.145 mol/L 0.200 mol/L KOH HCl + KOH ---  H 2 O + KCl 1 mol HCl = 1 mol KOH

13. Problem What is the pH at each point in the titration of 25.00-mL of 0.100 M HAc with 0.100 M NaOH? a) before NaOH added b) after 10.00-mL c) after 12.50-mL d) after 25.00-mL (these are volumes added to original 25 ml) a. Find initial [H 3 O + ]; use Ka (1.8*10 -5 ) HC 2 H 3 O 2 + H 2 O  -  H 3 O + + C 2 H 3 O 2 - Eq: 0.100-x x x = 1.3*10 -3 pH = -log(1.3*10 -3 ) = 2.89 b. 35-mL tot vol; amt neutralize= 0.025 L*(0.100 mol/L) = 0.0025 mol base added = 0.01 L*(0.100 M) = 0.001 mol HC 2 H 3 O 2 + OH  -  H 2 O + C 2 H 3 O 2 - I 0.0025 0 C -0.001 +0.001 E 0.0015 0.001 Convert to M using vol:.0015/.035 =.0429 M.001/.035 =.0286 M

14. c. 37.5-mL tot vol; NaOH added=0.0125 L*(0.100 mol/L) = 0.00125 mol HC 2 H 3 O 2 + OH  -  H 2 O + C 2 H 3 O 2 - I 0.0025 0 C -0.00125 +0.00125 E 0.00125 0.00125 Convert to M using vol:.00125/.0375 =.0333 M.00125/.0375 =.0333 M d. Add 25-mL at eq.pt, neutralization complete; 50.00-mL tot vol amt C 2 H 3 O 2 - present =0.0025 mol/0.050 L = 0.05 M C 2 H 3 O 2 - + H 2 O  -  HC 2 H 3 O 2 + OH - Eq: 0.05-x x x K b = K w /K a = 5.6*10 -10 K b = x 2 /0.05 x = 2.53*10 -6 pOH = 5.28 then pH = 8.72

15. e. Follow up. What happens beyond the equivalence pt? pH is determined from excess base present. - add 26.0 mL, what is the pH? tot vol 51.00 mL (0.026 L)*(0.100 M) = 0.0026 mol 0.0026 – 0.0025 = 0.0001 mol OH - excess M = 0.0001/.051 = 1.96*10 -3

16. Problem 25.0-mL of 0.145 M HCl soln is titrated by 0.200 M KOH. How many mL of base must be added to reach the end point. Want: mL KOH Data: 25.0-mL HCl @ 0.145 mol/L 0.200 mol/L KOH HCl + KOH ---  H 2 O + KCl 1 mol HCl = 1 mol KOH

17. Problem What is the pH at each point in the titration of 25.00-mL of 0.100 M HAc with 0.100 M NaOH? a) before NaOH added b) after 10.00-mL c) after 12.50-mL d) after 25.00-mL (these are volumes added to original 25 ml) a. Find initial [H 3 O + ]; use Ka (1.8*10 -5 ) HC 2 H 3 O 2 + H 2 O  -  H 3 O + + C 2 H 3 O 2 - Eq: 0.100-x x x = 1.3*10 -3 pH = -log(1.3*10 -3 ) = 2.89 b. 35-mL tot vol; amt neutralize=0.025L*(0.100 mol/L) = 0.0025 mol base added = 0.01 L*(0.100 M) = 0.001 mol HC 2 H 3 O 2 + OH  -  H 2 O + C 2 H 3 O 2 - I 0.0025 0 C -0.001 +0.001 E 0.0015 0.001 Convert to M using vol:.0015/.035 =.0429 M.001/.035 =.0286 M

18. c. 37.5-mL tot vol; NaOH added=0.0125 L*(0.100 mol/L) = 0.00125 mol HC 2 H 3 O 2 + OH  -  H 2 O + C 2 H 3 O 2 - I 0.0025 0 C -0.00125 +0.00125 E 0.00125 0.00125 Convert to M using vol:.00125/.0375 =.0333 M.00125/.0375 =.0333 M d. Add 25-mL at eq.pt, neutralization complete; 50.00-mL tot vol amt C 2 H 3 O 2 - present =0.0025 mol/0.050 L = 0.05 M C 2 H 3 O 2 - + H 2 O  -  HC 2 H 3 O 2 + OH - Eq: 0.05-x x x K b = K w /K a = 5.6*10 -10 K b = x 2 /0.05 x = 2.53*10 -6 pOH = 5.28 then pH = 8.72

19. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 17.7 Calculating pH for a Weak Acid–Strong Base Titration Solution Analyze We are asked to calculate the pH before the equivalence point of the titration of a weak acid with a strong base. Plan We first must determine the number of moles of CH 3 COOH and CH 3 COO – present after the neutralization reaction. We then calculate pH using K a, [CH 3 COOH], and [CH 3 COO – ]. Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH (K a = 1.8  10  5 ). Solve Stoichiometry Calculation: The product of the volume and concentration of each solution gives the number of moles of each reactant present before the neutralization: The 4.50  10  3 of NaOH consumes 4.50  10  3 of CH 3 COOH: The total volume of the solution is 45.0 mL + 50.0 mL = 95.0 mL = 0.0950 L

20. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward The resulting molarities of CH 3 COOH and CH 3 COO – after the reaction are therefore Equilibrium Calculation: The equilibrium between CH 3 COOH and CH 3 COO – must obey the equilibrium-constant expression for CH 3 COOH: Solving for [H + ] gives Comment We could have solved for pH equally well using the Henderson–Hasselbalch equation. Practice Exercise (a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (C 6 H 5 COOH, K a = 6.3  10  5. (b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH 3. Answers: (a) 4.20, (b) 9.26 Sample Exercise 17.7 Calculating pH for a Weak Acid–Strong Base Titration Continued

21. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 17.8 Calculating the pH at the Equivalence Point Solution Analyze We are asked to determine the pH at the equivalence point of the titration of a weak acid with a strong base. Because the neutralization of a weak acid produces its anion, which is a weak base, we expect the pH at the equivalence point to be greater than 7. Plan The initial number of moles of acetic acid equals the number of moles of acetate ion at the equivalence point. We use the volume of the solution at the equivalence point to calculate the concentration of acetate ion. Because the acetate ion is a weak base, we can calculate the pH using K b and [CH 3 COO – ]. Solve The number of moles of acetic acid in the initial solution is obtained from the volume and molarity of the solution: Moles = M  L = (0.100 mol)/L(0.0500 L) = 5.00  10  3 mol CH 3 COOH Hence, 5.00  10  3 mol of [CH 3 COO – ] is formed. It will take 50.0 mL of NaOH to reach the equivalence point (Figure 17.9). The volume of this salt solution at the equivalence point is the sum of the volumes of the acid and base, 50.0 mL + 50.0 mL = 100.0 mL = 0.1000 L. Thus, the concentration of CH 3 COO – is The CH 3 COO – ion is a weak base: Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH.

22. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Sample Exercise 17.8 Calculating the pH at the Equivalence Point The K b for CH 3 COO – can be calculated from the K a value of its conjugate acid, K b = K w /K a = (1.0  10  14 )/ (1.8  10  5 ) = 5.6  10  10. Using the K b expression, we have Making the approximation that and then solving for x, we have x = [OH – ] = 5.3  10  6 M, which gives pOH = 5.28 pH = 8.72. Check The pH is above 7, as expected for the salt of a weak acid and strong base. Practice Exercise Calculate the pH at the equivalence point when (a) 40.0 mL of 0.025 M benzoic acid (C 6 H 5 COOH, K a = 6.3  10  5 ) is titrated with 0.050 M NaOH; (b) 40.0 mL of 0.100 M NH 3 is titrated with 0.100 M HCl. Answers: (a) 8.21, (b) 5.28 Continued

23. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Strong Acid–Strong Base Titrations Consider adding a strong base (e.g., NaOH) to a solution of a strong acid (e.g., HCl). We can divide the titration curve into four regions. 1. Initial pH (before any base is added) The pH is given by the strong acid solution. Therefore, pH < 7. 2. Between the initial pH and the equivalence point. When base is added, before the equivalence point, the pH is given by the amount of strong acid in excess. Therefore, pH < 7. 3. At the equivalence point. The amount of base added is stoichiometrically equivalent to the amount of acid originally present. The cation of a strong base and the anion of a strong acid do not undergo hydrolysis. Therefore, pH = 7.00. 4. After the equivalence point. The pH is determined by the excess base in the solution. Therefore, pH > 7. The shape of a strong base-strong acid titration curve is very similar to a strong acid-strong base titration curve. Initially, the strong base is in excess, so the pH > 7. As acid is added, the pH decreases but is still greater than 7. At the equivalence point, the pH is given by the salt solution (i.e., pH = 7). After the equivalence point, the pH is given by the strong acid in excess, so pH is less than 7.

24. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Weak Acid-Strong Base Titration Consider the titration of acetic acid, HC 2 H 3 O 2, with NaOH. Again, we divide the titration into four general regions: 1. Before any base is added. The solution contains only weak acid. Therefore, pH is given by the equilibrium calculation. 2. Between the initial pH and the equivalence point. As strong base is added it consumes a stoichiometric quantity of weak acid: HC 2 H 3 O 2 (aq) + OH – (aq) C 2 H 3 O 2 – (aq) + H 2 O(l) However, there is an excess of acetic acid. Therefore, we have a mixture of weak acid and its conjugate base. Thus, the composition of the mixture is that of a buffer. The pH is given by the buffer calculation. First the amount of C 2 H 3 O 2 – generated is calculated, as well as the amount of HC 2 H 3 O 2 consumed (stoichiometry). Then the pH is calculated using equilibrium conditions (Henderson- Hasselbalch equation). 3. At the equivalence point, all the acetic acid has been consumed and all the NaOH has been consumed. However, C 2 H 3 O 2 – has been generated. Therefore, the pH depends on the C 2 H 3 O 2 – concentration. The pH > 7 at the equivalence point. More importantly, the pH of the equivalence point is NOT equal to 7 for a weak acid-strong base titration.

25. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 4. After the equivalence point. the pH is given by the concentration of the excess strong base. The pH curve for a weak acid-strong base titration differs significantly from that of a strong acidstrong base titration. For a strong acid-strong base titration: the pH begins at less than 7 and gradually increases as base is added. Near the equivalence point, the pH increases dramatically. For a weak acid-strong base titration: the initial pH rise is steeper than in the strong acid-strong base case. However, then there is a leveling off due to buffer effects. The middle section of the titration curve is not as steep for a weak acid-strong base titration. The shape of the two curves after the equivalence point is the same because pH is determined by the strong base in excess. The pH at the equivalence point differs also. The pH is 7.00 for the strong acid-strong base equivalence point due to the formation of a neutral salt. The pH is > 7.00 for the weak acid-strong base equivalence point due to the formation of a basic salt.

26. SOLUBILITY EQUILIBRA Principals: examines quantitative aspects of solubility & ppt process K sp = solubility constant Cr 2 (SO 4 ) 3 (s) --------> 2 Cr +3 (aq) + 3 SO 4 -2 (aq) Q c + solid = K sp

27. K sp 1. temp dependant 2. used to calc solubility of cmpd Comparing K sp cmpds w / same total # ions formed, then, higher K sp is greater the solubility Write dissoc eqn, # mols each ion Write K sp expression Find Molarity - convert solubility to molar CALCULATIONS Find K sp Solubility of silver dichromate @ 15 o C is 8.3*10 -3 g / 100 mL Ag 2 Cr 2 O 7 (s) --------> 2 Ag +1 (aq) + Cr 2 O 7 -2 (aq)

28. [Cr 2 0 7 -2 ] = 2 [Ag + ] 0.0192 = 2*0.0192 = 0.0384 Find Solubility K sp = [0.0384] 2 [0.0192] = 2.83*10 -5 What is the molar solubility of iron III hydroxide in water? K sp = 6.3*10 -10 Write dissoc eqn, ion expression, ICE table

29. K sp = [Fe +3 ] [OH - ] 3 K sp = [x] [3x] 3 = 27x 4 K sp = 27x 4 Fe(OH) 3 (s) --------> Fe +3 (aq) + 3 OH -1 (aq) I ------ 0 0 C -x +x +3x E -x x 3x Ignore: solid

30. PPT FORMATION - - will or not form ??? Q sp = K sp soln satur, no  Q sp < K sp soln unsat, no ppt Q sp > K sp ppt form till soln satur Will ppt form, if so, what is it? 0.20 L of 0.050 M Na 3 PO 4 & 0.10 L of 0.20 M Ca(NO 3 ) 2 Ions present: Na +1 PO 4 -3 Ca +2 NO 3 -1 solubility rules: Ca 3 (PO 4 ) 2 (s) the ppt, insoluble Ca 3 (PO 4 ) 2 (s) 3 Ca +2 (aq) +2 PO 4 -3 (aq) K sp = [Ca +2 ] 3 [PO 4 -3 ] 2 K sp = 1.2*10 -29

31. Q sp = [Ca +2 ] 3 [PO 4 -3 ] 2 = (0.067) 3 (0.033) 2 = 3.28*10 -7 mols Ca +2 = 0.10 L * 0.20 M = 0.020 mols [Ca +2 ] = 0.020 mols / 0.30 L = 0.067 M Q sp > K sp, so ppt will form Ca 3 (PO 4 ) 2 ppt till Q sp = 1.2*10 -29 mols PO 4 -3 = 0.20 L * 0.05 M = 0.10 mols [PO 4 -3 ] = 0.10 mols / 0.30 L = 0.033 M If Q sp = 6.7*10 -35 What then??? Q sp < K sp, no ppt

32. COMPLEX IONS Central METAL ion covalent bonded to 2+ LIGANDS Metal ion acts as Lewis Acid, Ligand acts as Base Can separate metal from its ore, isolate & remove toxic metal, convert metal ion to diff form

33. ION GROUPS Separation of 2 competing ppting metal ions thru diff properties of solubility of cmpds Add ppting ion till Q sp of more soluble ion close to its K sp This allows max amt of less soluble cmpd ppt out, while none of the more soluble Mixture of many diff metal ions, use various techniques to sep ions into characteristic groups pg 737

34. Group 1 Insol Chlorides ppt w / HCl Ag + Hg 2 +2 Pb +2 Group 2 Acid-insol Sulfides ppt w / H 2 S Cu +2 Cd +2 As +3 Sb +3 Bi +3 Sn +2 Sn +4 Hg +2 Pb +2 Group 3 Base-insol S -2 & OH -1 ppt w / NH 4 + -NH 3 buffer Zn +2 Mn +2 Ni +2 Fe +2 Co +2 as S -2 Al +3 Cr +3 as OH - Group 4 Insol PO 4 -3 ppt w / (NH 4 ) 2 HPO 4, NaCO 3 Mg +2 Ca +2 Ba +2 Group 5 Alkali & NH 4 + Ions generally speaking, what’s left Use various techniques to ppt out specific ions from each group

35. Calculate the pH of a soln that is 0.350 M pyridinium chloride (C 5 H 5 NHCl) and 0.210 M pyridine (C 5 H 5 N). WB C 5 H 5 N & C 5 H 5 NHCl contain common ion effect: C 5 H 5 NH + I 0.210 0.350 0.0 C -x +x +x E 0.210-x 0.350+x x C 5 H 5 N(aq) ) + H 2 O(l)  C 5 H 5 NH + (aq) + OH - (aq) assume 5% rule for WA

36. x = [OH - ] = 1.02*10 -9 pOH = -Log(1.02*10 -9 ) = 8.99 pH = 14 – 8.99 = 5.01 K b = 1.7*10 -9

37. The addition of bromide ion will decrease the water solubility of which of the following salts? a.BaSO 4 b. Li 2 CO 3 c. PbS d. AgBr AgBr Which pair of compounds will form a buffer solution when dissolved in water in equimolar amounts? a. HCl and KCl b. HNO 3 and NaNO 3 c. HCl and NH 4 Cl d. NH 3 and NH 4 Cl NH 3 and NH 4 Cl

38. The K a of HCN is 4.9 x 10 -10. What is the pH of a buffer solution that is 0.100 M in both HCN and KCN? a. 4.7 b. 7.0 c. 9.3 d. 14.0 9.3 same concentrations, -log(4.9*10 -10 ) The K a of HCN is 4.9 x 10 -10. What is the pH of a buffer solution that is 0.100 M in HCN and 0.200 M in KCN? a. 7.0 b. 9.0 c. 9.3 d. 9.6 9.6 pH = pKa + log[KCN]/[HCN]

39. The K a of HCN is 4.9 x 10 -10. What is the pH of a buffer solution that is 1.00 M in HCN and 0.100 M in KCN? a. 7.0 b. 8.3 c. 9.0 d. 9.3 8.3 pH = pKa + log[KCN]/[HCN]

40. In titrating a weak base with a strong acid, the best indicator to use would be: a.methyl red (changes color at pH = 5). b.bromothymol blue (changes at pH = 7). c.phenolphthalein (changes at pH = 9). d.none of the above. methyl red (changes color at pH = 5) Titrating a weak acid with a strong base, best indicator to use would be: a.methyl red (changes color at pH = 5). b.bromothymol blue (changes at pH = 7). c.phenolphthalein (changes at pH = 9). d.none of the above. phenolphthalein (changes at pH = 9)

41. The K sp of BaCO 3 is 5.0 x 10 -9. What is the concentration of barium ion in a saturated aqueous solution of BaCO 3 ? a. 7.1 x 10 -5 M b. 2.5 x 10 -9 M c. 5.0 x 10 -9 M d. 1.0 x 10 -8 M 7.1 x 10 -5 M 5.0*10 -9 = x*x 5.0*10 -9 = x 2 The K sp of BaF 2 is 1.7 x 10 -6. What is the concentration of barium ion in a saturated aqueous solution of BaF 2 ? a. 1.7 x 10 -6 M b. 3.4 x 10 -6 M c. 7.5 x 10 -3 M d. 1.5 x 10 -2 M 7.5 x 10 -3 M K sp = [Ba +2 ][F - ] 2 = (x)*(2x) 2 1.7*10 -6 = 4x 3

42. The K sp of BaF 2 is 1.7 x 10 -6. What is the concentration of fluoride ion in a saturated aqueous solution of BaF 2 ? a. 1.7 x 10 -6 M b. 5.7 x 10 -5 M c. 7.6 x 10 -3 M d. 1.5 x 10 -2 M 1.5 x 10 -2 M BaF 2 --  Ba +2 + 2 F - [F - ] =2x = 2(7.5*10 -3 )

43. Which of the following substances will be more soluble in acidic solution than in basic solution: (a)Ni(OH) 2 (s), (b) CaCO 3 (s), (c) BaF 2 (s), (d) AgCl(s)? So, which is more soluble at low pH than high pH Ni(OH) 2 more soluble in acidic, basicity of OH - CaCO 3 dissolves in acidic, as CO 3 - basic anion BaF 2 dissolves, as F - basic anion (d) The solubility of AgCl is unaffected by changes in pH because Cl – is the anion of a strong acid and therefore has negligible basicity.

44. Addition of _____ will increase the solubility of MgCO 3. MgCO 3 (s) Mg 2+ (aq) + CO 3 2– (aq) MgCl 2 Na 2 CO 3 NaOH HCl KHCO 3 HCl

45. Predict the order of precipitation of Ba 2+, Pb 2+, Ca 2+ with the addition of NaF. Ca 2+ then Ba 2+ then Pb 2+ Pb 2+ then Ba 2+ then Ca 2+ Pb 2+ then Ca 2+ then Ba 2+ Ca 2+ then Pb 2+ then Ba 2+ Ba 2+ then Pb 2+ then Ca 2+ K sp BaF 2 1.7 x 10 –6 PbF 2 3.6 x 10 –8 CaF 2 3.9 x 10 –11 Ca 2+ then Pb 2+ then Ba 2+

46. Predict the order of precipitation of Cl –, CO 3 2–, Br – upon the addition of AgNO 3. Br – then CO 3 2– then Cl – Br – then Cl – then CO 3 2– Cl – then CO 3 2– then Br – Cl – then Br – then CO 3 2– CO 3 2– then Cl – then Br – K sp AgCl1.8 x 10 –10 Ag 2 CO 3 8.1 x 10 –12 AgBr5.0 x 10 –13 Br – then Cl – then CO 3 2–

47. A solution contains 1.0 × 10 -2 M Ag + and 2.0 ×10 -2 M Pb 2+. When Cl – is added to the solution, both AgCl (K sp = 1.8× 10 -10 ) and PbCl 2 (K sp = 1.7×10 -5 ) precipitate from the solution. What concentration of Cl – is necessary to begin the precipitation of each salt? Which salt precipitates first? We are given K sp values for the two possible precipitates. Using these and the metal ion concentrations, we can calculate what concentration of Cl – ion would be necessary to begin precipitation of each. The salt requiring the lower Cl – ion concentration will precipitate first.