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Acid-Base and Solubility Equilibria

Acid-Base and Solubility Equilibria. Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis. common ion. CH 3 COONa ( s ) Na + ( aq ) + CH 3 COO - ( aq ).

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Acid-Base and Solubility Equilibria

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  1. Acid-Base and Solubility Equilibria • Common-ion effect • Buffer solutions • Acid-base titration • Solubility equilibria • Complex ion formation • Qualitative analysis

  2. common ion CH3COONa (s) Na+(aq) + CH3COO-(aq) CH3COOH (aq) H+(aq) + CH3COO-(aq) (*) I. Common ion effect • Common ion: an ion that is produced from more than one solute • Common ion effect: shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. Example: Consider CH3COOH (weak acid, partial ionizes) Add CH3COONa (strong electrolyte, completely ionizes) Le Chatelier's principle: equilibrium (*) shift This shift to the left will also result in a decrease of the [H+] The ionization of a weak electrolyte is decreased by adding a strong electrolyte (i.e. a salt) that produces an common ion

  3. CH3COOH (aq) H+(aq) + CH3COO–(aq) = 1.8 x 10-5 = 1.8 x 10-5 x(0.30+x) 0.30x Ka Ka = 0.20 - x 0.20 What is the the pH of a solution containing 0.20 M CH3COOH and 0.30 M CH3COONa at 25oC? (Ka=1.8 x10-5). Ex: Initial (M) 0.20 0.00 0.30 Change (M) -x +x +x Equilibrium (M) 0.20 - x x 0.30+ x Assume 0.20 – x 0.20 0.30 + x 0.30 x = 1.2 x 10-5M [H+] = [CH3COO– ] = 1.2 x 10-5M pH = -log [H+] = 4.92

  4. ? How does buffer work ? H+(aq) + HCOO-(aq) HCOOH (aq) OH-(aq) + HCOOH (aq) HCOO-(aq) + H2O (l) II. Buffer solution HF+NaF A. A solution that resist pH change when acids or base are added. B. In general, a buffer is made of • A weak acid and its salt • A weak base and its salt NH3+NH4Cl Consider an equal molar mixture of HCOOH + HCOONa Add strong acid Add strong base

  5. Which of the following can be used to prepare for buffer solutions? (a) NaF/HF (b) NaBr/HBr, (c)Na2SO3/NaHSO3, and (d) CH3COOH and NaOH. Ex. (a) HF is a weak acid and F- is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) SO32- is a weak base and HSO3- is it conjugate acid buffer solution (d) CH3COOH is a weak base and NaOH is a strong base CH3COOH + NaOH --> CH3COONa + H2O buffer solution

  6. XA (s) X+(aq) + A-(aq) HA (aq) H+(aq) + A-(aq) [H+][A-] Ka [HA] Ka = [H+] = [HA] [A-] = - log Ka + log -log [H+] = -log Ka - log [HA] [A-] [HA] [A-] As pKa , the acid strength______. [conjugate base] pH = pKa + log [acid] C. pH of a Buffer solution Consider a mixture of a weak acid (HA) and its salt (XA) Henderson-Hasselbalch equation pKa = -log Ka

  7. pH = -log(1.8 x10-5)+ log [0.25] 0.015 0.025 [0.15] 0.10 0.10 CH3COOH (aq) + OH-(aq) CH3COO–(aq) + H2O (l) pH = 4.74 + log [CH3COO–] = =0.25 M [CH3COOH] = =0.15 M [conjugate base] 0.30 pH = pKa + log 0.20 [acid] What is the the pH of a solution containing 0.20 M CH3COOH and 0.30 M CH3COONa at 25oC? (Ka=1.8 x10-5)? What is the pH after the addition of (a) 20.0 mL of 0.050 M NaOH or (b) 20.0 mL of 0.050 M HCl to 80.0 mL of this buffer solution? Ex: Henderson-Hasselbalch eq. = 4.74+ 0.18 = 4.92 (a) Addition of 20.0 mL of 0.050 M NaOH (0.001 mol NaOH) NaOH will react with the acid in the buffer start (moles) 0.016 0.001 0.024 end (moles) 0.015 0.00 0.025 final volume = 80.0 mL+ 20.0 mL=100.0 mL = 4.96

  8. [0.23] 0.017 0.023 [0.17] 0.10 0.10 CH3COO–(aq) + H+(aq) CH3COOH (aq) pH = 4.74 + log [CH3COO–] = =0.23 M [CH3COOH] = =0.17 M [conjugate base] pH = pKa + log [acid] Continued Henderson-Hasselbalch eq. pKa = 4.74 (b) Addition of 20.0 mL of 0.050 M HCl (0.001 mol HCl) HCl will react with the base in the buffer start (moles) 0.024 0.001 0.016 end (moles) 0.023 0.0 0.017 final volume = 80.0 mL+ 20.0 mL=100.0 mL = 4.87

  9. [HCOO–] [HCOO–] [HCOO–] [HCOOH] [HCOOH] [HCOOH] pH = 3.85 = -log(1.7 x10-4) + log =1.2 log =0.080 [conjugate base] pH = pKa + log [acid] D. Preparation of buffer • Select acid/base pair with pKa that is closest to the desired pH. • Adjust ratio of base/acid to have the desired pH. Ex. How to prepare a buffer with pH = 3.85? [H+] =10-3.85=1.4 x10-4 HCOOH seems to be a good choice, Ka=1.7 x10-4 The buffer can be prepared by mixing 1.0 M HCOOH and 1.2 M HCOONa

  10. E. Buffer capacity • Quantity of acid or base that can be added before pH changes significantly • Buffers work best when the ratio of [base]/[acid] is 1. • The greater the concentrations of both [base] and [acid], the ________ buffering capacity greater Ex. Which of the following has the greater buffering capacity? (a) 1.0 L of 1.0 M NaF + 1.0 M HF solution, or (b) 1.0 L of 0.1 M NaF + 0.1M HF solution

  11. III. Titration • Acid-base titration: to determine the concentration of unknown solution • Standard solution (titrant): a solution of known concentration • Equivalence point : the point at which the reaction is complete • Indicator : a substance changes color at/near the equivalence point.

  12. NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq) OH-(aq) + H+(aq) H2O (l) Titration curve: plot of pH as a function of mL A. Strong acid-strong base titration

  13. H+(aq) + OH-(aq) H2O (l) A 40.0 mL sample of 0.200 M HCl is titrated with 0.200 M NaOH. Calculate the pH of the soution after the following volume of NaOH have been added. (a) 35.0 mL, (b) 39.0 mL, (c) 41.0 mL, and (d) 45.0 mL. Q. Initial pH = 0.699 mol HCl: 0.0400 L x 0.200 M = 0.00800 mol mol NaOH: 0.0350 L x 0.200 M = 0.00700 mol excess HCl: 0.00800 mol - 0.00700 mol = 0.00100 mol Final volume = (40.0 + 35.0)mL =75.0 mL [H+]=0.00100 mol/ 0.0750 L = 0.0133 M pH= -log[H+]= 1.875 (b) mol HCl: 0.0400 L x 0.200 M = 0.00800 mol mol NaOH: 0.0390 L x 0.200 M = 0.00780 mol excess HCl: 0.00800 mol - 0.00780 mol = 0.00020 mol Final volume = (40.0 + 39.0)mL =79.0 mL [H+]=0.00020 mol/ 0.0790 L = 0.0025 M pH= -log[H+]= 2.60

  14. H+(aq) + OH-(aq) H2O (l) continued (c) mol HCl: 0.0400 L x 0.200 M = 0.00800 mol mol NaOH: 0.0410 L x 0.200 M = 0.00820 mol excess NaOH: 0.00820 mol - 0.00800 mol = 0.00020 mol Final volume = (40.0 + 41.0)mL =81.0 mL [OH–]=0.00020 mol/ 0.0810 L = 0.0025 M pOH= -log[OH–]= 2.61 pH=14.0 -pOH = 11.39 (d) mol HCl: 0.0400 L x 0.200 M = 0.00800 mol mol NaOH: 0.0450 L x 0.200 M = 0.00900 mol excess NaOH: 0.00900 mol - 0.00800 mol = 0.00100 mol Final volume = (40.0 + 45.0)mL =85.0 mL [OH–]=0.00100 mol/ 0.0850 L = 0.0118 M pOH= -log[OH–]= 1.929 pH=14.0 -pOH = 12.071

  15. CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l) CH3COOH (aq) + OH-(aq) CH3COO-(aq) + H2O (l) CH3COO-(aq) + H2O (l) OH-(aq) + CH3COOH (aq) Buffering region B. Weak acid-strong base titration At equivalence point (pH > 7):

  16. HNO2(aq) + OH-(aq) NO2-(aq) + H2O (l) Initial (M) 0.01 Change (M) [NO2-] = = 0.05 M NO2-(aq) + H2O (l) OH-(aq) + HNO2(aq) 0.200 Equilibrium (M) x2 [OH-][HNO2] = Kb = 0.05-x [NO2-] Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ? Q. start (moles) 0.01 0.01 0.0 end (moles) 0.0 0.01 Final volume = 200 mL 0.05 0.00 0.00 -x +x +x 0.05 - x x x pOH = 5.98 = 2.2 x 10-11 pH = 14 – pOH = 8.02 0.05 – x 0.05 x 1.05 x 10-6 = [OH-]

  17. Ka=2 x10-4 Ka=2 x10-6 Ka=2 x10-8 1. Initial pH of a weak acid 2. Before equivalence point: buffering region 3. At the equivalence point: a weak base 4. After the equivalence point: excess strong base raises pH B. Weak acid-strong base titration curve 50 mL of 0.1 M weak acid is titrated with 0.1 M NaOH pH=pKa equivalence point

  18. HCl (aq) + NH3(aq) NH4Cl (aq) H+(aq) + NH3(aq) NH4Cl (aq) NH4+(aq) + H2O (l) NH3(aq) + H+(aq) Buffering region C. Strong acid-weak base titration At equivalence point (pH < 7):

  19. HIn (aq) H+(aq) + In-(aq) [HIn] [HIn]  10  0.1 [In-] [In-] IV. Acid-base indicator Color of acid (HIn) predominates Color of conjugate base (In-) predominates

  20. The titration curve of a strong acid with a strong base

  21. Q: Which indicator(s) would you use for a titration of HCOOH with NaOH ? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein

  22. AgCl (s) Ag+(aq) + Cl-(aq) CaF2(s) Ca2+(aq) + 2F-(aq) Ag2SO4(s) 2Ag+(aq) + SO42-(aq) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq) A. Solubility product constant, Ksp V. Solubility equilibria Ksp= [Ag+][Cl-] Ksp= [Ca2+][F-]2 Ksp= [Ag+]2[SO42-] Ksp= [Ca2+]3[PO33-]2

  23. a. Dissolution of an ionic solid in aqueous solution: B. Solubility and KSP Unsaturated solution Q < Ksp No precipitate Q = Ksp Saturated solution Precipitate will form Q > Ksp Supersaturated solution b. Determine Kspfrom solubility or solubility from Ksp Molar solubility (mol/L): moles of solute in 1 L of a saturated solution Solubility (g/L):grams of solute dissolved in 1 L of a saturated solution.

  24. AgCl (s) Ag+(aq) + Cl-(aq) Initial (M) Change (M) Equilibrium (M)  s = Ksp 1.3 x 10-5 mol AgCl 1 L soln x 143.35g AgCl 1 mol AgCl What is the solubility of silver chloride in g/L ? Ksp=1.6x10-10 Q. 0.00 0.00 +s +s s s Ksp= [Ag+][Cl-] Ksp= s2 s = 1.3 x 10-5 [Ag+] = 1.3 x 10-5M [Cl-] = 1.3 x 10-5M Solubility of AgCl = = 1.9 x 10-3g/L

  25. SrF2(s) Sr2+(aq) + 2 F-(aq) Initial (M) Change (M) Equilibrium (M) 1.1 x10-2 g SrF2/(125.6 g/mol) 0.100 L soln The solubility of SrF2 is found to be 1.1 x 10-2 g in 100 mL of aqueous solution. Determine the value of Ksp of SrF2. Q. 0.00 0.00 -s +s +2s s 2s s: Solubility of SrF2 in M = = 8.74 x 10–4 M Ksp= [Sr2+][F–]2 = s(2s)2 = 4 s3 = 2.7 x 10-9

  26. Ca(OH)2(s) Ca2+(aq) + 2 OH-(aq) 2 Q = [Ca2+]0[OH-]0 If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a precipitate form? (Ksp = 8.0 x10-6) Q. Which ions are present in solution? Na+, OH-, Ca2+, Cl-. Only possible precipitate is _________(solubility rules). Ca(OH)2 Is Q > Kspfor Ca(OH)2? [Ca2+]0 = 0.100 M x 1.00 L/(1.00 L +0.00200 L)=0.100 M [OH-]0 = 0.00200 L x 0.200 M /(1.00L + 0.00200 L)=4.0 x 10-4M = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8 Ksp = [Ca2+][OH-]2 = 8.0 x 10-6 Q < Ksp No precipitate will form

  27. CaF2(s) Ca2+(aq) + 2F-(aq) CaF2(s) Ca2+(aq) + 2F-(aq) CaF2(s) Ca2+(aq)+ 2 F-(aq) Initial (M) (a) Change (M) Equilibrium (M) NaF (s) Na+ (aq) + F-(aq) C. Factors affecting solubility Common ion effect: common ion decreasesthe solubility. left If NaF is added, equilibrium shifts __________. Ex. What is the molar solubility of CaF2 in (a) pure water and (b) 0.010 M NaF? (Ksp=3.9 x10-11) s = 2.1 x 10-4 Ksp = 3.9 x 10-11 =s(2s)2 = 4s3 (b) [F-] = 0.010 M [F-] = 0.010 + 2s 0.010 0.00 0.010 Ksp = (0.010)2 x s -s +s +2s s = 3.9 x 10-7 s 0.010+2s

  28. Ex. Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) CaF2(s) Ca2+(aq) + 2 F-(aq) remove add Ex. CaF2(s) +2 H+(aq) Ca2+(aq) + 2 HF (aq) F–(aq) + H+(aq) HF (aq) 2. Effect of pH • Insoluble bases dissolve in acidic solutions • Insoluble acids dissolve in basic solutions increases In acidic solution, solubility_________. Why? Ex. Which salt will be more soluble in acidic solution than in pure water (a) ZnCO3, (b) AgCN, and (c) BiI3. (a) and (b) because CO32– and CN– react with H+

  29. AgCl (s) Ag+(aq) + Cl-(aq) AgBr (s) Ag+(aq) + Br-(aq) = 7.7 x 10-11M = 1.6 x 10-8M [Ag+] = [Ag+] = 7.7 x 10-13 1.6 x 10-10 = = 0.010 0.010 Ksp Ksp [Cl-] [Br-] What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br- and Cl- at a concentration of 0.010 M? Q. Ksp= 7.7 x 10-13 Ksp= [Ag+][Br-] Ksp= 1.6 x 10-10 Ksp= [Ag+][Cl-] 7.7 x 10-11M < [Ag+] < 1.6 x 10-8M

  30. Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) AgCl (s) Ag+(aq) + Cl–(aq) stability of complex Kf [Ag(NH3)2+ ] Kf = =1.7 x107 [Ag+][NH3]2 VI. Complex ion equilibria and solubility Complex ion: an ion containing a central metal cation (Lewis acid) bonded to one or more molecules or ions (Lewis base). Solubility of slightly soluble salts can be increased by complex ion formation By adding NH3, solubility of AgCl increases due to 3. The formation constant or stability constant (Kf) is the equilibrium constant for the complex ion formation.

  31. VII. Qualitative analysis

  32. Flow chart for separation of cations Solution containing ions of all cation groups Group 1 precipitates AgCl, Hg2Cl2, PbCl2 +HCl Filtration Solution containing ions of remaining groups +H2S Group 2 precipitates CuS, CdS, SnS, Bi2S3 Filtration Solution containing ions of remaining groups Group 3 precipitates CoS, FeS, MnS, NiS ZnS, Al(OH)3,Cr(OH)3 +NaOH Filtration Solution containing ions of remaining groups +Na2CO3 Group 4 precipitates BaCO3, CaCO3, SrCO3 Filtration Solution contains Na+, K+, NH ions + 4

  33. Flame test of cations sodium lithium potassium copper

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