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CHAPTER 17 Acid-Base Equilibriu and Solubility Equilibria

CHAPTER 17 Acid-Base Equilibriu and Solubility Equilibria. Neutralization Reaction A neutralization reaction is the reaction of an acid with a base, usually giving salt plus water as products. For three types of neutralization reactions the reaction goes essentially to completion:

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CHAPTER 17 Acid-Base Equilibriu and Solubility Equilibria

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  1. CHAPTER 17 Acid-Base Equilibriu and Solubility Equilibria

  2. Neutralization Reaction A neutralization reaction is the reaction of an acid with a base, usually giving salt plus water as products. For three types of neutralization reactions the reaction goes essentially to completion: strong acid + strong base HCl(aq) + NaOH(aq)  NaCl(aq) + H2O() strong acid + weak base HCl(aq) + NH3(aq)  NH4Cl(aq) weak acid + strong base HF(aq) + NaOH(aq)  NaF(aq) + H2O() For weak acid + weak base an equilibrium is established between reactants and products. This case is mathematically difficult.

  3. Buffers A buffer solution is a solution whose pH does not change significantly upon the addition of a small amount of a strong acid or strong base. Buffer solutions contain a substantial amount of a weak acid and a weak base, usually a weak acid-weak base conjugate pair. Buffer solutions are important for systems where a particular value for pH is required, such as in biological systems. Examples: CH3COOH + CH3COO- acetate buffer H2CO3 + HCO3- carbonate buffer NH3 + NH4+ ammonia buffer

  4. How Buffers Work Consider a buffer solution containing a weak acid “HA” and the conjugate base of the weak acid “A-”. What will happen if we add a small amount of a strong acid (HCl) or strong base (NaOH)? strong acid HCl(aq) + A-(aq)  HA(aq) + Cl-(aq) strong base NaOH(aq) + HA(aq)  Na+(aq) + A-(aq) + H2O() Buffer solution converts the strong acid or base into a weak acid or base. By doing so, they minimize the effect of these additions on pH.

  5. Example of Buffering Consider 1.000 L of each of the following two solutions. Solution A. 0.100 M CH3COOH and 0.200 M CH3COONa. Solution B. Pure water Find the initial pH and the pH after addition of 0.0010 moles HCl to each of the above solutions, at T = 25. C. The value for Ka for CH3COOH at this temperature is Ka = 1.8 x 10-5.

  6. Solution A. 0.100 M CH3COOH and 0.200 M CH3COONa. CH3COONa(aq)  Na+(aq) + CH3COO-(aq) CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq) Ka = [H3O+] [CH3COO-] = 1.8 x 10-5 [CH3COOH] Substance Initial Change Equilibrium H3O+ 0.000 x x CH3COO- 0.200 x 0.200 + x CH3COOH 0.100 - x 0.100 - x

  7. Ka = [H3O+] [CH3COO-] = (x) (0.200 + x) = 1.8 x 10-5 [CH3COOH] (0.100 - x) Assume x << 0.100 (x) (0.200) = 1.8 x 10-5 (0.100) x = (1.8 x 10-5) (0.100) = 9.0 x 10-6 (0.200) pH = - log10(9.0 x 10-6) = 5.05 Is 9.0 x 10-6 << 0.100? Yes (at least 10 times smaller).

  8. Now, consider the effect of adding 0.0010 moles of HCl (equi-valent to 0.0010 M HCl, since 1.00 L solution). HCl(aq) + CH3COO-(aq)  CH3COOH(aq) + Cl-(aq) initial 0.0010 M 0.2000 M 0.1000 M after rxn. 0.0000 M 0.1990 M 0.1010 M 0.0010 M CH3COOH(aq) + H2O()  H3O+(aq) + CH3COO-(aq) Ka = [H3O+] [CH3COO-] = 1.8 x 10-5 [CH3COOH] Substance Initial After HCl Change Equilibrium H3O+ 0.000 0.000 x x CH3COO- 0.200 0.199 x 0.199 + x CH3COOH 0.100 0.101 - x 0.101 – x

  9. Ka = (x) (0.199 + x) = 1.8 x 10-5 (0.101 – x) Assume x << 0.100, then (x) (0.199) = 1.8 x 10-5 (0.101) x = (0.101) (1.8 x 10-5) = 9.1 x 10-6 (0.199) pH = - log10(9.1 x 10-6) = 5.04 The pH decreased from 5.05 to 5.04, a change of 0.01 pH unit.

  10. Solution B (pure water) pure water. pH = 7.00 After addition of 0.0010 moles of HCl we simply have a 0.001 M solution of HCl, a strong acid. So pH = - log10(0.0010) = 3.00 We may summarize the results as follows: solution A solution B initial pH 5.05 7.00 pH after addn. of 5.04 3.00 0.0010 mol HCl The pH of solution A (the buffer solution) changed by only a small amount after addition of the HCl. The pH of solution B (unbuffered water) changed by 4 pH units (a change in H3O+ concentration of a factor of 10,000).

  11. Henderson Equation The Henderson (or Henderson-Hasselbalch) equation applies for solutions of a weak acid/conjugate base pair. We may develop the equation as follows. Consider a weak acid HA HA(aq) + H2O(l)  H3O+(aq) + A-(aq) Ka = [H3O+] [A-] so [H3O+] = Ka[HA] [HA] [A-] Therefore - log10[H3O+] = - log10Ka - log10([HA]/[A-]) pH = pKa- log10([HA]/[A-]) , or pH = pKa + log10([A-]/[HA]) , the Henderson equation.

  12. Properties of the Henderson Equation 1) The Henderson equation applies for solutions of a weak acid/conjugate base pair. 2) There must be appreciable amounts (at least 0.001 M) of both the weak acid and the conjugate base present in the system. 3) If the weak acid and conjugate base concentrations are the same ([HA] = [A-]), then pH = pKa + log10([A-]/[HA]) = pKa + log10(1) = pKa This suggests that the best buffers are those where pH ~ pKa. In fact, buffers work best if the pKa of the weak acid is within about 1 pH unit of the desired buffer pH.

  13. 4) While buffers are often prepared by combining a weak acid and conjugate base pair, they can also be prepared by adding a small amount of strong base to a weak acid solution, or a small amount of strong acid to a weak base solution. Example: Consider 1.000 L of a 0.100 M solution of acetic acid (CH3COOH). If we add 0.050 moles of NaOH, then CH3COOH(aq) + NaOH(s)  CH3COO-(aq) + Na+(aq) + H2O() initial (.100 mol) (.050 mol) (.000 mol) final (.050 mol) (.000 mol) (.050 mol) The neutralization reaction is used to produce the conjugate base needed for the buffer. Note that we require that there be a larger number of moles of weak acid than strong base.

  14. Buffer Calculation (Example) We have 0.500 M stock solutions of the following: CH3COOH HF HOCl NaCH3COO NaF NaOCl Ka = 1.8 x 10-5 Ka = 3.5 x 10-4 Ka = 3.5 x 10-8 pKa = 4.74 pKa = 3.46 pKa = 7.46 How would you prepare 1.000 L of a pH = 3.75 buffer solution?

  15. Example: We have 0.500 M stock solutions of the following: CH3COOH HF HOCl NaCH3COO NaF NaOCl Ka = 1.8 x 10-5 Ka = 3.5 x 10-4 Ka = 3.5 x 10-8 pKa = 4.74 pKa = 3.46 pKa = 7.46 How would you prepare 1.000 L of a pH = 3.75 buffer solution? Which acid/conjugate base pair should we use? We should use the one where the pKa is as close as possible to the pH of the buffer we wish to make. That would be the HF/NaF system. Now, let x = mL of the HF stock solution (1000 - x) = mL of the NaF stock solution

  16. Then, in the 1.000 L buffer solution (using MiVi = MfVf) [HF] = (0.500 M) x [F-] = (0.500 M) (1000 - x) 1000 1000 pH = pKa + log10([A-]/[HA]) 3.75 = 3.46 + log10 (0.500)[(1000 - x)/1000] (0.500) (x/1000) 3.75 = 3.46 + log10[(1000 - x)/x] 3.75 - 3.46 = 0.29 = log10[(1000 - x)/x] 100.29 = 1.95 = (1000 - x)/x 1.95 x = 1000 - x 2.95 x = 1000 ; x = 1000/2.95 = 339. So 339 mL of the HF solution and 661 mL of the NaF solution.

  17. Titration By titration, we mean the reaction of a fixed amount of one sub-stance by the slow addition of a second substance. In an acid-base titration we slowly add an acid to a fixed amount of base, or we slowly add a base to a fixed amount of acid. We use titrations to determine the concentration of acid or base in a stock solution. We may also be interested in the titration curve, a plot of pH vs mL of added titrant (solution being slowly added). Equivalence point. The equi-valence point in a titration is the point where sufficient titrant has been added to have a complete reaction with no excess acid or base.

  18. Indicator An indicator is a substance (usually a weak acid) that changes colors at a particular pH. HInd(aq) + H2O(l)  H3O+(aq) + Ind-(aq) color 1 color 2 We often use indicators in titrations. We choose an indicator so that the end point (point where color change occurs) is as close as possible to the equivalence point for the titration. Based on the Henderson equation pH = pKa + log10([A-]/[HA]) we expect the color change will occur at pH  pKInd.

  19. In choosing an indicator we would like the color change to be as close as possible to the equivalence point for the titration.

  20. Phenolphthalein is a suitable indicator for both the strong acid+strong base titration (blue line) and weak acid+strong base titration (red line). Methyl red is a suitable indicator only for the strong acid+strong base titration. Other indicators would also be suitable choices.

  21. Titration of a Strong Base With a Strong Acid Consider the titration of 25.00 mL of a 0.0100 M solution of NaOH (a strong base) with a 0.0100 M solution of HCl (a strong acid). The reactions that will occur are as follows: NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) neutralization NaOH(aq)  Na+(aq) + OH-(aq) if excess base HCl(aq)  H+(aq) + Cl-(aq) if excess acid If there is excess NaOH the pH will be greater than 7.0. If there is excess HCl the pH will be less than 7.0. At the equivalence point the moles of NaOH and HCl will be equal, and the solution will be neutral (pH = 7.0).

  22. The equivalence point occurs at pH = 7.0. A good indicator would be one where the color change occurs near pH = 7.0 (such as bromthymol blue). Notice that the titration curve is steep (nearly vertical) near the equivalence point. Note that we can calculate the pH at every point in the titration (which was done in generating the above titration curve).

  23. pH Calculation During Titration Consider the titration of 25.00 mL of a 0.0100 M solution of NaOH (a strong base) with a 0.0100 M solution of HCl (a strong acid). The reactions that will occur are as follows: NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) neutralization NaOH(aq)  Na+(aq) + OH-(aq) if excess base HCl(aq)  H+(aq) + Cl-(aq) if excess acid What is the pH after addition of 10.00 mL of HCl solution? Strategy: 1) Find the moles of acid and base present, and the total volume. 2) Find the species present after the neutralization reaction. 3) Find the pH (as in a regular acid/base calculation)

  24. mol NaOH = 0.02500 L (0.0100 M) = 2.5 x 10-4 mol NaOH mol HCl = 0.0100 L (0.0100 M) = 1.0 x 10-4 mol HCl Volume - 0.0250 L + 0.0100L = 0.0350 L NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) neutralization After neutralization, excess NaOH is Excess NaOH = 2.5 x 10-4 mol - 1.0 x 10-4 mol = 1.5 x 10-4 mol [NaOH]excess = (1.5 x 10-4 mol)/(0.0350 L) = 0.00429 M NaOH(aq)  Na+(aq) + OH-(aq) if excess base pOH = - log10(.00429) = 2.37 pH = 14.00 - 2.47 = 11.63

  25. Titration of a Strong Acid With a Strong Base Consider the same system, but the case where 25.00 mL of a 0.0100 M solution of HCl is titrated with a 0.0100 M solution of NaOH. The reactions taking place are the same, but now the solution is initially acidic.

  26. Titration of a Weak Acid With a Strong Base Consider the titration of 25.00 mL of a 0.0100 M solution of CH3COOH (a weak acid; Ka = 1.8 x 10-5) with a 0.0100 M solution of NaOH (a strong base). The reactions that will occur are as follows: CH3COOH(aq) + NaOH(aq)  H2O(l) + Na+(aq) + CH3COO-(aq) neutralization CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq) if excess weak acid NaOH(aq)  Na+(aq) + OH-(aq) if excess base

  27. If there is excess CH3COOH then there will usually be a buffer solution with both CH3COOH and CH3COO- present. CH3COOH(aq) + NaOH(aq)  H2O(l) + Na+(aq) + CH3COO-(aq) If there is excess base then the pH will be controlled by the concentration of excess strong base. NaOH(aq)  Na+(aq) + OH-(aq) Note that at the half-equivalence point pH  pKa, since pH = pKa + log10([A-]/[HA])

  28. At the half-equivalence point (point where we have added half the titrant needed to reach the equivalence point) pH  pKa for the weak acid. Notice that there will be a region where the solution forms a buffer solution. Note that the pH at the equivalence point is greater than 7.0. Indicator - phenolphthalein

  29. Example of Weak Acid + Strong Base Calculation 25.00 mL of a 0.100 M solution of formic acid (HCOOH, Ka = 1.7 x 10-4) is titrated with a 0.100 M solution of KOH, a strong base. What is the pH after 10.00 mL of KOH solution has been added?

  30. 25.00 mL of a 0.100 M solution of formic acid (HCOOH, Ka = 1.7 x 10-4) is titrated with a 0.100 M solution of KOH, a strong base. What is the pH after 10.00 mL of KOH solution has been added? Moles HCOOH = 0.02500 L (0.100 mol/L) = 2.50 x 10-3 mol Moles NaOH = 0.01000 L (0.100 mol/L) 1.00 x 10-3 mol Volume = 25.00 mL + 10.00 mL = 35.00 mL HCOOH(aq) + NaOH(aq)  Na+(aq) + HCOO-(aq) + H2O(l) Initial moles Moles after Concentration after neutralization neutralization HCOOH 2.50 x 10-3 1.50 x 10-3 0.04286 HCOO- 0 1.00 x 10-3 0.02857 NaOH 1.00 x 10-3 0 0

  31. We can now use the reaction for a weak acid in equilibrium with its conjugate base. HCOOH(aq) + H2O(l)  H3O+(aq) + HCOO-(aq) Ka = [H3O+][HCOO-] = 1.7 x 10-4 [HCOOH] After neutralization Change Equilibrium HCOOH 0.04286 - x 0.04286 - x COOH- 0.02857 x 0.02857 + x H3O+ 0 x x (0.02857 + x)(x) = 1.7 x 10-4 If we assume x << 0.028, then (0.04286 - x) x = (0.04286)(1.7 x 10-4) = 2.6 x 10-4 pH = 3.59 (0.02857)

  32. Titration of a Weak Base With a Strong Acid Consider the titration of 25.00 mL of a 0.0100 M solution of NH3 (a weak base; Kb = 1.8 x 10-5) with a 0.0100 M solution of HCl (a strong acid). The reactions that will occur are as follows: NH3(aq) + HCl(aq)  NH4+(aq) + Cl-(aq) neutralization NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) if excess weak base HCl(aq)  H+(aq) + Cl-(aq) if excess acid If there is excess NH3 then there will usually be a buffer solution with both NH3 and NH4+ present. If there is excess acid then the pH will be controlled by the concentration of excess strong acid. As was the case for the titration of a weak acid with a strong base, at the half equivalence point pH  pKa (this will be for NH4+, the conjugate acid of NH3).

  33. At the half-equivalence point (point where we have added half the titrant needed to reach the equivalence point) pH  pKa for the conjugate acid of the weak base. Notice that there will be a region where the solution forms a buffer solution. Note that the pH at the equivalence point is less than 7.0. Indicator - chlorophenol red

  34. Titration of a Polyprotic Acid With a Strong Base Consider the titration of 1.000 L of a 1.000 M solution of a diprotic acid (H2A) with NaOH, a strong base.

  35. Solubility Product We previously divided ionic compounds into two general categories: insoluble - does not dissolve in water soluble - dissolves in water However, “insoluble” ionic compounds usually dissolve in water to a very small extent. The solubility product, Ksp, is defined as the equilibrium constant for the solubility reaction. Examples: AgCl AgCl(s)  Ag+(aq) + Cl-(aq) Ksp = [Ag+] [Cl-] CaF2 CaF2(s)  Ca2+(aq) + 2 F-(aq) Ksp = [Ca2+] [F-]2 Since our reactant is a solid, equilibrium requires only that there be some solid present in the system.

  36. Ksp(Al(OH)3) = [Al3+][OH-]3 = 1.8 x 10-33 Ksp(CaF) = [Ca2+][F-]2 = 4.0 x 10-11 Ksp(AgBr) = [Ag+][Br-] = 7.7 x 10-13 Ksp(AgCl) = [Ag+][Cl-] = 1.6 x 10-6 Ksp(ZnS) = [Zn2+][S2-] = 3.0 x 10-23 The solid compound does not appear in the expression for Ksp, but must be present in some amount for equilibrium for a saturated solution.

  37. Solubility and Molar Solubility There are two terms that are related to the solubility product. Solubility (solubility by mass) - The number of grams of a compound that will dissolve per 1.00 L of water. Molar solubility - The number of moles of a compound that will dissolve per 1.00 L of water. Of course, our previous methods for indicating solubility (such as grams per 100 grams solvent) are still valid. Example: The solubility product for lead II chloride (PbCl2, MW = 278.1 g/mol) is Ksp = 1.2 x 10-5 at T = 25. C. What are the solubility by mass and the molar solubility of lead II chloride?

  38. Example: The solubility product for lead II chloride (PbCl2, MW = 278.1 g/mol) is Ksp = 1.2 x 10-5 at T = 25. C. What are the solubility by mass and the molar solubility of lead II chloride? PbCl2(s)  Pb2+(aq) + 2 Cl-(aq) Ksp = [Pb2+] [Cl-]2 = 1.2 x 10-5 Substance Initial Change Equilibrium Pb2+ 0.00 + x x Cl- 0.00 + 2x 2x (x) (2x)2 = 4x3 = 1.2 x 10-5 x3 = 3.0 x 10-6 ; x = (3.0 x 10-6)1/3 = 1.44 x 10-2 molar solubility = 1.44 x 10-2 moles/L solubility = 1.44 x 10-2 mol 278.1 g = 4.01 g 1 L 1 mol L

  39. Solubility and the Reaction Quotient Consider a solution where only ions are present (no solid initially present). As a specific example consider a 1:1 ionic compound MX MX(s)  M+(aq) + X-(aq) Qsp = [M+][X-] There are three possibilities for Qsp, the reaction quotient If Qsp < Ksp the solution is unsaturated (more solid could dissolve). If there is no solid present, the solution is stable, but not at equilibrium. If Qsp = Ksp the solution is saturated (no more solid could dissolve). If Qsp > Ksp the solution is supersaturated (a precipitate will form as the system goes to equilibrium).

  40. Consider the solubility of lead II bromide, PbBr2. PbBr2(s)  Pb2+(aq) + 2 Br-(aq) Qsp = [Pb] [Br]2 Qsp < Ksp Qsp = Ksp Qsp > Ksp unsaturated saturated supersaturated “stable” equilibrium “unstable”

  41. Factors Affecting Solubility 1) Temperature. By Le Chatlier’s principle, we may make the following general statements If the solubility reaction is exothermic (H < 0) then solubility decreases as temperature increases. AB(s)  A+(aq) + B-(aq) + “heat” If the solubility reaction is endothermic (H > 0) then solubility increases as temperature increases. AB(s) + “heat”  A+(aq) + B-(aq) 2) pH. This will be true for hydroxide compounds (those containing OH- ion) or compounds containing weak base anions (F-, CN-, CO32-, S2-, PO33-, and other conjugate bases of weak acids). Example: What is the maximum concentration of Ni2+ ion in a buffer solution with pH = 10.00? Ksp(Ni(OH)2) = 5.5 x 10-16.

  42. Example: What is the maximum concentration of Ni2+ ion in a buffer solution with pH = 10.00? Ksp(Ni(OH)2) = 5.5 x 10-16. Ni(OH)2 Ni2+(aq) + 2 OH-(aq) Ksp = [Ni2+] [OH-]2 = 5.5 x 10-16 Since pH = 10.00, pOH = 14.00 - 10.00 = 4.00. So [OH-] = 10-4.00 = 1.0 x 10-4 M Ksp = [Ni2+] [OH-]2 [Ni2+] = Ksp = 5.5 x 10-16 = 5.5 x 10-8 M [OH-]2 (1.0 x 10-4)2 Note that it is possible for [Ni2+] to be  5.5 x 10-8 M.

  43. 3) Common ion effect. The common ion effect refers to the fact that in equilibrium reactions all that matters are the concentrations of the ions involved in the equilibrium. The source of the ions does not matter. If a common ion is already present in solution this can affect the solubility of a compound. Example: What is the molar solubility of AgCl in a) pure water and b) in a 0.100 M solution of NaCl. Ksp(AgCl) = 1.8 x 10-10.

  44. Example: What is the molar solubility of AgCl in a) pure water and b) in a 0.100 M solution of NaCl. Ksp(AgCl) = 1.8 x 10-10. AgCl(s)  Ag+(aq) + Cl-(aq) Ksp = [Ag+] [Cl-] = 1.8 x 10-10 a) pure water compound initial change equilibrium Ag+ 0.00 + x x Cl- 0.00 + x x (x) (x) = x2 = 1.8 x 10-10 x = (1.8 x 10-10)1/2 = 1.34 x 10-5 molar solubility = 1.34 x 10-5 mol/L.

  45. b) 0.100 M solution of NaCl NaCl(s)  Na+(aq) + Cl-(aq) compound initial change equilibrium Ag+ 0.00 + x x Cl- 0.100 + x 0.100 + x (x) (0.100 + x) = 1.8 x 10-10 Assume x << 0.100 (x) (0.100) = 1.8 x 10-10 x= (1.8 x 10-10) = 1.8 x 10-9 (0.100) molar solubility = 1.8 x 10-9 mol/L. This is much smaller than the molar solubility we found for silver chloride in pure water (1.34 x 10-5 mol/L).

  46. Selective Precipitation We may use differences in solubility to remove specific cations from a mixture by selective precipitation, the formation of a precipitate with one ion in the mixture. Example: Consider a solution containing 0.0100 M Mg2+ ion and 0.0100 M Cu2+ ion. Can we separate these ions?

  47. Example: Consider a solution containing 0.0100 M Mg2+ ion and 0.0100 M Cu2+ ion. Can we separate these ions? Consider adding OH- ion. Mg(OH)2(s)  Mg2+(aq) + 2 OH-(aq) Ksp = [Mg2+] [OH-]2 = 5.6 x 10-12 Cu(OH)2(s)  Cu2+(aq) + 2 OH-(aq) Ksp = [Cu2+] [OH-]2 = 1.6 x 10-19 Mg [OH-] = {Ksp/[Mg2+]}1/2 = {5.6 x 10-12/0.01}1/2 = 2.4 x 10-5 M Cu [OH-] = {Ksp/[Cu2+]}1/2 = {1.6 x 10-19/0.01}1/2 = 4.0 x 10-9 M So if OH- is added to the solution Cu(OH)2 will begin to precipitate when [OH-] reaches 4.0 x 10-9. No Mg(OH)2 will precipitate until [OH-] reaches 2.4 x 10-5 M. So we can separate most of the Cu2+ ion out of solution.

  48. Complex Ion Formation Some metal cations will react with small molecules or ions (NH3, CN-, OH-) to form complex ions. A complex ion is defined as a metal cation bonded to one or more small molecules or ions. For example Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) The equilibrium constant for the formation of a complex ion is called the formation constant (or stability constant), Kf Kf = [Ag(NH3)2+] [Ag+] [NH3]2 Formation constants are usually large numbers (meaning complex ion formation is favored when the reactants are present). We may do equilibrium calculations in the same way we do with other systems at equilibrium.

  49. Kf(Cu(CN)42-) = [Cu(CN4)2-] Kf(CO(NH3)63+) = [CO(NH3)63+] [Cu2+][CN-]4 [Co3+][NH3]6

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