2 Aspects of Aqueous Equilibria: The Common Ion EffectSalts like sodium acetate are strong electrolytesNaC2H3O2(aq) Na+(aq) + C2H3O2-(aq)The C2H3O2- ion is a conjugate base of a weak acidHC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)[H3O+] [C2H3O2-]Ka =[HC2H3O2]
3 Now, lets think about the problem from the The Common Ion EffectNow, lets think about the problem from theperspective of LeChatelier’s PrincipleWhat would happen if the concentration of theacetate ion were increased?[H3O+] [C2H3O2-]Ka =[HC2H3O2]Q > K and the reaction favors reactantAddition of C2H3O2- shifts equilibrium, reducing H+
4 HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq) [H3O+] [C2H3O2-]The Common Ion EffectKa =[HC2H3O2]HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)Since the equilibrium has shifted to favor the reactant, it would appear as if the dissociation of the weak acid(weak electrolyte) had decreased.
5 Aspects of Aqueous Equilibria: The Common Ion Effect HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)So where might the additional C2H3O2-(aq) come from? Remember we are not adding H+. So it’s not like we can add more acetic acid.How about from thesodium acetate?
6 NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq) HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)In general, the dissociation of a weak electrolyte (acetic acid) is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyteThe shift in equilibrium which occurs is called theCOMMON ION EFFECT
7 HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq) Let’s explore the COMMON ION EFFECT in a little more detailSuppose that we add 8.20 g or mol sodium acetate, NaC2H3O2, to 1 L of a M solution of acetic acetic acid, HC2H3O2. What is the pH of the resultant solution?NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq)HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)
8 Calculate the pH of a solution containing 0.06 M formic acid (HCH2O, Ka = 1.8 x 10-4)and 0.03 M potassium formate, KCH2O.Now you try it!
9 HCl (aq) + H2O H3O+(aq) + Cl-(aq) Calculate the fluoride ion concentration and pH of a solution containing 0.10 mol of HCl and 0.20 mol HF in 1.0 LHCl (aq) + H2O H3O+(aq) + Cl-(aq)HF (aq) + H2O H3O+ (aq) + F-(aq)
10 Addition of NH4+ shifts equilibrium, reducing OH- Now, lets think about the problem from the perspective of LeChatelier’s PrincipleBut this time lets deal with a weak base and a salt containing its conjugate acid.NH3(aq) + H2O NH4+(aq) + OH- (aq)[NH4+] [OH-]Q > K and the reaction favors reactantKb =[NH3]Addition of NH4+ shifts equilibrium, reducing OH-
11 Calculate the pH of a solution produced by mixing 0 Calculate the pH of a solution produced by mixing 0.10 mol NH4Cl with 0.40 L of 0.10 M NH3(aq), pKb = 4.74?NH4Cl(aq) NH4+(aq) + Cl-(aq)NH3(aq) + H2O NH4+(aq) + OH-
12 HC2H3O2 (aq) + OH-(aq) H2O + C2H3O2-(aq) Common Ions Generated by Acid-Base ReactionsThe common ion that affects a weak-acid or weak-base equilibrium may be present because it is added as a salt, or the common ion can be generated by reacting an acid and base directly (no salt would be necessary….which is kind of convenient if you think about it)Suppose we react 0.20 mol of acetic acid (weak) with mol of sodium hydroxide strong)HC2H3O2 (aq) + OH-(aq) H2O + C2H3O2-(aq)0.20 mol0.10 mol-0.10 mol-0.10 mol+0.10 mol0.10 mol0.10 mol
13 Suppose we react 0.20 mol of acetic acid with 0.10 mol of sodium hydroxideHC2H3O2(aq) + OH- (aq) H2O + C2H3O2-aq)0.20 mol0.10 mol-0.10 mol-0.10 mol0.10 mol0.10 mol0.10 molLet’s suppose that all this is occurring in 1.0 L of solutionHC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)0.10 M0.10 M
14 NH4Cl(aq) NH4+(aq) + Cl- (aq) Sample problem: Calculate the pH of a solution produced by mixing 0.60 L of 0.10 M NH4Cl with 0.40 L of 0.10 M NaOHNH4Cl(aq) NH4+(aq) + Cl- (aq)NH OH NH H2O0.06 mol0.04 mol-0.04 mol-0.04 mol0.04 mol0.02 mol0.04 molDon’t forgetto convert toMOLARITIESNH H2O H3O NH30.04 M0.02 M
15 Calculate the pH of a solution formed by mixing 0. 50 L of 0 Calculate the pH of a solution formed by mixing 0.50 L of M NaOH with0.50 L of 0.30 M benzoic acid(HC7H5O2, Ka = 6.5 x 10-5)Now you try it!
16 adding acid or basecalculate the pH of a solution that has .2 mol of NaOH added to a solution that is .25 M HC2H3O2 and .32M NaC2H3O2 HC2H3O2(aq) + OH- (aq) H2O + C2H3O2-aq
17 HC2H3O2(aq) H C2H3O2-aq-X X XX (.52+X) =1.8 x 10-5.05-X
18 BUFFERED SOLUTIONS Suppose we have a salt: MX M+(aq) + X-(aq) A buffered solution is a solution that resists change in pH upon addition of small amounts of acid or base.Suppose we have a salt: MX M+(aq) + X-(aq)And we’ve added the salt to a weak acid containing the same conjugate base as the salt, HX:HX +H2O H3O+ + X-And the equilibrium expression for this reaction is[H+ ] [ X-]Ka =[HX]Note that the concentration of the H+ is dependent upon the Ka and the ratio between the HX and X- (the conjugateacid-base pair)[HX][H +] =Ka[X-]
19 to change to an appreciable degree. Two important characteristics of a buffer are buffering capacity and pH. Buffering capacity is the amount of acid or base the buffer can neutralize before the pH beginsto change to an appreciable degree.The pH of the buffer depends upon the KaThis capacity depends on the amount of acid and base from which the buffer is madeThe greater the amounts of the conjugate acid-base pair, the more resistant the ratio of their concentrations, and hence the pH, to change[HX][H +] =Ka[X-]Henderson-Hasselbalch Equation[HX]-log[H +] =-log Ka[X-]pH =pKalog[X-]+[HX]
20 NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq) 0.1 M0.1 M0.1 MHC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq)0.1 M0.1-xxx0.1 - xx0.1 + xx(0.1 + x )pH = 4.74x = 1.8 x 10-51.8 x 10-5 =0.1 - xUsing the Henderson-Hasselbalch EquationNote that these are initialconcentrations[.1]pH =4.74+log[.1]
21 A liter of solution containing 0. 100 mol of HC2H3O2 and 0 A liter of solution containing mol of HC2H3O2 and mol NaC2H3O2 forms a buffered solution of pH Calculate the pH of this solution (a) after mol NaOH is added, (b) after adding mol HCl is added.NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq)0.1 M0.1 M0.1 MStep 1HC2H3O2(aq) + OH- H2O C2H3O2- (aq)0.1 M0.02 M0.1 M0.02 M-0.02 M-0.02 M0.08 M0.00 M0.12 MStep 2Henderson-Hasselbalch EquationNote that these areinitial concentrations[.12]4.74+logpH =[.08]pH = 4.92
22 initial concentrations pH = 4.74 log [.08] + [.12] pH = 4.56 A liter of solution containing mol of HC2H3O2 and mol NaC2H3O2 forms a buffered solution of pH Calculate the pH of this solution (a) after mol NaOH is added, (b) after adding mol HCl is added.C2H3O2-(aq) + H+ HC2H3O20.10 M0.02 M0.10 MStep 1-0.02 M-0.02 M0.02 M0.08M0.00 M0.12 MHenderson-Hasselbalch EquationNote that these areinitial concentrationsStep 2pH =4.74log[.08]+[.12]pH = 4.56
23 HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-aq Now consider, for a moment, what would have happened if I had added mol of NaOH or 0.02 mol HCl to .1 M HC2H3O2.HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-aq0.1 M-xxx0.1 - xxxx21.8 x 10-5 =0.1 - xx =pH = 2.9
25 pH = -log [0.02] pH = 1.7 HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-aq 0.100.02 MH+ from HCl Completely dissociates therefore thepH is calculated without regard for the weak acidpH = -log [0.02]pH = 1.7
26 add 2 ml10 M HClto .1M HC2H3O2+ .1M NaC2H3O2pH = 4.74add 2 ml10 M HClto 1.8 x 10-5M HClpH = 4.74to .12M HC2H3O2+ .06M NaC2H3O2pH = 4.56a drop of .18.02 M HClpH= 1.7a drop of 3.04
27 before any acid or base are added, Sample exercise: Consider a litter of buffered solution that is M in formic acid (HC2H3O) and M in sodium formate (NaC2H3O ). Calculate the pH of the bufferbefore any acid or base are added,(b) after the addition of mol HNO3,(c) after the addition of mol KOH
28 Titration Curves End points Stoichiometrically equivalent quantities of acid and base have reactedHCl(aq) + NaOH(aq) H2O + NaCl
29 Titration of a weak acid and a strong base results in pH curves that look similar to those of a strong acid-strong base curve except that the curve (a) begins at a higher pH, (b) the pH rises more rapidly in the early part of the titration, but more slowly near the equivalence point, and (3) the equivalence point pH is not 7.0
31 Calculating pH’s from Titrations Calculate the pH in the titration of acetic acid by NaOH after 30.0 ml of M NaOH has been added to 50 mL of acetic acidHC2H3O2(aq) + OH- H2O(l) + C2H3O2-0.005 mol0.003 molmol0.003 molmolmol0.003 mol[.0370]logpH =4.74+[.0250]pH = 4.91
32 Determining the Ka From the Titration Curve pKa = pH = 4.74
33 Solubility Equilibria Ksp The equilibrium expression for the following reaction is CaF2(s) Ca2+(aq) + 2F-(aq)[Ca2+] [F-]2[CaF2]Look at the table on page 759 or the appendices A26, where you will find the value of the ksp to be 4.1 x 10-11= Ksp
34 Calculating Ksp from solubility The solubility of Bi2S3 is 1.0 x 10-15M what would be the Ksp?Bi2S3(s) Bi3+(aq) S2-(aq)1.0 x (1.0 x 10-15) 3(1.0x 10-15)Ksp = [Bi3+]2 [S2-]3Ksp = (2.0 x 10-15)2(3.0 x 10-15)3 = 1 x 10-73
35 Calculating solubility from Ksp The Ksp of Cu(IO3)2 is 1.4 x 10-7 what would be the solubility?Cu(IO3)2(s) Cu2+(aq) IO31-(aq)X X XKsp= 1.4 x 10-7 = (X)(2X)2 = 4X3X = 3.3 x 10-3 mol/L
36 Drop the 2X as insignificant Common ion effectWhat would be the solubility of Ag2CrO4 in a solution that is .1M AgNO3,the Ksp = 9.0 x 10-12Ag2CrO4 2Ag1+ + CrO42-X X X9.0 x = (2X+.1)2(X)Drop the 2X as insignificantX = 9.0 x mol/L
37 pH and Solubility Mg(OH)2 Mg2+ + 2OH- If the pH is raised (the OH- is raised) then we have the common ion effect and the solubility is decreased.If the pH is lowered (the H+ is raised) then OH- is reacted with H+ to make water and the solubility is increased.
38 pH and solubility of salts Ag3PO4 3Ag1+ + PO43-If the pH is lowered (the H+ is raised) the PO43- reacts with H+ to make HPO42- , which essentially removes PO43-from the equation, shifting the reaction to the right and the solubility is increased.
39 Will a precipitate form? If 750 mL of 4.00E-3M Ce(NO3)3 is mixed with 300 mL of 2.00E-2M KIO3, will a precipitate form?Ce(NO3)3(aq) + KIO3 (aq) → Ce(IO3)3(s) + KNO3(aq)(750)(4 x 10-3) = (1050) X; X= 2.86 x 10-3M for Ce3+(300)(2 x 10-2) = (1050) Y; Y= 5.71 x 10-3 M for IO31-Ce(IO3)3(s) Ce3+ (aq) + 3IO31- (aq)Ksp = 1.9 x 10-10Q = (2.86 x 10-3)(5.71 x 10-3)3 = 5.32 x 10-10Q is greater than K so a precipitate will form