Additional Aspects of Aqueous Equilibria Aspects of Aqueous Equilibria: The Common Ion Effect Salts like sodium acetate are strong electrolytes NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq) The C2H3O2- ion is a conjugate base of a weak acid HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq) Ka = [H3O+] [C2H3O2-] [HC2H3O2] The Common Ion Effect Now, lets think about the problem from the perspective of LeChatelier’s Principle What would happen if the concentration of the acetate ion were increased? +] [C H O -] [H O 3 2 3 2 Ka = [HC2H3O2] Q > K and the reaction favors reactant Addition of C2H3O2- shifts equilibrium, reducing H+ The Common Ion Effect K = a [H3O+] [C2H3O2-] [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. 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 the sodium acetate? 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 electrolyte The shift in equilibrium which occurs is called the COMMON ION EFFECT Let’s explore the COMMON ION EFFECT in a little more detail Suppose that we add 8.20 g or 0.100 mol sodium acetate, NaC2H3O2, to 1 L of a 0.100 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) 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! Calculate the fluoride ion concentration and pH of a solution containing 0.10 mol of HCl and 0.20 mol HF in 1.0 L HCl + H2O H3O+(aq) + Cl-(aq) (aq) HF (aq) + H2O H3O+ (aq) + F-(aq) Now, lets think about the problem from the perspective of LeChatelier’s Principle But this time lets deal with a weak base and a salt containing its conjugate acid. NH3(aq) + H2O NH4+(aq) + OH- (aq) +] [OH-] Q > K and the [NH 4 Kb = reaction favors [NH3] reactant Addition of NH4+ shifts equilibrium, reducing OH- 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- Common Ions Generated by Acid-Base Reactions The 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 0.10 mol of sodium hydroxide strong) HC2H3O2 (aq) + OH-(aq) H2O + C2H3O2-(aq) 0.20 mol 0.10 mol -0.10 mol -0.10 mol 0 0.10 mol 0 +0.10 mol 0.10 mol Suppose we react 0.20 mol of acetic acid with 0.10 mol of sodium hydroxide HC2H3O2(aq) + OH- (aq) H2O + 0.20 mol -0.10 mol 0.10 mol 0.10 mol -0.10 mol 0 C2H3O2-aq) 0 0.10 mol 0.10 mol Let’s suppose that all this is occurring in 1.0 L of solution HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq) 0 0.10 M 0.10 M 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 NaOH NH4Cl(aq) NH4+(aq) + Cl- (aq) NH4+ + OH- NH3 + H2O 0 0.06 mol 0.04 mol -0.04 mol -0.04 mol 0.04 mol 0 0.02 mol 0.04 mol Don’t forget to convert to MOLARI TIES NH4+ + 0.02 M H2O H3O+ 0 + NH3 0.04 M Calculate the pH of a solution formed by mixing 0.50 L of 0.015 M NaOH with 0.50 L of 0.30 M benzoic acid (HC7H5O2, Ka = 6.5 x 10-5) Now you try it! adding acid or base calculate 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 .20 .32 ..25 -.20 -.20 +.20 .05 0 .52 HC2H3O2(aq) .05 -X H+ 0 X X (.52+X) =1.8 x 10-5 .05-X + C2H3O2-aq .52 X BUFFERED SOLUTIONS 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 HX +H2O H3O+ + Xsalt, HX: And the equilibrium [H+ ] [ X-] expression for this reaction Ka = [HX] is Note that the concentration of the H+ is dependent upon the Ka and the ratio between the [HX] HX and X (the conjugate [H +] = Ka [X-] acid-base pair) 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 begins to change to an appreciable degree. •The pH of the buffer depends upon the Ka •This capacity depends on the amount of acid and base from which the buffer is made •The greater the amounts of the conjugate acid-base pair, the more resistant the ratio of their concentrations, and [HX] hence the pH, to change + K [H ] = a [X-] Henderson-Hasselbalch Equation [HX] + -] -log[H ] = -log Ka [X pH = pKa log + [X-] [HX] NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq) 0.1 M 0.1 M 0.1 M HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-(aq) 0.1 M 0 -x 0.1 - x x x 0.1 x 0.1 + x x(0.1 + x ) -5 pH = 4.74 -5 x = 1.8 x 10 1.8 x 10 = 0.1 - x Using the Henderson-Hasselbalch Equation Note that these [.1] are initial pH = 4.74 + log [.1] concentrations A liter of solution containing 0.100 mol of HC2H3O2 and 0.100 mol NaC2H3O2 forms a buffered solution of pH 4.74. Calculate the pH of this solution (a) after 0.020 mol NaOH is added, (b) after adding 0.020 mol HCl is added. NaC2H3O2(aq) Na+(aq) + C2H3O2-(aq) 0.1 M 0.1 M 0.1 M Step 1 HC2H3O2(aq) + OH- H2O + C2H3O2- (aq) 0.1 M 0.02 M 0.1 M -0.02 M 0.08 M -0.02 M 0.00 M 0.02 M 0.12 M Step 2 Henderson-Hasselbalch Equation Note that these are [.12] pH = 4.74 + log initial concentrations [.08] pH = 4.92 A liter of solution containing 0.100 mol of HC2H3O2 and 0.100 mol NaC2H3O2 forms a buffered solution of pH 4.74. Calculate the pH of this solution (a) after 0.020 mol NaOH is added, (b) after adding 0.020 mol HCl is added. C2H3O2-(aq) + H+ HC2H3O2 0.10 M 0.02 M 0.10 M Step 1 -0.02 M -0.02 M 0.02 M 0.12 M 0.08M 0.00 M Henderson-Hasselbalch Equation Step 2 pH = 4.74 + log [.08] Note that these are pH = 4.56 [.12] initial concentrations Now consider, for a moment, what would have happened if I had added 0.020 mol of NaOH or 0.02 mol HCl to .1 M HC2H3O2. HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-aq 0 0.1 M 0 -x x x 0.1 - x x x 2 x 1.8 x 10-5 = 0.1 - x x = 0.0013 pH = 2.9 HC2H3O2(aq) + OH- H2O + C2H3O2-(aq) 0.1 M 0.02 M 0.00 -0.02 M -0.02 M 0.02 M 0.08 M 0.00 M 0.02 M Henderson-Hasselbalch Equation pH = 4.74 + log [.02] [.08] pH = 4.13 HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2-aq 0.10 0.02 M 0 H+ from HCl Completely dissociates therefore the pH is calculated without regard for the weak acid pH = -log [0.02] pH = 1.7 add 2 ml 10 M HCl to 1.8 x 10-5M HCl add 2 ml 10 M HCl to .1M HC2H3O2 + .1M NaC2H3O2 pH = 4.74 pH = 4.74 .02 M HCl pH= 1.7 a drop of 3.04 to .12M HC2H3O2 + .06M NaC2H3O2 pH = 4.56 a drop of .18 Sample exercise: Consider a litter of buffered solution that is 0.110 M in formic acid (HC2H3O) and 0.100 M in sodium formate (NaC2H3O ). Calculate the pH of the buffer (a)before any acid or base are added, (b) after the addition of 0.015 mol HNO3, (c) after the addition of 0.015 mol KOH Titration Curves End points Stoichiometrically equivalent quantities of acid and base have reacted HCl(aq) + NaOH(aq) H2O + NaCl 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 Calculating pH’s from Titrations Calculate the pH in the titration of acetic acid by NaOH after 30.0 ml of 0.100 M NaOH has been added to 50 mL of 0.100 acetic acid HC2H3O2(aq) + OH- H2O(l) + C2H3O20.005 mol 0.003 mol 0 0.003 mol -0.003 mol -0.003 mol -0.002 mol 0.003 mol 0 [.0370] pH = 4.74 + log [.0250] pH = 4.91 Determining the Ka From the Titration Curve pKa = pH = 4.74 Solubility Equilibria Ksp The equilibrium expression for the following reaction is CaF2(s) Ca2+(aq) + 2F-(aq) [Ca2+] [F-]2 = Ksp [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 Calculating Ksp from solubility The solubility of Bi2S3 is 1.0 x 10-15M what would be the Ksp? Bi2S3(s) 2Bi3+(aq) + 3S2-(aq) 1.0 x 10-15 2(1.0 x 10-15) 3(1.0x 10-15) Ksp = [Bi3+]2 [S2-]3 Ksp = (2.0 x 10-15)2(3.0 x 10-15)3 = 1 x 10-73 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) + 2IO31-(aq) X X 2X Ksp= 1.4 x 10-7 = (X)(2X)2 = 4X3 X = 3.3 x 10-3 mol/L Common ion effect What would be the solubility of Ag2CrO4 in a solution that is .1M AgNO3, the Ksp = 9.0 x 10-12 Ag2CrO4 2Ag1+ + CrO42X 2X + .1 X 9.0 x 10-12 = (2X+.1)2(X) Drop the 2X as insignificant X = 9.0 x 10-10 mol/L pH and Solubility Mg(OH)2 Mg2+ + 2OHIf 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. pH and solubility of salts Ag3PO4 3Ag1+ + PO43If the pH is lowered (the H+ is raised) the PO43- reacts with H+ to make HPO42- , which essentially removes PO43from the equation, shifting the reaction to the right and the solubility is increased. 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 IO31Ce(IO3)3(s) Ce3+ (aq) + 3IO31- (aq) Ksp = 1.9 x 10-10 Q = (2.86 x 10-3)(5.71 x 10-3)3 = 5.32 x 10-10 Q is greater than K so a precipitate will form