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) [H3O+] [C2H3O2-] Ka = [HC2H3O2]

Now, lets think about the problem from the 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? [H3O+] [C2H3O2-] Ka = [HC2H3O2] Q > K and the reaction favors reactant Addition of C2H3O2- shifts equilibrium, reducing H+

HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq) [H3O+] [C2H3O2-] The Common Ion Effect Ka = [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

HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq) 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!

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 L HCl (aq) + H2O  H3O+(aq) + Cl-(aq) HF (aq) + H2O  H3O+ (aq) + F-(aq)

Addition of NH4+ shifts equilibrium, reducing OH- 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) [NH4+] [OH-] Q > K and the reaction favors reactant Kb = [NH3] Addition of NH4+ shifts equilibrium, reducing OH-

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-

HC2H3O2 (aq) + OH-(aq)  H2O + C2H3O2-(aq) 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 +0.10 mol 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 + C2H3O2-aq) 0.20 mol 0.10 mol -0.10 mol -0.10 mol 0.10 mol 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) 0.10 M 0.10 M

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 NaOH NH4Cl(aq)  NH4+(aq) + Cl- (aq) NH4+ + OH-  NH3 + H2O 0.06 mol 0.04 mol -0.04 mol -0.04 mol 0.04 mol 0.02 mol 0.04 mol Don’t forget to convert to MOLARITIES NH4+ + H2O  H3O+ + NH3 0.04 M 0.02 M

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 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 ..25 .20 .32 -.20 -.20 +.20 .05 0 .52

HC2H3O2(aq)  H+ + C2H3O2-aq .05 0 .52 -X X X X (.52+X) =1.8 x 10-5 .05-X

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 conjugate acid-base pair) [HX] [H +] = Ka [X-]

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 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 hence the pH, to change [HX] [H +] = Ka [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 0.1 -x x x 0.1 - x x 0.1 + x x(0.1 + x ) pH = 4.74 x = 1.8 x 10-5 1.8 x 10-5 = 0.1 - x Using the Henderson-Hasselbalch Equation Note that these are initial concentrations [.1] pH = 4.74 + log [.1]

A liter of solution containing 0. 100 mol of HC2H3O2 and 0 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 -0.02 M -0.02 M 0.08 M 0.00 M 0.12 M Step 2 Henderson-Hasselbalch Equation Note that these are initial concentrations [.12] 4.74 + log pH = [.08] pH = 4.92

initial concentrations pH = 4.74 log [.08] + [.12] pH = 4.56 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 0.08M 0.00 M 0.12 M Henderson-Hasselbalch Equation Note that these are initial concentrations Step 2 pH = 4.74 log [.08] + [.12] pH = 4.56

HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-aq 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 0.1 M -x x x 0.1 - x x x x2 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 = [.02] 4.74 + log [.08] pH = 4.13

pH = -log [0.02] pH = 1.7 HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-aq 0.10 0.02 M 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 .1M HC2H3O2 + .1M NaC2H3O2 pH = 4.74 add 2 ml 10 M HCl to 1.8 x 10-5M HCl pH = 4.74 to .12M HC2H3O2 + .06M NaC2H3O2 pH = 4.56 a drop of .18 .02 M HCl pH= 1.7 a drop of 3.04

before any acid or base are added, 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 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) + C2H3O2- 0.005 mol 0.003 mol -0.003 mol 0.003 mol -0.003 mol -0.002 mol 0.003 mol [.0370] log pH = 4.74 + [.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 [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

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

Drop the 2X as insignificant 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+ + CrO42- X 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+ + 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.

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.

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 10-10 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