Additional Aspects of Aqueous Equilibria Aspects of Aqueous

Additional Aspects of Aqueous Equilibria

Aspects of Aqueous Equilibria: The Common Ion Effect Salts like sodium acetate are strong electrolytes Na. C 2 H 3 O 2(aq) Na+(aq) + C 2 H 3 O 2 -(aq) The C 2 H 3 O 2 - ion is a conjugate base of a weak acid HC 2 H 3 O 2(aq) + H 2 O H 3 O+(aq) + C 2 H 3 O 2 -(aq) Ka = [H 3 O+] [C 2 H 3 O 2 -] [HC 2 H 3 O 2]

The Common Ion Effect Now, lets think about the problem from the perspective of Le. Chatelier’s Principle What would happen if the concentration of the acetate ion were increased? +] [C H O -] [H O 3 2 Ka = [HC 2 H 3 O 2] Q > K and the reaction favors reactant Addition of C 2 H 3 O 2 - shifts equilibrium, reducing H+

The Common Ion Effect K = a [H 3 O+] [C 2 H 3 O 2 -] [HC 2 H 3 O 2] HC 2 H 3 O 2(aq) + H 2 O H 3 O+(aq) + C 2 H 3 O 2 -(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 HC 2 H 3 O 2(aq) + H 2 O H 3 O+(aq) + C 2 H 3 O 2 -(aq) So where might the additional C 2 H 3 O 2 -(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?

Na. C 2 H 3 O 2(aq) Na+(aq) + C 2 H 3 O 2 -(aq) HC 2 H 3 O 2(aq) + H 2 O H 3 O+(aq) + C 2 H 3 O 2 -(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, Na. C 2 H 3 O 2, to 1 L of a 0. 100 M solution of acetic acid, HC 2 H 3 O 2. What is the p. H of the resultant solution? Na. C 2 H 3 O 2(aq) Na+(aq) + C 2 H 3 O 2 -(aq) HC 2 H 3 O 2(aq) + H 2 O H 3 O+(aq) + C 2 H 3 O 2 -(aq)

Calculate the p. H of a solution containing 0. 06 M formic acid (HCH 2 O, Ka = 1. 8 x 10 -4) and 0. 03 M potassium formate, KCH 2 O. Now you try it!

Calculate the fluoride ion concentration and p. H of a solution containing 0. 10 mol of HCl and 0. 20 mol HF in 1. 0 L HCl + H 2 O H 3 O+(aq) + Cl-(aq) HF (aq) + H 2 O H 3 O+ (aq) + F-(aq)

Now, lets think about the problem from the perspective of Le. Chatelier’s Principle But this time lets deal with a weak base and a salt containing its conjugate acid. NH 3(aq) + H 2 O NH 4+(aq) + OH- (aq) +] [OH-] Q > K and the [NH 4 Kb = reaction favors [NH 3] reactant Addition of NH 4+ shifts equilibrium, reducing OH-

Calculate the p. H of a solution produced by mixing 0. 10 mol NH 4 Cl with 0. 40 L of 0. 10 M NH 3(aq), p. Kb = 4. 74? NH 4 Cl(aq) NH 4+(aq) + Cl-(aq) NH 3(aq) + H 2 O NH 4+(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) HC 2 H 3 O 2 (aq) + OH-(aq) H 2 O + C 2 H 3 O 2 -(aq) 0. 20 mol 0. 10 mol -0. 10 mol 0 +0. 10 mol

Suppose we react 0. 20 mol of acetic acid with 0. 10 mol of sodium hydroxide HC 2 H 3 O 2(aq) + OH- (aq) H 2 O + C 2 H 3 O 2 -aq) 0. 20 mol -0. 10 mol -0. 10 mol 0 0 0. 10 mol Let’s suppose that all this is occurring in 1. 0 L of solution HC 2 H 3 O 2(aq) + H 2 O H 3 O+(aq) + C 2 H 3 O 2 -(aq) 0 0. 10 M

Sample problem: Calculate the p. H of a solution produced by mixing 0. 60 L of 0. 10 M NH 4 Cl with 0. 40 L of 0. 10 M Na. OH NH 4 Cl(aq) NH 4+(aq) + Cl- (aq) NH 4+ + OH NH 3 + H 2 O 0 0. 06 mol 0. 04 mol -0. 04 mol 0 0. 02 mol 0. 04 mol Don’t forget to convert to MOLARI TIES NH 4+ + 0. 02 M H 2 O H 3 O+ 0 + NH 3 0. 04 M

Calculate the p. H of a solution formed by mixing 0. 50 L of 0. 15 M Na. OH with 0. 50 L of 0. 30 M benzoic acid (HC 7 H 5 O 2, Ka = 6. 5 x 10 -5) Now you try it!

BUFFERED SOLUTIONS A buffered solution is a solution that resists change in p. H 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 +H 2 O H 3 O+ + 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 p. H. Buffering capacity is the amount of acid or base the buffer can neutralize before the p. H begins to change to an appreciable degree. • The p. H 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 p. H, to change + K [H ] = a [X-] Henderson-Hasselbalch Equation [HX] + -] -log[H ] = -log Ka [X p. H = p. Ka + log [X-] [HX]

Na. C 2 H 3 O 2(aq) Na+(aq) + C 2 H 3 O 2 -(aq) 0. 1 M HC 2 H 3 O 2(aq) + H 2 O H 3 O+(aq) + C 2 H 3 O 2 -(aq) 0. 1 M 0 -x 0. 1 - x x x 0. 1 + x x(0. 1 + x ) -5 p. H = 4. 74 -5 x = 1. 8 x 10 = 0. 1 - x Using the Henderson-Hasselbalch Equation Note that these [. 1] are initial p. H = 4. 74 + log [. 1] concentrations

Now consider, for a moment, what would have happened if I had added 0. 020 mol of Na. OH or 0. 02 mol HCl to. 1 M HC 2 H 3 O 2(aq) + H 2 O H 3 O+(aq) + C 2 H 3 O 2 -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 p. H = 2. 9

HC 2 H 3 O 2(aq) + H 2 O H 3 O+(aq) + C 2 H 3 O 2 -aq 0. 10 0. 02 M 0 H+ from HCl Completely dissociates therefore the p. H is calculated without regard for the weak acid p. H = -log [0. 02] p. H = 1. 7

add 2 ml 10 M HCl to 1. 8 x 10 -5 M HCl add 2 ml 10 M HCl to. 1 M HC 2 H 3 O 2 +. 1 M Na. C 2 H 3 O 2 p. H = 4. 74 . 02 M HCl p. H= 1. 7 a drop of 3. 04 to. 12 M HC 2 H 3 O 2 +. 06 M Na. C 2 H 3 O 2 p. H = 4. 56 a drop of. 18

Titration Curves End points Stoichiometrically equivalent quantities of acid and base have reacted HCl(aq) + Na. OH(aq) H 2 O + Na. Cl

Titration of a weak acid and a strong base results in p. H curves that look similar to those of a strong acid-strong base curve except that the curve (a) begins at a higher p. H, (b) the p. H rises more rapidly in the early part of the titration, but more slowly near the equivalence point, and (3) the equivalence point p. H is not 7. 0

Calculating p. H’s from Titrations Calculate the p. H in the titration of acetic acid by Na. OH after 30. 0 ml of 0. 100 M Na. OH has been added to 50 m. L of 0. 100 acetic acid HC 2 H 3 O 2(aq) + OH- H 2 O(l) + C 2 H 3 O 20. 005 mol 0. 003 mol 0 0. 003 mol -0. 003 mol 0. 002 mol 0. 003 mol 0 [. 0375] p. H = 4. 74 + log [. 0250] p. H = 4. 92

Determining the Ka From the Titration Curve p. Ka = p. H = 4. 74

Solubility Equilibria Ksp The equilibrium expression for the following reaction is Ca. F 2(s) Ca 2+(aq) + 2 F-(aq) [Ca 2+] [F-]2 = Ksp [Ca. F 2] Look at the table on page 759 or the appendices A 26, where you will find the value of the ksp to be 4. 1 x 10 -11

Calculating Ksp from solubility The solubility of Bi 2 S 3 is 1. 0 x 10 -15 M what would be the Ksp? Bi 2 S 3(s) 2 Bi 3+(aq) + 3 S 2 -(aq) 1. 0 x 10 -15 2(1. 0 x 10 -15) 3(1. 0 x 10 -15) Ksp = [Bi 3+]2 [S 2 -]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(IO 3)2 is 1. 4 x 10 -7 what would be the solubility? Cu(IO 3)2(s) Cu 2+(aq) + 2 IO 31 -(aq) X X 2 X Ksp= 1. 4 x 10 -7 = (X)(2 X)2 = 4 X 3 X = 3. 3 x 10 -3 mol/L

Common ion effect What would be the solubility of Ag 2 Cr. O 4 in a solution that is. 1 M Ag. NO 3, the Ksp = 9. 0 x 10 -12 Ag 2 Cr. O 4 2 Ag 1+ + Cr. O 42 X 2 X +. 1 X 9. 0 x 10 -12 = (2 X+. 1)2(X) Drop the 2 X as insignificant X = 9. 0 x 10 -10 mol/L

p. H and Solubility Mg(OH)2 Mg 2+ + 2 OHIf the p. H is raised (the OH- is raised) then we have the common ion effect and the solubility is decreased. If the p. H is lowered (the H+ is raised) then OH- is reacted with H+ to make water and the solubility is increased.

p. H and solubility of salts Ag 3 PO 4 3 Ag 1+ + PO 43 If the p. H is lowered (the H+ is raised) the PO 43 - reacts with H+ to make HPO 42 - , which essentially removes PO 43 from the equation, shifting the reaction to the right and the solubility is increased.

Will a precipitate form? If 750 m. L of 4. 00 E-3 M Ce(NO 3)3 is mixed with 300 m. L of 2. 00 E-2 M KIO 3, will a precipitate form? Ce(NO 3)3(aq) + KIO 3 (aq) → Ce(IO 3)3(s) + KNO 3(aq) (750)(4 x 10 -3) = (1050) X; X= 2. 86 x 10 -3 M for Ce 3+ (300)(2 x 10 -2) = (1050) Y; Y= 5. 71 x 10 -3 M for IO 31 Ce(IO 3)3(s) Ce 3+ (aq) + 3 IO 31 - (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