ACIDS AND BASES: Dissociation Constants

C12-5-09 ACID/BASE EQUILIBRIUM OUTCOME QUESTION(S): C12-5-09 ACID/BASE EQUILIBRIUM Describe the relationship between the hydronium and hydroxide ion concentrations in water Include: the ion product of water, Kw Write the equilibrium expression (Ka or Kb) and solve problems involving pH, pOH, and percent dissociation. These problem types use skills seen during the equilibrium unit and should be recognizable… Vocabulary & Concepts 

HA H+(aq) + A–(aq) HA + H2O H3O+ + A- BOH B+(aq) + OH–(aq) Strong Acid HA + H2O H3O+ + A- Weak Acid Ka = [H+][A-] [HA] Acid dissociation constant: BOH B+(aq) + OH–(aq) Strong Base B + H2O BH+ + OH- Weak Base Kb = [B+][OH-] [B] Base dissociation constant:

Ka = 1.7 x 10-5 HC2H3O2 + H2O H3O+ + C2H3O2- [I] [C] Initially a 0.10 M solution of acetic acid, it reaches equilibrium with a [H3O+] = 1.3 x 10-3 M. What is the acid dissociation constant, Ka? ICE table… HC2H3O2 + H2O H3O+ + C2H3O2- [I] 0.10 0 0 [C] -1.3 x 10-3 +1.3 x 10-3 +1.3 x 10-3 0. 0987 1.3 x 10-3 [E] 1.3 x 10-3 Ka = [H3O+][C2H3O2-] [HC2H3O2] Ka = 1.7 x 10-5 Ignore the units for K Ka = [1.3 x 10-3][1.3 x 10-3] [0.0987]

HA(aq) + H2O(l) H3O+(aq) + A-(aq) The weak acid HA has a [HA]i = 0.50 M and an equilibrium constant of Ka = 7.3 x 10-8. What are the equilibrium concentrations? HA(aq) + H2O(l) H3O+(aq) + A-(aq) [I] 0.50 0 0 -x +x +x [C] [E] 0.5 - x x x Type IV Problem: Given only initial values and Keq Solve for the variable x Ka = [H+][A-] [HA]

HA(aq) + H2O(l) H3O+(aq) + A-(aq) 0.5 - x x x [E] 0.49981 1.9 x 10-4 1.9 x 10-4 ~ 0.5 *see…it was OK to ignore it Ka = [H+][A-] [HA] 7.3 x 10-8 = [x][x] 0.50 Based on the very small Ka value – 7.3 x 10-8 – assume that x will be negligible compared to the large 0.50 initial concentration and ignore it (always) - x (7.3 x 10-8)(0.50) = x2 √ 3.65 x 10-8 = x2 √ 1.9 x 10-4 = x

If you can’t make the equation – find it on the Acid Strength Chart Calculate the pH of a 0.10 mol/L hydrogen sulfide solution. (Ka=1.0 x 10-7) If you can’t make the equation – find it on the Acid Strength Chart H2S(g) + H2O(l) H3O+(aq) + HS-(aq) 0.10-x x x [E] [I] 0.10 0 0 1.0 x 10-4 [C] -x +x +x Ka = [H+][HS-] [H2S] √ 1.0 x 10-8 = x2 √ [E] 0.10-x x x 1.0 x 10-4 = x 1.0 x 10-7 = [x][x] 0.10 Equation 1: pH = - log [H+] = - log(1.0 x 10-4) = 4.00 - x (1.0 x 10-7)(0.10) = x2

How much product has formed Percent Dissociation Ka / Kb represent the degree of dissociation How much product has formed Percent dissociation (PD) is another way to represent how much has ionized / dissociated in solution (acidic or basic): Percent dissociation = [H+] [HA] • 100% Percent dissociation = [OH-] [B] • 100%

Weak acid <<100% dissociated Calculate the PD of a solution of formic acid (CH2OOH) if the hydronium ion concentration is . 0.100 M 4.21 x 10-3 M CH2O2H + H2O H3O+ + CH2O2- [E] 0.100 4.21 x 10-3 4.21 x 10-3 Percent dissociation = [H+] [HA] • 100% Weak acid <<100% dissociated PD = [4.21 x 10-3] [0.100] • 100% PD = 4.21% of the original acid dissociated

HPO42- + H2O H2PO4- + OH- [E] 0.25 - x +x +x Calculate the Kb and pH of a 0.25 M solution of a hydrogen phosphate if the dihydrogen phosphate ion (H2PO42-) dissociates 0.080%. 1. Construct the solution equation H+ HPO42- + H2O H2PO4- + OH- base [E] 0.25 - x +x +x Unknown Equation Find the compound structure Figure out if the solution is acidic or basic Move the proton and create the products Fill in the necessary ICE table information You may not need to write the ICE table down if you recognize the pattern…

This is the kind of compound question to expect on a test… HPO42- + H2O H2PO4- + OH- [E] 0.25 +x +x - x 2.0 x 10-4 2.0 x 10-4 2. Use the PD formula to complete the ICE table PD = [OH-] 100% • [B] (0.080%)(0.25) = [OH-] 100% [OH-]= 2.0 x 10-4 M

HPO42- + H2O H2PO4- + OH- Kb = [H2PO4-][OH-] [HPO42-] 3. Use ICE table to find Kb and pH HPO42- + H2O H2PO4- + OH- [E] 0.25 +x +x 2.0 x 10-4 2.0 x 10-4 Kb = [H2PO4-][OH-] [HPO42-] pH pOH [OH-] Kb= [2.0 x 10-4][2.0 x 10-4] 0.25 Equation 3: pOH = -log[OH-] pOH = 4.00 Equation 5: pH = 14.00 - 4.00 pH = 10.00 Kb = 1.6 x 10-7 Be careful with pH – the solution is basic so we have to make conversions

CAN YOU / HAVE YOU? C12-5-09 ACID/BASE EQUILIBRIUM Describe the relationship between the hydronium and hydroxide ion concentrations in water Include: the ion product of water, Kw Write the equilibrium expression (Ka or Kb) and solve problems involving pH, pOH, and percent dissociation. Vocabulary & Concepts