A Different Look At Buffers Write the equation for the dissociation of hydrogen fluoride. (B-L perspective) HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq) Write the Ka expression: Ka = [H3O+] [F-] / [HF] = 7.2 x 10-4 What is the Ka value? Small What does that mean about the [F-]? Small What about the [H3O+]? Small Can you calculate those concentrations with the info given? NO! If you added a quantity of strong acid to the solution would LCP kick in? Yes, but not too much! Would the pH of the solution change? Yes, pH ↓

A Different Look At Buffers (cont’d) HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq) How would you describe the substance NaF in terms of this system? A salt of the conjugate base! If I added some NaF, would LCP kick in? Yes, but not too much!! What would happen to the pH of the solution as a result of the addition? pH ↑ Rearrange the mass action expression to solve for [H3O+]? [H3O+] = Ka ([HF] / [F-]) If I added enough NaF to get to the [F-] to be the same as the [HF] that would be significant…how come? [H3O+] = Ka Another look… (take –log of both sides of this): -log [H3O+] = -log (Ka ([HF] / [F-])) pH = -log Ka + -log ([HF] / [F-]) pH = pKa + -log ([HF] / [F-]) pH = pKa + log [F-] Henderson-Hasselbalch Eq. [HF]

A Different Look At Buffers (cont’d) HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq) What is the pH of a solution of 1.5 M HF? 7.2 x 10-4 = [H3O+] [F-] / [HF] 7.2 x 10-4 = (x) (x) / (1.5-x) x = [H3O+] = 0.033 pH = 1.48 What would the pH of a solution described as 1.5 M HF and 1.3 M NaF? Use Henderson-Hasselbalch: neglect these at eq: [F-] = 1.3 + x pH = pKa + log [F-] [HF] [HF] = 1.5 - x = -log 7.2 x 10-4 + log ([1.3] / [1.5]) = 3.1 – 0.062 = 3.0

A Different Look At Buffers (cont’d) HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq) What if 3.00 mL of KOH (0.90 M) was added to 75.0 mL of the buffer (1.50 M HF; 1.30 M NaF). What would happen to the pH of the solution? First do the stoichiometry… Moles of OH- added = 0.0027 mol Moles of HF initially = 0.113 mol Moles of F- initially = 0.0975 mol Moles of HFafter reaction with OH-= 0.110 mol Moles of F-after reaction with OH-= 0.100 mol [HF] before equilibrium = 0.110 mol/0.078 L = 1.41 M [F-] before equilibrium = 0.100 mol/0.078 L = 1.28 M Use Henderson-Hasselbalch: at eq: [F-] = 1.28 + x pH = pKa + log [F-] [HF] [HF] = 1.41 - x neglect these = -log 7.2 x 10-4 + log ([1.28] / [1.41]) = 3.1 – 0.042 = 3.1

A Different Look At Buffers (cont’d) Summary pH of Weak Acid = 1.48 pH of Buffer = 3.0 pH of Buffer = 3.1 (after addition of strong base) pH = pKa + log [B-] [HB] Buffers work best when: The [ ]’s are large for both acid and the conjugate base The ratio of base to acid is close to 1. Teacher note: follow this up with video on buffering found in chapter 14 folder