1. Chem 1310: Introduction to physical chemistry Part 4a: Doing pH calculations A recipe with examples. Don't use without thinking!

2. only strong acids? C A >>  K w ? yes assume [H + ]=C A, calc pH, [OH - ] from there use ICE for H 2 O equilibrium, solve quadratic equation only weak acid? C A K a >>  K w ? no yes no yes C A >> K a ? use ICE for HA equilibrium, assume [H + ] << C A use ICE for HA equilibrium, solve quadratic equation no calculation gets too difficult

3. salt of weak acid /weak base? yes consider the conjugated acid and base! acidic if K a (conj acid) > K b (conj base) calculation too difficult (?) use ICE for HA equilibrium, assume [H + ] << C A,C B no yes no mixture of acid and conjugate base? calculation gets too difficult buffer this produces the buffer formula (Henderson-Hasselbalch)

4. Strong acid, "high" concentration C A >>  K w Example: 0.022 M HCl [H + ] = 0.022 M pH = - log 0.022 = -(-1.66) = 1.66

5. Strong acid, low concentration C A   K w Example: 2*10 -7 M HNO 3 x = 4.1*10 -8 [H + ] = 1.41*10 -7 M pH = - log 1.41*10 -7 = 6.85 H+H+ OH - ICACA 0 C+x+x+x+x ECA+xCA+xx

6. Weak acid, "high" concentration C A >> K a 0.07 M CH 3 COOH, K a = 1.8*10 -5 x = [H + ] = 1.11*10 -3 M (exact); 1.12*10 -3 M (approx) pH = - log 1.11*10 -3 = 2.95 HAA-A- H+H+ ICACA 00 C-x-x+x+x+x+x ECA-xCA-xxx

7. Salt of weak acid, weak base HAB-B- A-A- HBH+H+ OH - ICC0000 EC-aC-aC-bC-babxy

8. Salt of weak acid, weak base 0.013 M NH 4 F, K a = 5.6*10 -10, K b = 1.4*10 -11 x = [H+] = 6.16*10 -7 M (exact); 6.33*10 -7 M (approx) pH = - log 6.16*10 -7 = 6.21 (6.20 approx)

9. Buffer, "high" concentration C A,C B >> K a 0.28 M H 3 PO 4, 0.15 M NaH 2 PO 4, K a = 7.5*10 -3 x = [H + ] = 1.24*10 -2 M (exact); 1.40*10 -2 M (approx) pH = - log 1.24*10 -2 *10 -3 = 1.91 (1.85 approx) HAA-A- H+H+ ICACA 00 C-x-x+x+x+x+x ECA-xCA-xxx