1. Ch. 14: Acids and Bases 14.6 Bases

2. Strong Base Weak Base

3. Bases Strong Bases – Group 1A and 2A hydroxides – Ex: NaOH or Ca(OH) 2 Find pH of 5.0x10 -2 M NaOH solution – [OH-] = 5.0x10 -2 M – [H+]=1.0x10 -14 / 5.0x10 -2 M = 2.0x10 -13 – pH = -log(2.0x10 -13 ) = 12.70 What would be another way to get pH? What if it was a Ca(OH) 2 solution?

4. Bases Weak Bases – do not contain OH - – accepts H + from H 2 O to make OH - – usually have at least one unshared pair of electrons – used to form bond with hydrogen

5. Weak Bases K b – refers to reaction of base with water to make conjugate acid and OH - – for weak bases only B(aq) +H 2 O (l)  BH + (aq) + OH - (aq)

6. Strengths of bases: K b, K a, K w The higher the K b, the stronger the base. If you only have the K a, for the conjugate acid of the base you want: K w = K a K b Solve for K b Compare K b values

7. Details of Base Strengths Strong bases have infinite K b Conjugates of Strong Acids have K b = zero Water’s K b = K w

8. Example Find the pH for 15.0 M solution of NH 3 (K b = 1.8x10 -5 ) – NH 3 will create more OH- than water so autoionization can be ignored NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - I15.000 C-x+x+x E15.0-xxx

9. Example

10. Codeine (C 18 H 21 NO 3 ) is a weak organic base. A 5.0x10 -3 M solution of codeine has a pH of 9.95. Calculate the K b for this substance. What is the chemical reaction? C 18 H 21 NO 3 + H 2 O ⇄ HC 18 H 21 NO 3 + + OH - Find [OH-] first from pH – pOH= 14.00-9.95 = 4.05 – [OH-] = 10 -4.05 = 8.9x10 -5

11. Example C 18 H 21 NO 3 + H 2 O ⇄ HC 18 H 21 NO 3 + + OH - C 18 H 21 NO 3 + H 2 O ⇄ HC 18 H 21 NO 3 + + OH - I 5.0x10 -3 00 C-x+x+x E 5.0x10 -3 -x xx x = [OH-] = 8.9x10 -5

12. Example C 18 H 21 NO 3 + H 2 O ⇄ HC 18 H 21 NO 3 + + OH - C 18 H 21 NO 3 + H 2 O ⇄ HC 18 H 21 NO 3 + + OH - I 5.0x10 -3 00 C-x+x+x E 5.0x10 -3 -x xx x = [OH-] = 8.9x10 -5

13. Ch. 14: Acids and Bases 14.7 Polyprotic Acids

14. Polyprotic Acids acids that can donate more than one H+ always dissociate one at a time Example: carbonic acid – H 2 CO 3  H + + HCO 3 - K a1 = 4.3x10 -7 – HCO 3 -  H + + CO 3 2- K a2 = 5.6x10 -11 K a1 > K a2 > K a3 for typical weak polyprotic acid Why?

15. Important Polyprotic Acids phosphoric acid: H 3 PO 4 – only 1 st step creates important amount of H+ sulfuric acid: H 2 SO 4 – 1 st step is strong (H 2 SO 4 is a strong acid) – 2 nd step is weak (HSO 4 - is a weak acid)

16. Practice Polyprotic acid problem Find the pH of a 1.00x10 -2 M sulfuric acid solution

17. Example: H 2 SO 4 Find the pH of a 1.00x10 -2 M sulfuric acid solution First H+ is strong, second H+ is weak. Treat like a mixture of strong and weak acid. (add parts together) – H +, HSO 4 -, H 2 O Species after first dissoc – From 1 st dissociation, [H+]= 1.00x10 -2 M 2 nd dissoc HSO 4 -  SO 4 2- + H + 2 nd dissoc HSO 4 -  SO 4 2- + H + I 1.00x10 -2 0 C-x+x+x E 1.00x10 -2 -x x 1.00x10 -2 +x

18. Example