1. ACIDS AND BASES Dissociation Constants

2. weaker the acid, the stronger its conjugate base stronger the acid, the weaker its conjugate base

3. stronger the base, the weaker its conjugate acid weaker the base, the stronger its conjugate acid

4.  Write equilibrium expression for an acid or base  Calculate the acid/base dissociation constant  Calculate the percent dissociation

5. HA (aq) H + (aq) + A - (aq) K a - acid dissociation constant HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) Strong Acid Weak Acid Larger K a : strong acid: more product : more H +.

6. BOH (aq) B + (aq) + OH - (aq) Larger K b : strong base : more product : more OH -. K b - base dissociation constant Strong Base B (aq) + H 2 O (l) BH + (aq) + OH - (aq) Weak Base

7. Initially a 0.10 M solution of acetic acid, it reached equilibrium with a [H 3 O + ] = 1.3 x 10 -3 M. What is the acid dissociation constant, K a ? CH 3 COOH (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 ¯ (aq) I 0.10 0 0 C -1.3 x 10 -3 +1.3 x 10 -3 +1.3 x 10 -3 E 0.987 1.3 x 10 -3 1.3 x 10 -3

8. K a = 1.7 x 10 -5 There are no units for the Ka value

9. HA is a weak acid with a K a of 7.3 x 10 -8. What are the equilibrium concentrations (HA, H 3 O + and A¯) if the initial concentration of HA is 0.50 mol/L? I 0.50 0 0 C -x +x+x E 0.5-x +x +x HA (aq) + H 2 O (l) H 3 O + (aq) + A¯ (aq)

10. *K a is small - assume that x is negligible compared to 0.50

11. [H 3 O + ] = [A¯] = x = 1.9 x 10¯ 4 mol/L [HA] = 0.50 - x = 0.50 - 1.9 x 10¯ 4 = 0.49981 mol/L [HA] = 0.50 mol/L *K a is small - assume that x is negligible compared to 0.50

12. Calculate the pH of a 0.10 mol/L hydrogen sulfide solution. (K a =1.0 x 10 -7 ) H 2 S (aq) + H 2 O (l) H 3 O + (aq) + HS - (aq) I 0.10 00 C -x+x+x E 0.10 - xxx

13. [H 3 O + ] = x = 1.0 x 10 -4 mol/L pH = -log [H 3 O + ] = -log(1.0 x 10 -4 ) pH = 4.00

14. Each acid/base has K associated with it. Diprotic/triprotic acids lose their hydrogens one at a time - Each ionization reaction has separate K a. Sulfuric acid H 2 SO 4 H 2 SO 4(aq) H + (aq) + HSO 4 ¯ (aq) HSO 4 ¯ (aq) H + (aq) + SO 4 -2 (aq) K a1 K a2

15. Percent Dissociation The dissociation constants represent the acid / base degree of dissociation. Another way to describe the amount of dissociation is by percent dissociation.

16. Calculate the percent dissociation of a 0.100 M solution of formic acid (CH 2 OOH) if the hydronium ion concentration is 4.21 x 10 -3 M. CH 2 O 2 H (aq) + H 2 O (l) H 3 O + (aq) + CH 2 O 2 ¯ (aq)

17. Calculate the K b of hydrogen phosphate ion (HPO 4 2 ¯) if a 0.25 mol/L solution of hydrogen phosphate is dissociated is 0.080%. HPO 4 2 ¯ + H 2 O H 2 PO 4 ¯ + OH¯ [OH - ] = [H 2 PO 4 - ] = 2.0 x 10 -4 mol/L

18. HPO 4 2 ¯ + H 2 O H 2 PO 4 ¯ + OH¯

19.  The smaller the K a or K b, the weaker the acid / base  The percent dissociation also describes the amount of acid/base dissociated  The percent dissociated is calculated by