1. Acid Dissociation Constant

2. Dissociation Constants For a generalized acid dissociation, the equilibrium expression would be This equilibrium constant is called the acid-dissociation constant, K a. [H 3 O + ] [A - ] [HA] K c = HA (aq) + H 2 O (l) A - (aq) + H 3 O + (aq)

3. Dissociation Constants The greater the value of K a, the stronger is the acid.

4. Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25  C is 2.38. Calculate K a for formic acid at this temperature. We know that [H 3 O + ] [COO - ] [HCOOH] K a =

5. Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25  C is 2.38. Calculate K a for formic acid at this temperature. To calculate K a, we need the equilibrium concentrations of all three things. We can find [H 3 O + ], which is the same as [HCOO - ], from the pH.

6. Calculating K a from the pH pH = -log [H 3 O + ] 2.38 = -log [H 3 O + ] -2.38 = log [H 3 O + ] 10 -2.38 = 10 log [H 3 O + ] = [H 3 O + ] 4.2  10 -3 = [H 3 O + ] = [HCOO - ]

7. Calculating K a from pH Now we can set up a table… [HCOOH], M[H 3 O + ], M[HCOO - ], M Initially0.1000 Change - 4.2  10 -3 + 4.2  10 -3 At Equilibrium 0.10 - 4.2  10 -3 = 0.0958 = 0.10 4.2  10 -3

8. Calculating K a from pH [4.2  10 -3 ] [0.10] K a = = 1.8  10 -4

9. Calculating Percent Ionization Percent Ionization =  100 In this example [H 3 O + ] eq = 4.2  10 -3 M [HCOOH] initial = 0.10 M [H 3 O + ] eq [HA] initial Percent Ionization =  100 4.2  10 -3 0.10 = 4.2%

10. Calculating pH from K a Calculate the pH of a 0.30 M solution of acetic acid, HC 2 H 3 O 2, at 25  C. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) K a for acetic acid at 25  C is 1.8  10 -5.

11. Calculating pH from K a The equilibrium constant expression is [H 3 O + ] [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] K a =

12. Calculating pH from K a We next set up a table… [C 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2 - ], M Initially0.3000 Change-x+x At Equilibrium 0.30 - x  0.30 xx We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

13. Calculating pH from K a Now, (x) 2 (0.30) 1.8  10 -5 = (1.8  10 -5 ) (0.30) = x 2 5.4  10 -6 = x 2 2.3  10 -3 = x

14. Calculating pH from K a pH = -log [H 3 O + ] pH = -log (2.3  10 -3 ) pH = 2.64