1. The pK a Scale

2. Acid Dissociation Constant, K a K a, is a measure of how readily H + ions are released HA (aq) + H 2 O (aq) H 3 O + (aq) + A - (aq) K a = [H + ][A - ] [HA]

3. K a Values Stronger acids – higher K a values Weak acids – lower K a values AcidKaKa HCl1.0 x 10 7 HNO 3 4.0 x 10 1 HF5.6 x 10 -4 HCOOH1.6 x 10 -4 CH 3 COOH1.7 x 10 -5

4. pK a Since K a values for weak acids are very small it is convenient to use the pK a scale where: pK a = -log 10 K a If K a = 1.70 x 10 -5 mol dm -3 Then pK a = -log 10 (1.70 x 10 -5 ) pK a = 4.77

5. K a Values & pK a Values Explain the trend in strength of these acids Calculate the pK a values of these acids AcidKaKa CCl 3 COOH0.2 HCOOH1.5 x 10 -4 CH 3 COOH1.6 x 10 -5 CH 3 CH 2 COOH1.3 x 10 -5

6. pK a Values Stronger acids – higher K a values Stronger acids – lower pK a values AcidKaKa pK a CCl 3 COOH0.20.7 HCOOH1.5 x 10 -4 3.8 CH 3 COOH1.6 x 10 -5 4.8 CH 3 CH 2 COOH1.3 x 10 -5 4.9

7. Calculate the pH of a 0.1M solution of a weak acid given that it has a pK a value of 4.76 at 298K. pK a = -log 10 K a 4.76 = -log 10 Ka K a = 1.738 x 10 -5 [H + ] = √ 1.738 x 10 -5 x 0.1 K a ≈ [H + ] 2 [HA] total

8. Calculate the pH of a 0.1M solution of a weak acid given that it has a pK a value of 4.76 at 298K. pH = -log 10 [H + ] pH = 2.88 [H + ] = 1.318 x 10 -3 mol dm -3

9. A 0.01M solution of a weak acid, X, has a pH of 3.37 at 200K. Calculate the pK a value for X at this temperature. pH = -log 10 [H + ] [H + ] = 4.266 x 10 -4 K a ≈ [H + ] 2 [HA] total K a ≈ (4.266 x 10 -4 ) 2 0.01

10. A 0.01M solution of a weak acid, X, has a pH of 3.37 at 200K. Calculate the pK a value for X at this temperature. pK a = -log 10 K a pK a = 4.74 K a ≈ 1.8197 x 10 -5