1. The Relationship between K a and K b : For the ionization of the weak acid HA ; HA + H 2 O ↔ H 3 O + + A -, K a = [H 3 O + ] [A - ] [HA] For the ionization of bases A - + HOH ↔ HA + OH -, K b = [HA] [OH - ] [A - ] Thus [H 3 O + ] = [HA] K a, [A] [OH - ] = [A - ] K b [HA]

2. Since K w = [OH - ] [H + ], thus substituting from former equations ; K w = [HA]K a x [A - ] K b or K w = K a x K b [A - ] [HA] Taking logarithms : log K a + log K b = log K w Multyplying -1, - log K a - log K b = log K w pk a + pK b = 14 The Relationship between pH and pK a : From the dissociation of the weak acid HA, K a = [H + ] [A - ] = [H + ] 2 (since [H + ] = [A - ] ) [HA] [HA] [H + ] 2 = K a [HA] thus [H + ] = K a [HA] Log [H + ] = ½ log K a [HA] = ½ ( log K ab + log[HA]) -log[H + ] = -log K a - log[HA] 2

3. PH = pK a + p[HA] 2 Similarly pOH = pK b + p[A - ] 2 Example ; The weak acid HA is o.1%ionized in a 0.2Msolution. a)what is the equilibrium constant for the dissociation of the acid b)what is the pH of the solution ? c)compare the acidity of the HA solution with that of a 0.2M HCL Solution. Solution: HA ↔ H + + A - 0.2M 0 0 The amount of HA dissociated = 0.2 x 0.1 = 2 x 10 -4 M 100 At equilibrium 0.2- 2x10 -4 2x10 -4 2 x 10 -4

4. When the amount dissociated is small (10% or less) it can be ignored and thus not subtracted from the denominator, K a = (2 x 10 -4 ) ( 2 x 10 -4 ) = 4 x 10 -8 = 2 x 10 -7 o.2 0.2 b) pH = - log[H + ], pH = - log 2 x 10 -4, pH = 3.7. c) A 0.2Msolution of HCL would be 100% ionized and produce 0.2M H +, thus its pH = -log 0.2 = 0.7. Henderson – Hasselbalch equation : From the dissociation of the weak acid HA, K a = [H + ] [A - ] [HA] [H + ][A - ] = K a [HA], thus [H + ] = K a [HA] [A - ]

5. Taking logarithms of both sides ; Log [H + ] = log K a + log [HA] [A - ] Multiplying both sides by -1, - log [H + ] = - log K a – log [HA] [A - ] PH = pK a + log [A - ] [HA] Similarily for weak bases ; pOH = pK b + log [M + ] [MOH] Titration of a weak acid l; When a weak acid is titrated with a strong base, For the following weak acid HA ↔ H + + A - The OH - ions of the base are neutralized by the H + as seen in The following equation ; OH -- + H + → H 2 O

6. The removal of H + disturbs the equilibrium between the weak acid And iits ions,therefore more HA ionizes to reestablish the Equilibrium and the newly produced H + is then neutralized by More OH - and this continues until all the hydrogen originally Present is neutralized thus behaving as a strong acid). 1-The pH of the weak acid before the titration begins can be Calculated from pH= pK a +p[HA] 2 2-The pH during the titration can be calculated from the Henderson – Hasselbalch equation ; pH = pKa + log [A - ] [HA] 3-The pH at the equivalence point is not 7 but higher because of The hydrolysis of A, - when HA is consumed the A - reacts with H 2 O To produce OH - and the undissociated weak acid HA as seen Seen in the Following equation ; HA + OH - ↔ H 2 O + A - The pH at this stage is calculated from pOH = pK b + p[A - ] 2

7. Thus pH is calculated from pK w = pH + pOH Example ; Calculate the appropriate values and draw the curve for the titration of 500ml of 0.1M weak acid HA, with 0.1M KOH. K a = 10 -5 (pK a = 5.0). Solution; The values needed to draw the curve are calculated using the General outline given above ; a)At start, the pH depends on the concentration of HA and the Value of K a only. pH= pK a + p[HA] = 5 + 1 = 3 2 2 b) At any point during the titration the pH can be calculated using the Henderson – Hasselbalch equation; PH = pK a + log [A - ] [HA]

8. For example after adding 80ml of 0.1M KOH Number of moles of KOH added = number of moles of OH — added since KOH is a monohydroxyl strong base. Number of moles of OH - added = 0.08 liter x 0.1M = 8 x 10 -3 mol Thus 0.008 moles of HA have been converted to 0.008 moles of A -, Moles of remaining HA = moles HA originally present – moles of HA titrated to A -, Original number of moles present = 0.5 liter x 0.1M = 0.05moles Moles of HA remaining = 0.05 – 0.008 = 0.042 moles Since the ratio of A - moles / HA moles is the same as the ratio Of [A - ] / [HA]. Thus pH = 5 + log 0.008 = 5 + log 0.1904= 5 – 0.72 0.042 = 4.279 c)when 250ml of 0.1M KOH is added, here half the weak acid is titrated and the number of moles of A - = HA moles remaining

9. - Thus [A - ] = [HA], so pH = pK a + log 1 PH = pK a = 5.0 d)After adding 400ml of KOH ; Number of moles of OH - added = 0.4 liter x 0.1M = 0.04 mole Number of moles of A - produced = 0.04mole Thus number of moles of HA remaining = 0.05-0.04 = 0.01mol PH = pK a + log [A - ] = 5 + log 0.04 = 5 + log 4 = 5.6 [HA] 0.01 c) When exactly 500ml of 0.1M KOH is added (note although This resembles the end point of the titration the pH is not 7 Due reasons mentioned earlier) At this point pOH is calculated first from the following POH = pK b + p[A - ] 2 PK b = pK w - pK a = 14 – 5 = 9 Number of moles A - = 0.1M x 0.5 = 0,0.5mole -

10. [A ] number of moles / total volume in liters (note the total volume is 500ml 0f acid + 500 ml of base added Thus a total volume of 1000ml) [A - ] = 0.05/ 1.0 = 0.05M Thus pOH = 9 + (-log0.05) = 9 + 1.3 = 5.15 2 2 PH = pK w – pOH = 14 – 5.15 = 8.85. -