1. Acid-Base Titration Things to learn : - strong acid – strong base titration - weak acid – strong base titration - strong acid – weak base titration - prediction of titration curve - acid-base indicator

2. Strong Acid - Strong Base Titration In strong acid – strong base titration, there are three regions of the titration curve that represent different kinds of calculations : - before equivalence point - at equivalence point - after equivalence point Example : Consider the titration of 50.00 ml of 0.020 M KOH with 0.10 M HBr KOH + HBrH 2 O + KBr Before titration : Moles of HO - present = (0.050 l)(0.020 mol/l) = 0.001 moles

3. What happens when 3.00 ml of HBr is added ? No. of moles in 3.00 ml HBr = (0.003)(0.10) = 0.0003 Moles of HO - unconsumed = 0.001 – 0.0003 = 0.007 Since the volume in the flask in now 53 ml, [HO - ] in the flask = (0.007 mol)/(0.053 l) = 0.0132 M pH of solution =-log [H + ] = -log = 12.12 KwKw [HO - ] What happens at the equivalence point ?

4. At the equivalence point, the H + is just sufficient to react with all the HO - to form water. The pH is determined by the dissociation of water. Since there is 0.001 mol HO - in the flask, 0.001 mol H + must be added to reach equivalence point Volume of H + added ==10 ml Let : x = [H + ] = [HO - ] K w = [H + ] [HO - ] = x 2 x = 1.00 x 10 -7 pH = 7.00 0.001 mol 0.10 mol/l pH at the equivalence point in any strong acid – strong base titration will be 7.00 at 25 o C What happens after the equivalence point ?

5. When 10.50 ml of HBr is added to the solution : moles of excess H + = (0.0005 l)(0.10 mol/l) = 0.00005 mol Concentration of excess H + = =8.26 x 10 -4 M pH = -log (8.26 x 10 -4 ) = 3.08 0.00005 mol 0.0605 l

6. Weak Acid - Strong Base Titration Strong + weakcomplete reaction In weak acid – strong base titration, the titration curve consists of four regions : - before any base is added HAH + + A - - before the equivalence point : solution consists of a mixture of unreacted HA and A - - at the equivalence point : all the HA has been converted to A - A - + H 2 O HA + HO - - beyond the equivalence point : excess strong base is added and the pH of the solution is determined by the amount of strong base KaKa KbKb

7. Example : Consider a 50.0 ml solution of 0.020 M HA with pK a = 6.15 which is treated with 0.10 M NaOH Before the addition of NaOH : HAH + + A - K a = 10 -6.15 Let x = [H + ] =[A - ] K a ===10 -6.15 x = 1.19 x 10 -4 pH = -log (1.19 x 10 -4 ) = 3.93 [H + ][A - ] [HA] x2x2 0.020 - x 2 What happens when 3.0 ml of 0.10 M NaOH is added ?

8. When NaOH is added, a mixture of HA and A - is created a buffer Moles of NaOH added = (0.003 l)(0.10 M) = 0.0003 Concentration of A - = = 5.66 x 10 -3 M Moles of HA left = (0.050 l)(0.020 M) – 0.0003 =0.0007 Concentration of HA ==0.0132 M Using the Henderson-Hasselbalch equation pH = pK a + log = 6.15 + log = 5.78 0.0003 0.053 0.0007 0.053 [A - ] [HA] 5.66 x 10 -3 0.0132 What happens at the equivalence point ?

9. At the equivalence point, sufficient amount of NaOH has been added to react with all the HA Moles of HA present = (0.050 l)(0.020 M) = 0.0010 Volume of NaOH added = (0.0010)/(0.10 M) =0.01 l = 10 ml Concentration of A - == 0.0167 M HA + HO - H 2 O + A - HA + HO - Since K w = K a K b K b = =1.43 x 10 -8 Let x = [HA] = [HO - ] 0.0010 mol 0.060 l KbKb KaKa KwKw KaKa

10. K b ===1.43 x 10 -8 x = 1.54 x 10 -5 M Using K w = [H + ][HO - ] = 1x 10 -14 [H + ] = pH = -log[H + ] = 9.18 The pH at the equivalence point in not 7.00. In weak acid – strong base titration, the pH at the equivalence point is always higher than 7 [HO - ][HA] [A - ] x2x2 0.0167 - x 1x 10 -14 1.54 x 10 -5 What happens after the equivalence point ?

11. Now there is excess NaOH in the solution Since NaOH is a strong base, we can say that the pH is determined by the concentration of the excess NaOH When 10.10 ml of NaOH is added there is an excess of 0.10 ml of NaOH. [HO - ] == 1.66 x 10 -4 M pH = -log[H + ] = -log = 10.22 KwKw [HO - ] (0.10 M)(0.0001 l) 0.06010 l

12. Weak Base - Strong Acid Titration B + H + BH + Since it is a strong acid, therefore reaction goes essentially to completion In weak base – strong acid titration, the titration curve consists of four regions : - before any acid is added: B + H 2 OBH + + HO - - before the equivalence point – solution is a buffer B + HABH + + A - BH + + H 2 OB + H 3 O + pH = pK a (for BH + ) + log KbKb KaKa [B] [BH + ]

13. - at the equivalence point B + HA BH + + A - BH + + H 2 O B + H 3 O + pH is obtained by considering the acid dissociation [BH + ]  original [B] because of dilution Since the solution at equivalence point contains BH +, thus the pH should be below 7 - after the equivalence point There is excess HA in the solution. Since HA is a strong acid, it determines the pH (contribution from the hydrolysis of BH + is comparatively small and can be neglected)

14. Titration in Diprotic Systems Treatment is an extension from the monoprotic system Example : Consider the titration of 10 ml of 0.10 M base, B, with 0.10 M HCl. The base is dibasic with pK b1 = 4.00 and pK b2 = 9.00. Calculate the pH at each point along the titration curve Before an acid is added : B + H 2 OBH + + HO - Let x = [BH + ] = [HO - ]. Thus K b1 = == 1.00 x 10 -4 x = 3.11 x 10 -3 [H + ] =pH=11.49 K b1 [BH + ] [HO - ] [B] x2x2 0.10 - x KwKw [HO - ]

15. When acid is added and before the first equivalence point, we have a buffer containing B and BH + B + H + BH + BH + + H + BH 2 2+ BH 2 2+ BH + + H + BH + B + H + If 1.5 ml of HCl has been added : Moles of HCl added = (0.0015 l)(0.10 M) = 1.5 x 10 -4 = moles of BH + formed [ BH + ] = = 1.304 x 10 -2 M 1.5 x 10 -4 0.0115 K b1 K b2 K a1 K a2

16. [B] [BH + ] Moles of B left =(0.010)(0.10) – (1.5 x 10 -4 ) =0.00085 [B] == 7.39 x 10 -2 M Using the Henderson-Hasselbalch equation: pH = pK a2 + log = p( )+ log = 10.00 + log (5.667) = 10.75 0.00085 0.0115 KwKw K b1 7.39 x 10 -2 1.304 x 10 -2

17. At the first equivalence point, all the B has been converted to BH + which is both an acid and a base Moles of B present = (0.010)(0.10) = 0.001 Volume of acid used == 10 ml [BH + ] ==0.05 M Using : [H + ] = where K 1 = K a1 and K 2 = K a2 [H + ] =3.16 x10 -8 pH = 7.50 K 1 K 2 [BH + ] + K 1 K w K 1 +[BH + ] 0.001 mol 0.10 M 0.001 mol 0.020 l

18. At regions between the first and second equivalence points, a buffer containing BH + and BH 2 + is formed : BH + + H + BH 2 + BH 2 + BH + + H + When 15 ml of HCl is added : Moles of BH 2 2+ = (0.005 l)(0.10 M) = 0.0005 [BH 2 2+ ] == 0.02 M Moles of BH + left = 0.001 – 0.0005 = 0.0005 [BH + ] == 0.02 M K a1 0.0005 mol 0.025 l 0.0005 mol 0.025 l

19. Using the Henderson-Hasselbalch equation: pH = pK a1 + log = 5.00 + log 1 = 5.00 [BH + ] [BH 2 2+ ] At the second equivalence point, all the BH + has been converted to BH 2 + BH + + H + BH 2 2+ BH 2 2+ BH + + H + pH of the solution is determined by the acid dissociation Moles of BH + present at first equivalence point = 0.001 Volume of acid added between first and second equivalence point = (0.001 mol)/((0.10 M) = 10 ml K a1

20. Moles of BH 2 2+ = moles of BH + = 0.001 [BH 2 2+ ] == 0.033 M K a1 = = x = 5.72 x 10 -4 pH = -log (5.72 x 10 -4 ) = 3.24 0.001 mol 0.030 l [BH + ][H + ] [BH 2 2+ ] KwKw K b2 x2x2 0.033 - x Beyond the second equivalence point, the pH is determined by the excess HCl If the total volume of HCl added is 25.0 ml, Moles of excess HCl = (0.005 l)(0.10 M) = 0.0005 [H + ]== 1.43 x 10 -2 M pH= 1.85 0.0005 0.035

21. (a)Titration of 10.0 ml of 0.100 M base (pK b1 = 4.00, pK b2 = 9.00) with 0.100 M HCl (b) Titration of 10.0 ml of 0.100 M nicotine (pK b1 = 6.15, pK b2 = 10.85) with 0.100 M HCl

22. Finding the End Point Titrations are commonly performed to determine : - how much analyte is present - the equilibrium constants of the analyte How would one determine the end point ? (i) Autotitrator -pH is measured by electrodes immersed in the analyte solution  pH/  V and (  pH/  V)/  V are computed -when  pH/  V is maximum and (  pH/  V)/  V = 0 end point

23. (ii) Indicator An acid-base indicator is itself an acid or base whose various protonated species have different colours : eg phenolphthalein HlnIn - + H + pH = pK 1 + log The indicator changes color over a pH range Generally only one color is observed if the ratio of the concentrations of the two forms is 10:1 [In - ] [Hln] acid colorbase color

24. So the pH in going from one color to the other has changed from (pK 1 - 1) to (pK 1 + 1) most indicators require a transition range of about two pH units The pH range over which the color changes is called the transition range 0.7< pH< 2.78.0< pH< 9.6 When only the color of the unionized form is seen : = pH = pK 1 + log = pK 1 - 1 When only the color of the ionized form is seen : = pH = pK 1 + 1 1 10 1 [In - ] [Hln] 10 1 [In - ] [Hln]

25. Choosing an Indicator An indicator with a color change near pH 5.54 would be useful in determining the end point of the titration. The closer the point of color change is to pH 5.54 the more accurate will be the end point The difference between the observed end point (when there is a color change) and the true equivalence point is called the indicator error Use only a few drops of dilute indicator solution for each titration