1. Chemistry 102(60) Summer 2002 Instructor: Dr. Upali Siriwardane e-mail:upali@chem.latech Office: CTH 311 Phone 257-4941 Office Hours: 8:30-10;30 M, W, Tu,Th, F (Test 1): Chapter 12: July 11, 2002 (Test 2): Chapter 13: July 18, 2002 (Test 3): Chapter 14: July 25, 2002 (Test 4): Chapter 15: July 31, 2002 (Test 5): Chapter 17: Aug 8, 2002 (Homework): Chapter 18: Aug 8, 2002 Make-up test:Comprehensive: Aug 9, 2002

2. Acid-base Equilibria (weak acid/base solutions) K a and K b % dissociation of weak acid/bases pH of weak acid/base solutions pH of salt solutions Buffers pH of buffer solutions Titrations: selecting indicators

3. How do you calculate pH of weak acids/bases From % dissociation From K a or K b What is % dissociation Amount dissociated % Dissoc. = ------------------------- x 100 Initial amount

4. How do you calculate % dissociation from K a or K b 1.00 M solution of HCN; K a = 4.9 x 10 -10 What is the % dissociation for the acid?

5. 1.00 M solution of HCN; K a = 4.9 x 10 -10 First write the dissociation equilibrium equation: HCN(aq) + H 2 O(l) H 3 + O(aq) + CN - (aq) [HCN] [H + ] [CN - ] Ini. Con. 1.00 M 0.0 M 0.00 M Cha. Con -x x x Eq. Con. 1.0 - x x x [H 3 + O ][CN - ] x 2 K a = ----------- =---------------- [HCN] 1.0 - x 1.00 M solution of HCN; K a = 4.9 x 10 -10

6. 1.0 - x ~ 1.00 since x is small x 2 K a = -----------; K a =4.9 x 10 -10 = x 2 1.0 x = sqrt 4.9 x 10 -10 = 2.21 x 10 -5 Amount disso. 2.21 x 10 -5 ----------------- x 100 =- ------------- x 100 Ini. amount 1.00 % Diss.=2.21 x 10 -5 x 100 = 0.00221 % 1.00 M solution of HCN; K a = 4.9 x 10 -10

7. % Dissociation gives x (amount dissociated) need for pH calculation Amount dissociated % Dissoc. = ------------------------- x 100 Initial amount/con. x % Dissoc. = --------------------------- x 100 concentration

8. 1 M HF, 2.7% dissociated Notice the conversion of % dissociation to a fraction: 2.7/100=0.027) x=0.027 Calculate the pH of a weak acid from % dissociation

9. HF(aq) + H 2 O(l) H 3 + O(aq) + F - (aq) [H + ][F - ] K a = ----------- [HF] [HF] [H + ] [F - ] Ini. Con. 1.00 M 0.0 M0.00 M Chg. Con -x x x Eq.Con. 1.0-0.027 0.0270.027 pH = -log [H + ] pH = -log(0.027) pH = 1.57 Calculate the pH of a weak acid from % dissociation

10. Weak acid equilibria ExampleExample Determine the pH of a 0.10 M benzoic acid solution at 25 o C if K a = 6.28 x 10 -5 HBz (aq) + H 2 O (l) H 3 O + (aq) + Bz -(aq) The first step is to write the equilibrium expression K a = [H 3 O + ][Bz - ] [HBz]

11. Weak acid equilibria HBz H 3 O + Bz - Initial conc., M 0.10 0.00 0.00 Change,  M -x x x Eq. Conc., M0.10 - x x x [H 3 O + ] = [Bz - ] = x We’ll assume that [Bz - ] is negligible compared to [HBz]. The contribution of H 3 O + from water is also negligible.

12. Weak acid equilibria Solve the equilibrium equation in terms of xSolve the equilibrium equation in terms of x K a = 6.28 x 10 -5 = x = (6.28 x 10 -5 )(0.10) H 3 O + = 0.0025 M pH= 2.60 x 2 0.10

13. pH from K a or K b 1.00 M solution of HCN; K a = 4.9 x 10 -10 First write the dissociation equilibrium equation: HCN(aq) + H 2 O(l) H 3 + O(aq) + CN - (aq) [HCN] [H + ] [CN - ] Ini. Con. 1.00 M 0.0 M 0.00 M Chg. Con -x x x Eq. Con. 1.0 - x x x

14. [H 3 + O ][CN - ] x 2 K a = ---------------=---------------- [HCN] 1.0 - x 1.0 - x ~ 1.00 since x is small x 2 K a = -----------; K a =4.9 x 10 -10 = x 2 1.0 x = sqrt 4.9 x 10 -10 = 2.21 x 10 -5 pH = -log [H + ] pH = -log(2.21 x 10 -5 ) pH = 4.65 Weak acid Equilibria

15. What salt solutions would be acidic, basic and neutral? 1)strong acid + strong base = neutral 2)weak acid + strong base = basic 3)strong acid + weak base = acidic 4)weak acid + weak base = neutral, basic or an acidic solution depending on the relative strengths of the acid and the base.

16. What pH? Neutral, basic or acidic? a)NaCl neutral b) NaC 2 H 3 O 2 basic c) NaHSO 4 acidic d) NH 4 Cl acidic

17. Reaction of a basic anion with water is an ordinary Brønsted-Lowry acid-base reaction. CH 3 COO - (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) This type of reaction is given a special name.Hydrolysis The reaction of an anion with water to produce the conjugate acid and OH -. The reaction of a cation with water to produce the conjugate base and H 3 O +. Hydrolysis

18. How do you calculate pH of a salt solution? Find out the pH, acidic or basic? If acidic it should be a salt of weak base If basic it should be a salt of weak acid if acidic calculate K a from K a = K w /K b if basic calculate K b from K b = K w /K a Do a calculation similar to pH of a weak acid or base

19. What is the pH of 0.5 M NH 4 Cl salt solution? (NH 3 ; K b = 1.8 x 10 -5 ) Find out the pH, acidic if acidic calculate K a from K a = K w /K b K a = K w /K b = 1 x 10 -14 /1.8 x 10 -5 ) K a = 5.56. X 10 -10 Do a calculation similar to pH of a weak acid

20. Continued NH 4 + + H 2 O H 3 + O + NH 3 [NH 4 + ] [H 3 + O ] [NH 3 ] Ini. Con. 0.5 M 0.0 M0.00 M Eq. Con. 0.5 - x x x [H 3 + O ] [NH 3 ] K a (NH 4 + ) = --------------------= [NH 4 + ] x 2 ---------------- ; appro.:0.5 - x. 0.5 (0.5 - x)

21. x 2 K a (NH 4 + ) = ----------- = 5.56 x 10 -10 0. 5 x 2 = 5.56 x 10 -10 x 0.5 = 2.78 x 10 -10 x= sqrt 2.78 x 10 -10 = 1.66 x 10 -5 [H + ] = x = 1.66 x 10 -5 M pH = -log [H + ] = - log 1.66 x 10 -5 pH = 4.77 pH of 0.5 M NH 4 Cl solution is 4.77 (acidic ) Continued

22. This is an example of Le Châtelier’s principle. Common ion effect The shift in equilibrium caused by the addition of an ion formed from the solute. Common ion An ion that is produced by more than one solute in an equilibrium system. Adding the salt of a weak acid to a solution of weak acid is an example of this. Common Ion Effect

23. Weak acid and salt solutions E.g. HC 2 H 3 O 2 and NaC 2 H 3 O 2 Weak base and salt solutions E.g. NH 3 and NH 4 Cl. H 2 O + C 2 H 3 O 2 - OH - + HC 2 H 3 O 2 (common ion) H 2 O + NH 4 + H 3 + O + NH 3 (common ion)

24. Solutions that resist pH change when small amounts of acid or base are added. Two types weak acid and its salt weak base and its salt HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) Add OH - Add H 3 O + shift to right shift to left Based on the common ion effect. Buffers

25. The pH of a buffer does not depend on the absolute amount of the conjugate acid-base pair. It is based on the ratio of the two. Henderson-Hasselbalch equation. Easily derived from the K a or K b expression. Starting with an acid pH = pK a + log Starting with a base pH = 14 - ( pK b + log ) [HA] [A - ] [HA] Buffers

26. Calcualtion of pH of Buffers Henderson Hesselbach Equation [ACID] pH = pK a - log --------- [BASE] [BASE] pH = pK a + log --------- [ACID]

27. Control of blood pHControl of blood pH Oxygen is transported primarily by hemoglobin in the red blood cells. CO 2 is transported both in plasma and the red blood cells. CO 2 (aq) + H 2 OH 2 CO 3 (aq) H + (aq) + HCO 3 - (aq) The bicarbonate buffer is essential for controlling blood pH The bicarbonate buffer is essential for controlling blood pH Buffers and blood

28. Acid-base indicators are highly colored weak acids or bases. HIn In - + H + color 1color 2 color 1 color 2 They may have more than one color transition. Example. Example. Thymol blue Red - Yellow - Blue One of the forms may be colorless - phenolphthalein (colorless to pink) Indicators

29. What is an Indicator? Indicator is an weak acid with different K a, colors to the acid and its conjugate base. E.g. phenolphthalein HIn H + + In - colorless pink Acidic colorless Basic pink

30. Selection of an indicator for a titration a) strong acid/strong base b) weak acid/strong base c) strong acid/weak base d) weak acid/weak base Calculate the pH of the solution at he equivalence point or end point

31. Acid-base indicators are weak acids that undergo a color change at a known pH. phenolphthalein pH Indicator examples

32. pH range of Indicators litmus (5.0-8.0) bromothymole blue (6.0-7.6) methyl red (4.8-6.0) thymol blue (8.0-9.6) phenolphthalein (8.2-10.0) thymolphthalein (9.4-10.6)

33. Indicator examples methyl red bromthymol blue

34. Titrations revisited Methods based on measurement of volume. If the concentration of an acid is known, the amount of a base can be found. If we know the concentration of the base, then we can determine the amount of acid. All that is needed is some calibrated glassware and either an indicator or pH meter.

35. Titrations Buret Buret - volumetric glassware used for titrations. It allows you to add a known amount of your titrant to the solution you are testing. If a pH meter is used, the equivalence point can be measured. An indicator will give you the endpoint.

36. Titrations

37. Note the color change which indicates that the ‘endpoint’ has been reached. StartEnd

38. Titration curves Acid-base titration curveAcid-base titration curve A plot of the pH against the amount of acid or base added during a titration. Plots of this type are useful for visualizing a titration. It also can be used to show where an indicator undergoes its color change.

39. Titration curves pH Equivalence Point % titration or ml titrant Buffer region Overtitration Indicator Transition

40. 50 mL of 0.1 M HCl and 50 mL 0.1M NaOH solution 25 mL 0.5 M HC 2 H 3 O 2 and 25 mL 0.5 M NaOH 50 mL of 1.0 M HCl and 50 mL of 1.0 NH 4 OH (NH 3 ) solution. What is the pH of Solutions?