1. C. Y. Yeung (CHW, 2009) p.01 Acid-Base Eqm (4): Buffer Solutions Q.:What is the new pH after addition of 0.001 mol HCl at 298K? (assume no volume change) 100 cm 3 H 2 O 0.001 mol HCl [H 3 O + ] = 0.001100/1000 = 0.010 M  pH = - log(0.010) = 2  pH would decrease from 7 to 2 after adding 0.001 mol HCl at 298K!

2. p.02 Q.:What is the new pH after addition of 0.001 mol HCl at 298K? (assume no volume change) 100 cm 3 0.1M CH 3 COOH K a = 1.76  10 -5 M 0.001 mol HCl  pH would decrease from 2.88 to 1.99 after adding 0.001 mol HCl at 298K! Before addition of HCl: x2x2x2x2 0.1 – x = 1.76  10 -5 x = 1.33  10 -3 pH = 2.88 After addition of HCl: extra [H 3 O + ] = 0.001100/1000 = 0.01M y(0.01+y) 0.1 – y = 1.76  10 -5 y = 1.76  10 -4 pH = -log (0.01+y) = 1.99

3. p.03 Q.:What is the new pH after addition of 0.001 mol HCl at 298K? (assume no volume change) 100 cm 3 0.1M CH 3 COOH and 0.1M CH 3 COO - Na + 0.001 mol HCl  pH would decrease from 4.75 to 4.67 after adding 0.001 mol HCl at 298K! Before addition of HCl: x (0.1+x) 0.1 – x = 1.76  10 -5 x = 1.76  10 -5 pH = 4.75 After addition of HCl: extra [H 3 O + ] = 0.01M [CH 3 COO - ] remained = 0.1 – 0.01 = 0.09M (0.09+y) y 0.11 – y = 1.76  10 -5 y = 2.15  10 -5 pH = -log (y) = 4.67

4. p.04 Note the difference … 100cm 3 H 2 O 100cm 3 0.1M CH 3 COOH 100cm 3 0.1M CH 3 COOH and 0.1M CH 3 COO - Na + Initial pH 72.884.75 pH after adding 0.001 mol HCl 21.994.67 tends to resist the changes in pH when small amounts of acid is added i.e. it is a “BUFFER” solution!

5. p.05 100 cm 3 0.1M CH 3 COOH and 0.1M CH 3 COO - Na + It is an ACIDIC BUFFER solution! (i.e. weak acid + conjugate base) It tends to resist changes in pH when small amounts of acid or base are added, and their pH is not affected by dilution. When acid is added, the additional H 3 O + would combine with the conjugate base so that the eqm shifts BW. i.e. [H 3 O + ] does not change much, pH remains almost unchanged. When base is added, some H 3 O + ions are reacted. Some acid molecules dissociate to produce H 3 O + so that the eqm shifts FW. i.e. [H 3 O + ] does not change much, pH remains almost unchanged. CH 3 COOH + H 2 O CH 3 COO - + H 3 O +

6. p.06 100 cm 3 0.1M NH 3 and 0.1M NH 4 Cl It is an BASIC BUFFER solution! (i.e. weak base + conjugate acid) When acid is added, some OH - are neutralized, some NH 3 molecules would dissociate so that the eqm shifts FW. i.e. [OH - ] does not change much, pH remains almost unchanged. When base is added, additional OH - ions would combine with NH 4 + so that the eqm shifts BW. i.e. [OH - ] does not change much, pH remains almost unchanged. NH 3 + H 2 O NH 4 + + OH -

7. p.07 pH of buffer is NOT affected by dilution.  WHY? pK a = [A - (aq)] [HA(aq)] - log pH conjugate base acid constant at constant temp. During dilution, both [A - ] and [HA] are diluted to the same extent. Thus the ratio [A - ]/[HA] is unchanged. Therefore, pH of buffer remains unchanged in dilution! pH of the buffer could be adjusted by varying the ratio of HA and A -.

8. p.08 Q.1:Lactic acid ( 乳酸 ), CH 3 CH(OH)COOH, is produced in muscle tissues during physical exertion. At body temperature, human blood pH is 7.40. Whether the lactic acid produced exists mainly in form of “undissociated molecule” or “dissociated lactate ions” in blood at body temperature? Whether the lactic acid produced exists mainly in form of “undissociated molecule” or “dissociated lactate ions” in blood at body temperature? (Given: K a of lactic acid at body temperature = 8.50  10 -4 mol dm -3 ) pK a = [CH 3 CH(OH)COO - ] [CH 3 CH(OH)COOH] pH – log 2.14  10 4 = [CH 3 CH(OH)COOH] [CH 3 CH(OH)COO - ]  Lactic acid formed is mainly in form of dissociated lactic ions. With a given K a, the ratio of [salt] and [acid] could be found from the pH value.

9. p.09 Q.2:Calculate the pH value of the resultant solution for 10cm 3 of 0.20M CH 3 COOH are mixed with 10cm 3 of 0.40M CH 3 COONa. (K a = 1.75  10 -5 mol dm -3 ) x(0.20+x) 0.10 – x 1.75  10 -5 = x = 8.75  10 -6  pH = 5.06 After mixing, new [CH 3 COOH] = 0.10M new [CH 3 COONa] = 0.20M An acid-salt mixture (acid and salt have comparable conc.)  ACIDIC BUFFER! pH of acidic buffer could be adjusted by [acid] and [salt].

10. p.10 Q.3:Calculate the pH value of the resultant solution for 10cm 3 of 1.00M NH 3 are mixed with 10cm 3 of 1.00M NH 4 Cl. (K b = 1.78  10 -5 mol dm -3 ) x(0.50+x) 0.50 – x 1.78  10 -5 = x = 1.78  10 -5  pOH = 4.75 After mixing, new [NH 3 ] = 0.50M new [NH 4 + ] = 0.50M A base-salt mixture (base and salt have comparable conc.)  BASIC BUFFER!  pH = 9.25 pH of basic buffer could be adjusted by [base] and [salt].

11. p.11 Article Reading … about Medical Science! Acid-Base Equilibria & pH Regulation in Blood Plasma

12. Assignment p.12 Next …. Acid-Base Indicators (Book 2 p. 164 – 168) Study the examples in p. 158 - 163, p.163 Check point 18.2, p.179 Q.1 - 6 [due date: 23/4(Thur)]