Presentation about CH600 Buffers PART1.pptx

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1. Buffer Systems CH600 Buffer Systems edits by CPGarcia 2017

2. concepts to tie together Acids & Bases [H+] ion concentration K a pH pK a Henderson- Hasselbach Equation Buffer solutions

3. 1} Acids & Bases Lewis Acids are e- pair acceptors Lewis Bases are e-pair donors 1) HCl Acid 2) HC 2 H 3 O 2 Acid 3) H 2 CO 3 Acid 4) CH 4 Acid 5) NH 4 +1 ammonium ion Acid 6) NH 3 a mmonia Base 7) Cl -1 chloride Base 8) C 2 H 3 O 2 -1 Base 9) H 2 O 10) CH 3 OH Acid or Base?

4. 1} Acids & Bases + H 2 O ⇋ + acid base conjugate base conjugate acid Acids - Bases equilibrium

5. 2} [H + ] ion concentration P ure water [H + ]=10 -7 and thus pH=7 A cids = high [H + ] and thus a low pH B ases = low [H + ] and thus a high pH

6. K a = an equilibrium ratio extent in which reactant & product are formed [H + ][A - ] [HA] 3} Equilibrium constant K a strong acids vs weak acids

7. problem solving ( K a ) 1. A 0.120 M solution of a weak acid (HA) has a pH of 3.26. what is the K a ? Step1 : Write the acid dissociation equation HA ⇌ H +  + A - Step2: Write the equilibrium expression: K a   = ( [H + ] [A - ] ) / [HA] Step3 : Since K a  is our unknown. Use the pH to calculate the [H + ] pH = -log [H + ], therefore [H + ] = 10¯ pH = 10¯ 3.26  = 5.49 54 x 10¯ 4  M a) Pay attention to the 1:1 molar ratio of [H + ] & [A¯]. Therefore : [ H + ] = [A¯] = 5.49 54 x 10 -4   M b) Some will use [ HA] = 0.120 ; others would subtract would use [ HA] = 0.120 - 5.4954 x 10 -4   M = method 1: K a  = [(5.4954 x 10¯ 4 ) (5.4954 x 10¯ 4 )] / 0.120 = 2.52 x10 -1 method 2: K a  = [(5.4954 x 10¯ 4 ) (5.4954 x 10¯ 4 )] / (0.120 - 5.4954 x 10¯ 4 ) = 2.53 x10 -1 in reality, it makes very little difference

8. problem solving ( K a ) 2. HC 9 H 7 O 4  (MW = 180. g/ mol ) is prepared by dissolving 3.60 g into a 1.00 L solution with a resulting pH 2.60 . what is the K a ? First, you must calculate the molarity: Moles HC 9 H 7 O 4 = 3.60g / 180. g/ mol = 0.0200 mole Molarity (M) = 0.0200 mol / 1.00L = 0.0200M Then apply the usual equilibrium formula: [( 2.5 x 10 -3 ) (2.5 x 10 -3 )] K a   = = 3.2 x 10 -4 0.0200

9. problem solving ( K a ) 3. A prepared 0.20M aqueous solution of propanoic acid has a pH of 4.88. what is the K a ? C 2 H 5 COOH + H 2 O ⇋ H 3 O + + C 2 H 5 COO -1 since therefor C 2 H 5 COOH H 3 O + C 2 H 5 COO -1

10. 4} pH = – log [ H + ] It approximates the negative log (base 10) of the molar concentrations of hydrogen ions H+ (H 3 +) in solution So a solution of HCl with a pH of 2.0 has a concentration of hydronium ions of 1x 10 -2 (1/100 !!) Compared to a more dilute solution of HCl with a pH of 5.0, which has a hydronium ions concentration of 1 x 10 -5 (1/100,000 ). pure water has [H + ]=10 -7 and thus pH=7 acids have a high [H + ] and thus a low pH bases have a low [H + ] and thus a high pH Because it’s a logarithmic scale, it doesn’t make “sense” to our brains. every factor of 10 difference in [H + ] ⋍ 1.0 pH units , Every factor of 2 difference in [H + ] ⋍ 0.3 pH units . Therefore, even numerically small differences in pH, can have profound biological effects…

11. 4} pH = – log [ H + ] It approx. the negative log (base 10) of the molar concentrations of H + (H 3 +) in solution. So anHCl with soln. pH=2.0 has a concentration of H + of 1x 10 -2 (1/100 !!) Compared to a more dilute solution of HCl with a pH of 5.0, which has a hydronium ions concentration of 1 x 10 -5 (1/100,000 ). pure water has [H + ]=10 -7 and thus pH=7 acids have a high [H + ] and thus a low pH bases have a low [H + ] and thus a high pH

12. 4} pH = – log [ H + ] problem solving (pH & pOH) 1. If the H +   conc. is 0.0001 M/l, what is the pH ?  (can be rewritten as 1.0 x 10 -4   M/l)   since pH = - log [H + ] pH = - (-4) =  4 2. If the H +   conc. is 0.00001 M/l, what is the OH -1 conc. ?  since pH = - log [ H + ]; & pH + pOH = 14   pH = -(-5) = 5  pOH = 14 - 5 = 9

13. Acid Normality pH Acetic N 2.4 Acetic 0.1 N 2.9 Acetic 0.01 N 3.4 Hydrochloric N 0.1 Hydrochloric 0.1 N 1.1 Hydrochloric 0.01 N 2.0 Sulfuric N 0.3 Sulfuric 0.1 N 1.2 Sulfuric 0.01 N 2.1 Molarity is the fractions of a mole in solution; Normality is a measure of the concentration of reactive groups (example H+) which may affect pH. 4} pH and “Normality”

14. 5} pK a = – log [ K a ] Low K a High pK a

15. 6} Henderson- Hasselbach 1) K a = [H + ][A - ] [HA] 2) [H + ] = K a [HA] [A - ] 3) -log[H + ] = -log K a -log [HA] [A - ] 4) -log[H + ] = -log K a +log [A - ] [HA] 5) pH = pK a +log [A - ] [ HA]

16. 6} Henderson- Hasselbach

17. “main function: resists changes in pH when acids or bases are added or when the solution is diluted .” {A} {B} is the pair a buffer solution ? Ex1 Acetic acid Sodium acetate HC 2 H 3 O 2 NaC 2 H 3 O 2 Yes , weak acid & base differ by 1 H + Ex2 Ammonium ion ammonia NH 4 +1 NH 3 Yes , weak acid & base differ by 1 H + Ex3 Hydrochloric acid Chloride HCl Cl -1 No , strong acid HCl Ex4 Carbonic acid Carbonate ion H 2 CO 3 CO 3 -2 No , the acid and base differ by 2H + 7} Buffer solution mixture of a weak acid & its conjugate base

18. buffers

19. 7} Buffer solution mixture of a weak acid & its conjugate base OH -1 HA ⇋ H + + A -1 HA ⇋ H + + A -1 H + HA ⇋ H + + A -1 buffer solution resists drastic change in pH resists drastic change in pH The ability of a buffer to resist a change of pH is due to the buffer capacity . The HIGHER the concentration of buffer components the LARGER the buffering capacity. The more concentrated the buffer is the more it will resist pH change.

20. 7} Buffer solution mixture of a weak acid & its conjugate base 1. The concentration of the acid and base are equal? B uffer prepared with 0.10M acetic acid and 0.10M acetate pH = pKa - log 10 [acid]/[base] pH = pKa - log 10 [0.10]/[0.10] pH = pK the pH where equal concentrations of acid & base are present is defined as the pKa 2. Buffer works most effectively at pH values that are ± 1 pH unit from the pKa (the buffer range)

21. What to report when writing about a buffer: Examples for your logbooks or formal report: “We used a 0.5M Tris buffer, pH 8.0 .” “ The reaction was carried out in a 0.1M boric acid – sodium hydroxide buffer adjusted to pH 9.2.” Name-Molarity-pH of the buffer

22. problem solving ( Buffers ) I. A buffer is made by mixing 0.250 mol of acetic acid and 0.250 mol of sodium acetate in 1 L of solution. What is the pH of 100.0 mL of the buffer ?

23. problem solving ( Buffers ) II. Instructions for making a buffer say to mix 60.0 mL of
0.100 M NH 3 with 40.0 mL of 0.100 M NH 4 Cl. What is the pH of this buffer? The combined volume is 60.0 mL + 40.0 mL = 100.0 mL Moles of NH 3 . = V NH 3 x M NH 3 = 0.060 L x 0.100 M = 0.0060 mol Moles of NH 4 + = V NH 4 Cl x M NH 4 Cl = 0.040 L x 0.100 M = 0.0040 mol

24. problem solving ( Buffers ) Instructions for making a buffer say to mix 60.0 mL of
0.100 M NH 3 with 40.0 mL of 0.100 M NH 4 Cl. What is the pH of this buffer? II. Instructions for making a buffer say to mix 60.0 mL of
0.100 M NH 3 with 40.0 mL of 0.100 M NH 4 Cl. What is the pH of this buffer? [OH-] = 2.7 x10 -5 -; pOH = - log [OH - ] = - log (2.7 x10 -5 ) = 4.57 pH = 14.00 – pOH = 14.00 – 4.57 = 9.43 pH = 14.00 – pOH = 14.00 – 4.57 = 9.43

25. Quick Questions B. The ammonia-ammonium ion buffer has a pH of about 9.2 and can be used to keep solutions in the basic pH range. What mass of ammonium chloride must be added to 400.0 mL of a 3.00 M ammonia solution to prepare a buffer? Kb (NH3) = 1.8 x 10-5 1 . 1.20g 2 . 20.44g 3 . 21.65g 4 . 62.94g 5 . 64.19g A. Which of the following statements is true? 1. HF dissociates more readily in a solution of NaF than in pure water. 
 2. HF dissociates equally well in a solution of KCl and in pure water . 3. HF does not dissociate in water to any extent

26. Quick Questions C. When acid is added to pure water, Kw, the ion‐ product constant of water, changes. True or Fal se D. A solution of 10 ‐8 M HCl and 10 ‐8 M acetic acid contains H + which is supplied mostly by the 1) strong acid. 2) weak acid. 3) both the strong and the weak acids. 4 ) wa ter. E. Compare two solutions, A (pH=4) to B (pH=6). 1) The hydronium ion conc. in A is twice that in solution B. 2) Solution A has greater buffering capacity than solution B. 3 ) C o ncentration of H + in A is 100 times that in B . 4) The OH -1 concentrations are equal in the two solutions since pH only measures the concentration of H + .

27. Quick Questions F. Two weak acids, A ( pKa =4) & B ( pKa =6). Which is true? 1 ) Ac id A dissociates to a greater extent in water than acid B. 2) Solutions of equal concentration, acid B will have a lower pH. 3) B is the conjugate base of A. 4) Acid A is more likely to be a polyprotic acid than acid B . 5) The equivalence point of acid A is higher than that of acid B. G. R atio of the concentration of a ___ over ___ describes the proportions of forms of a weak acid necessary to satisfy the Henderson‐Hasselbalch equation. 1) conjugate acid; conjugate base 2) conjugate base; conjugate acid 3) proton donor; proton acceptor 4) proton acceptor; proton donor 5 ) B and D H. At the midpoint of a titration curve 1) the conc. of a conjugate base is equal to the conc. of a conjugate acid. 2) the pH equals the pKa . 3) the ability of the solution to buffer is best. 4 ) A ll of the above. 5) A and B only.

28. bicarbonate buffer system