Solution equilibria Effect of common ion Ex. HF K a = For a 1.0 M solution of HF: about ~2.7% dissociation

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1 15,16 Solution equilibria Effect of common ion Ex. F K a = For a 1.0 M solution of F: about ~.7% dissociation With the addition of 1.0 M NaF F (aq) + (aq) + F (aq) final 1.0-x x 1.0+x (M) K a x 1.0 x ( 1.0 x) x ~ M (0.0007) 0.07% dissociated Equilibrium has been shifted to the left Buffered solutions esists a change in p Ex. Blood p 7.5~7.45 A(aq) (aq) + A (aq) ini [A] 0 [A ] (M) final [A]-x x [A ]+x _ x([a ] x) K a [A] x Assume x is small: K a x = [ ] [A ] [ ] [A] enderson-asselbalch equation [A] p pka log( ) pk [A ] a [A] [ ] K a [A ] [A ] log( ) [A] 1

2 Ex. Ex. Initial [A ] 0.50M [A] 0.50M 1.0 With addition of 0.01 M [A ] [A] 0.49 Small change With addition of 0.10 M p change ~0.18 Initial [A ] 0.5M 0. (log0. = 0.48) [A] 0.75M With addition of 0.10 M [A ] [A] 0.65 ptimum buffering is to keep [A ]/[A] = 1 (log0.54 = 0.7) p change = 0.1 p = pk a Buffering capacity: determined by the real magnitude of [A ] and [A] Buffer system in blood (l) + (l) (aq) + (aq) + (aq) Serves to control the conc. pk a1 = 6.1 at physiological T For a buffer p of 7.4 [ ] [ ] p pka1 log( ) 0 [ ] [ ] In reality [ ] ~ 0.04 M [ ] ~ M igher conc. More effective to neutralize extra acid Low capacity to neutralize extra base

3 rgans to regulate p lung and kidney eceptor in brain: sensitive to [ + ] and [ ] in body [ + ] too high: increase breath rate remove more shift equilibrium to the left Kidney: removes + and p for urine ~ ther useful buffer systems Phosphate number 1 at 5 o, p 6.87 Phosphate number at 5 o, p 7.41 ( P 4, K a = , pk a = 7.1) K P 4, Na P 4 Titrations and p curves Strong acid-strong base Ex. N (50 ml, 0.00 M) titrated with Na (0.100 M) ( ) ( ) M [ ] p 1.0 Determined by [ ] Determined by the strong acid Equivalence point Na added (ml)

4 Weak acid-strong base titration Before equivalence point dissociation of the weak acid After equivalence point controlled by Na Ex. Titration of acetic acid (Ac; 50 ml, 0.10 M) with Na (0.10 M) Addition of 5.0 ml of 0.10 M Na(aq) final Ac(aq) Ac (aq) + (aq).5mmol x 75mL.5 75 x Assume x is small K a = = x = [ + ] p = 4.74 x p 1.0 Equivalence point ontrolled by added buffer zone At equivalence point Start calculation from 5.0 mmol Ac in 100 ml = M Na added (ml) Ac (aq) + (l) Ac(aq) + (aq) final x x x 10 x x Kb x = x M = [ ] p = 8.7 4

5 p omparison weak acid Titration of weak bases with strong acids p strong acid Vol of Na 7.0 equivalence point acidic Vol of l(aq) Acid-base indicators Detection limit for color change [In ] [In] 1 10 observable In(aq) (aq) + In (aq) [In ] p pk a log( ) [In] _ In used in acid titrated by base Point with detection: p pk a 1 log( ) pk 10 a 1 [In ] [In]

6 Ex. Bromthymol blue, K a = , pk a = 7 [In ] blue [In] yellow p for color change: pk a 1 = 6 In base titrated by acid Starts with In point with detection: [In ] [In] 10 1 For bromthymol blue, color change at p = pk a + 1 = 8 verall, the useful range is p = pk a ± 1 Ex. Phenolphthalein: p ~ 810 hoice of indicator Start of color change (end point): within the vertical area around the equivalence point p phenolphthalein methyl red Ac l Na added (ml) (not suitable for the titration of Ac) 6

7 Polyprotic acids General expression A(aq) (aq) + A (aq) A (aq) (aq) + A (aq) A (aq) (aq) + A (aq) At the 1st equivalence point A is converted to A A (aq) (aq) + A (aq) A (aq) + (aq) A(aq) Dominates before the 1st equivalence point A (aq) + A (aq) A (aq) + A(aq) [A ] = [ A] K a1 [ ][A ] [ A] K a [ ][A ] [ A ] K a1 pk a1 K a pk [ ] [A [ A] a [ A ] p log [ A] Since [A ] = [ A] at equivalence point pk a1 pk pk p a1 a p pk a ] 7

8 Titration curve p A + A + [A ]/[A ] pk a + pk a pk a1 + pk a [ A ]/[A ] [ A]/[ A ] A + + A V of Na The acid-base properties of amino acids N Amphoteric both an acid and base K a1 K a N N N P Z N (positive) (neutral) (negative) K a1 = [Z][+ ] [P] K a = [N][ + ] [Z] 8

9 K a = [N][ + ] [Z] Titration curve When [Z] = [N] K a = [ + ] pk a = p p K a1 = [Z][+ ] [P] When [Z] = [P] K a1 = [ + ] pk a1 = p pi equivalent of base ( ) Isoelectric point N K a1 K a1 = [Z][+ ] [P] pk a1 =log[ + ] log [Z] [P] N K a N + + [N][ ] K a = [Z] pk a1 + pk a = p log [P] pk a =log[ + ] log [N] [N] [Z] When [N] = [P] pk a1 + pk a = p The net charge of amino acid is neutral The isoelectric point (the dipolar ion has the highest concentration) pk a1 + pk At this point a p = = pi * different amino acid has different pi 9

10 In general: When is neutral N pk a1 ~- pk a ~9-10 N N When is acidic N aspartic acid pi ~6 pk a1 =.09 pk a =.86 N + + pi = pk a = N N When is basic ( ) N lysine N pk a1 = pk a = 8.95 ( ) ( ) N N N N + + pk a = 10.5 Electrophoresis pi = 9.74 ( ) N N lys gly asp pi 10

11 Solubility equilibria and the solubility product Ex. af (s) a + (aq) + F (aq) K sp = [a + ][F ] Solubility product constant Ex. Bi S (s) Bi + (aq) + S (aq) solubility = M at 5 o K sp = [Bi + ] [S ] = ( ) ( ) = Solubility = x for Bi S K K sp = (x) (x) = 108x 5 5 sp x 108 elative solubility us(s) K sp = Ag S(s) Bi S (s) K sp = K sp = solubility K sp solubility K sp / 4 Solubility Bi S (s) > Ag S> us(s) ommon ion effect Decrease solubility 11

12 p and solubility Ex. Mg() (s) Mg + (aq) + (aq) Addition of + removes drives equilibrium to the right solubility increases 1