Applications of Aqueous 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 8 Applications of Aqueous Equilibria Effect of common ion Ex. F 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) a x 1.0 x ( 1.0 x) x ~ M (0.0007) 0.07% dissociated Equilibrium has been shifted to the left Buffered solutions Resists 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) a x = [ ] [A] x [A ] Assume x is small: [ a ] [A] enderson-asselbalch equation [A] p pa log( ) p [A ] a [A] [ ] 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 ] (log1.04 = 0.017) [A] 0.49 Small change With addition of 0.10 M p change ~0.18 Initial [A ] 0.5M [A] 0.75M 0. With addition of 0.10 M [A ] [A] 0.65 ptimum buffering is to keep [A ]/[A] = 1 (log0. = 0.48) (log0.54 = 0.7) p change = 0.1 p = p a Buffering capacity: determined by the real magnitude of [A ] and [A] Buffer system in blood (l) + (g) (aq) + (aq) + (aq) Serves to control the conc. p a1 = 6.1 at physiological T For a buffer p of 7.4 [ ] [ ] p pa1 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 Receptor in brain: sensitive to [ + ] and [ ] in body [ + ] too high: increase breath rate remove more shift equilibrium to the left idney: 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, a = , p a = 7.1) P 4, Na P 4 The exact treatment A pair of A/A (from NaA) A (aq) + (aq) + A (aq) a [ ][ A ] [ A] (aq) + (aq) + (aq) w = [ + ][ ] harge balance [Na + ] + [ + ] = [A ] + [ ] ([Na + ] = [A ] o ) Mass balance [A] o + [A ] o = [A] + [A ] Four relationships, four unknowns: [ + ], [ ], [A], [A ] [ ] = w / [ + ] From charge balance [A ] o + [ + ] = [A ] + w / [ + ] [A ] = [A ] o + [ + ] w / [ + ]

4 From mass balance [A] o + [A ] o = [A] + [A ] o + [ + ] w / [ + ] [ ] w [ A] [A] o [ ] a [ ] [ ]([A ] o [ ][A ] [ ] [A] [ ] w [A] o [ ] In fact, when [ w ] [ ] w ) is small a [ ][A ] [A] o o Ex. A buffered solution with M l M Nal l (aq) + (aq) + l (aq) a 8 x( x) x Assume x << x ~ M Assumption But this is a very small value 7 [ ] w ( ) heck 7 [ ] Small compared with 10 4 onclusion: 14 autoionization of has no effect unless [A] or [A ] are very small In the range of

5 Titrations and p curves Strong acid-strong base Ex. N (50.0 ml, 0.00 M) titrated with Na (0.100 M) With the addition of 10.0 ml of the titrant ( ) ( ) M [ ] Equivalence point: Addition of stoichiometric amount of the titrant p 1.0 Determined by [ ] 7.0 Equivalence point Determined by the strong acid Na added (ml) haracteristics: with sharp p change around equivalence point 5

6 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; 50.0 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).5 mmol x 75 ml.5 75 x Assume x is small 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 b x = x M = [ ] p = 8.7 6

7 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 p a log( ) [In] _ In used in acid titrated by base Point with detection: p p a 1 log( ) p 10 a 1 [In ] [In]

8 Ex. Bromthymol blue, a = , p a = 7 [In ] blue [In] yellow p for color change: p 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 = p a + 1 = 8 verall, the useful range is p = p 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) 8

9 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] a1 [ ][A ] [ A] a [ ][A ] [ A ] a1 p a1 a p [ ] [A [ A] a [A ] p log [ A] Since [A ] = [ A] at equivalence point ] p a1 p a p p p a1 p a 9

10 Titration curve p A + A + [A ]/[A ] p a + p a p a1 + p a [ A ]/[A ] [ A]/[ A ] A + + A V of Na The acid-base properties of amino acids R N Amphoteric both an acid and base R a1 R a R N A diprotic acid Exists in strongly acidic media N N P Z N (positive) (neutral) (negative) In strongly basic media a1 = [Z][+ ] [P] a = [N][ + ] [Z] 10

11 Isoelectric point ( 等電點 ) R N a1 a1 = [Z][+ ] [P] a R R N N + + [N][ + ] + + a = [Z] p a1 =log[ + ] log [Z] p a =log[ + ] log [P] [N] p a1 + p a = p log [P] [N] [Z] When [N] = [P] p a1 + p a = p The net charge of amino acid is neutral The isoelectric point (the dipolar ion has the highest concentration) p a1 + p At this point a p = = pi * Different amino acid has different pi a = [N][ + ] [Z] Titration curve When [Z] = [N] a = [ + ] p a = p p a1 = [Z][+ ] [P] When [Z] = [P] a1 = [ + ] p a1 = p pi equivalent of base ( ) 11

12 In general: When R is neutral R N p a1 ~- R p a ~9-10 N R N When R is acidic N aspartic acid pi ~6 p a1 =.09 p a =.86 N + + pi = p a = N N When R is basic ( ) N lysine N p a1 = p a = 8.95 ( ) ( ) N N N N + + p a = 10.5 pi = 9.74 Electrophoresis ( 電泳 ) ( ) N N Buffered at p = 6.0 lys gly asp pi 1

13 Solubility equilibria and the solubility product Ex. af (s) a + (aq) + F (aq) sp = [a + ][F ] Solubility product constant Ex. Bi S (s) Bi + (aq) + S (aq) solubility = M at 5 o sp = [Bi + ] [S ] = ( ) ( ) = Solubility = x for Bi S sp = (x) (x) = 108x 5 5 sp x 108 Relative solubility us(s) sp = Ag S(s) Bi S (s) sp = sp = solubility sp solubility sp / 4 Solubility Bi S (s) > Ag S> us(s) ommon ion effect Decrease solubility 1

14 omplications Ion pairing (A + )(B )(aq) The sp calculated from solubility may not be correct omplex formation Agl soluble p and solubility Ex. Mg() (s) Mg + (aq) + (aq) Addition of + removes drives equilibrium to the right solubility increases Ex. Fe(), sp = Q: Solubility in water Soln: Fe() (s) Fe + (aq) + (aq) Assume derived from Fe() is very small [ ] = M At equilibrium: [Fe + ]( ) = [Fe + ] = M Small indeed The exact solution: [Fe + ][ ] = [ + ][ ] = [Fe + ] + [ + ]=[ ] [ + ]=[ ] [Fe + ] ([ ] [Fe + ])([ ]) = [ ] [Fe + ][ ] =

15 [ ] [Fe + ][ ] = [ ] 4 [Fe + ][ ] = ( )[ ] [ ] 4 ( ) = ( )[ ] [ ] 4 ( )[ ] = 0 [ ] = [ ] = M Qualitative analysis Formation of precipitate (ppt) ion product Q > sp Equilibrium value Initial value Separate through selective precipitation Ex. A solution of M u M Pb + + I PbI sp = = (.0 10 )([I ]) [I ] =.6 10 M Minimum value to precipitate Pb + ui sp = = ( )([I ]) [I ] = M u + will precipitate first 15

16 Sulfides are often used to separate metal ions S(aq) (aq) + S (aq) a1 = S (aq) (aq) + S (aq) a ~ [S ] can be controlled by p Ex. Mn +, Ni +, u +, g + S, p ~ us, gs ppt Mn +, Ni + add p = 8 MnS, NiS ppt Qualitative analysis Mixture of ions l g l, Agl, Pbl Group I ppt gs, ds, Bi S, us, SnS Group II ppt os, ZnS, MnS, NiS, FeS r(), Al() Group III ppt solution S solution a, Ba Mg Group IV ppt Na solution Na Group V Ions: alkalai metal and N

17 omplex ion equilibria Ag + (aq) + N (aq) Ag(N ) + (aq) 1 =.1 10 Ligand oordination number: 1 AgN + (aq) + N (aq) Ag(N ) + (aq) = The major equilibrium: Ag + (aq) + N (aq) Ag(N ) + (aq) [Ag(N ) ] [Ag(N ) ][Ag(N 7 f [Ag ][N ] [Ag ][N ][N ][Ag(N ) ] With excess N all Ag + will be converted to Ag(N ) + ) ] Ex. Mixing ml of.0 M N with ml of M AgN After mixing: [N ] = 1.0 M [N ] >> [Ag + ] = M At equilibrium [N ] = 1.0 M [Ag(N ) ] [Ag + ] = M [Ag(N ) + ] = M [N ][Ag(N ) ] (1.0)[Ag(N ) ] [Ag(N ) + ] = M 4 [Ag(N ) ] f [Ag ][N [Ag ](1.0) ] [Ag + ] = M 7 17

18 omplex ion and solubility Ex. Agl(s) Ag + (aq) + l (aq) sp = Ag + (aq) + N (aq) Ag(N ) + (aq) 1 =.1 10 Ag(N ) + (aq) + N (aq) Ag(N ) + (aq) = verall Agl(s) + N (aq) Ag(N ) + (aq) + l (aq) f = sp 1 =.8 10 Q: Solubility of Agl in 10.0 M N Soln: x f.8 10 x = 0.48 M (10.0 x) f. solubility of Agl in water = = sp M More examples Q: Solubility of AgN(s) in a soln with [ + ] = 1.0 M? AgN(s) Ag + (aq) + N (aq) sp = N(aq) + (aq) + N (aq) a = Soln AgN(s) Ag + (aq) + N (aq) + (aq) + N (aq) N(aq) AgN(s) + + (aq) Ag + (aq) + N(aq) 1 sp a 6.10 AgN(s) + + (aq) Ag + (aq) + N(aq) Ini Equil 1.0 x x x x x x = M = [Ag + ] 18

19 Q: At 5 o, water saturated with 0.10 M S, p =.0 Maximum possible [Ni + ]? NiS(s) Ni + (aq) + S (aq) S(aq) + (aq) + S (aq) S (aq) + (aq) + S (aq) Soln [ + ] = 1 10 M controls [S ] S(aq) + (aq) + S (aq) = a1 a = sp = 10 1 a1 = a = very small [ ] [S ] (1 10 ) [S ] 6 [ S] [S ] = M [Ni + ][S ] = 10 1 = [Ni + ] ( ) [Ni + ]= M Q: Solubility of ubr(s) in 1.0 L of 1.0 M NaN? ubr(s) u + (aq) + Br (aq) sp = u + (aq) + N (aq) u(n) (aq) f = Soln ubr(s) + N (aq) Br (aq) + u(n) (aq) = sp f = very large Essentially 1.0 M N is all converted to u(n) [u(n) ] = [Br ] = 1 / (1.0 M) = 0. M sp = [u + ][Br ] = = [u + ][0.] [u + ] = M ] [u(n) 0. f [u ][N ] (.0 10 )[N ] [N ] = M 11 19

20 Q: With 0.1 M S, [u + ] = [Mn + ] = M, p range where us precipitates but MnS doesn t? Soln us(s) u + (aq) + S (aq) sp = MnS(s) Mn + (aq) + S (aq) sp = [u + ][S ] = = [ ][S ] [S ] = M for precipitation of us [Mn + ][S ] = = [ ][S ] [S ] = M for precipitation of MnS S(aq) + (aq) + S (aq) Assume [ S] doesn t change [ ] [S ] [ ] [S ] 6 [ S] = [ ] [S ] 7 Q: In a buffer with p =.00, solubility of SrF (s)? SrF (s) Sr + (aq) + F (aq) sp = F(aq) + (aq) + F (aq) a = Soln [ + ] = SrF (s) + (aq) F(aq) + Sr (aq) Equil x x [F] [Sr [ ] ] sp a ( ) (x) (x) [ ] x = x =.4 10 M 0