Acid-Base Equilibria and Solubility Equilibria

Transcription

1 ACIDS-BASES COMMON ION EFFECT SOLUBILITY OF SALTS Acid-Base Equilibria and Solubility Equilibria Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2 The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The presence of a common ion suppresses the ionization of a weak acid or a weak base. Consider a mixture of CH 3 COOH and HCl 3 1

2 The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The presence of a common ion suppresses the ionization of a weak acid or a weak base. Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). CH 3 COONa (s) CH 3 COOH (aq) Na + (aq) + CH 3 COO - (aq) H + (aq) + CH 3 COO - (aq) common ion 4 Consider mixture of salt NaA and weak acid HA. NaA (s) HA (aq) Na + (aq) + A - (aq) H + (aq) + A - (aq) K a = [H+ ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a -log [HA] [A - ] -log [H + ] = -log K a + log [A- ] [HA] ph = pk a + log [A- ] [HA] Henderson-Hasselbalch equation ph = pk a + log pk a = -log K a [conjugate base] [acid] 5 Henderson-Hasselbalch equation is an approximation. It can t be used at the beginning and at the equivalence point of the titration. An acceptable buffer region is considered to be within one ph unit on either side of the pka value. 6 2

3 What is the ph of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Mixture of weak acid and conjugate base! HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) x x +x x Equilibrium (M) x x x Common ion effect 0.30 x x 0.52 HCOOH pk a = 3.77 ph = pk a + log [HCOO- ] [HCOOH] ph = log [0.52] [0.30] = A buffer solution is a solution of: 1. A weak acid or a weak base and 2. The salt of the weak acid or weak base Both must be present! A buffer solution has the ability to resist changes in ph upon the addition of small amounts of either acid or base. Consider an equal molar mixture of CH 3 COOH and CH 3 COONa Add strong acid H + (aq) + CH 3 COO - (aq) Add strong base OH - (aq) + CH 3 COOH (aq) CH 3 COOH (aq) CH 3 COO - (aq) + H 2 O (l) 8 HCl H + + Cl - HCl + CH 3 COO - CH 3 COOH + Cl - 9 3

4 Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) KF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO 2-3 is a weak base and HCO 3- is its conjugate acid buffer solution 10 Calculate the ph of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the ph after the addition of 20.0 ml of M NaOH to 80.0 ml of the buffer solution? NH 4 + (aq) H + (aq) + NH 3 (aq) ph = pk a + log [NH 3] [NH 4+ ] pk a = 9.25 ph = log [0.30] [0.36] = 9.17 start (moles) end (moles) NH + 4 (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) final volume = 80.0 ml ml = 100 ml [NH 4+ ] = [NH 3 ] = ph = log [0.25] [0.28] = What mass of NaC 7 H (benzoic acid) must be dissolved in L of 0.30 M HC 7 H 5 O 2 to produce a solution with ph = 4.78? (Assume solution volume is constant at 0.200L) 12 4

5 Solution: HC 7 H 5 O 2 +H 2 0 H 3 O + +C 7 H 5 O 2- (all aqueous) Ka = Ka = [H 3 O + ][C 7H 5O 2 2] ]/[HC 7H 5O 2 2] = [H 3 O+]=10 ph = = M[HC 7 H 5 O 2 ]=0.30M[C 7 H 5 O 2 ] =1.14M = / mol benzoate 13 Mass = L x 1.14 mol C 7 H 5 O 2- /L = mol C 7 H 5 O 2-1 mol benzoate = 1 mol sodium benzoate, so mol C 7 H 5 O 2- = mol NaC 7 H 5 O mol NaC 7 H 5 O 2 x 144 g NaC 7 H 5 O 2 / mol NaC 7 H 5 O 2 = g NaC 7 H 5 O 2 See Petrucci, et al, Example 17.5, page Buffer Capacity: The quantity of acid or base than can be added to a volume of buffer solution before its ph changes significantly. 15 5

6 Chemistry In Action: Maintaining the ph of Blood Red blood cells in a capillary 16 Titrations (Review) In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point the point at which the reaction is complete Indicator substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink) 17 Alternative Method of Equivalence Point Detection monitor ph 18 6

7 Titrations Equivalence point the point at which the reaction is complete End point the point at which a detector detects a change in an indicator Indicator substance that changes color at (or near) the equivalence point The indicator changes color (pink) 19 phenolphthalein 20 Strong Acid-Strong Base Titrations NaOH (aq) + HCl (aq) OH - (aq) + H + (aq) H 2 O (l) H 2 O (l) + NaCl (aq) 21 7

8 Strong Acid & Strong Base Weak Acid-Strong Base Titrations CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) At equivalence point (ph > 7): CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) 23 Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) NH 4 Cl (aq) H + (aq) + NH 3 (aq) NH 4 Cl (aq) At equivalence point (ph < 7): NH + 4 (aq) + H 2 O (l) NH 3 (aq) + H + (aq) 24 8

9 Weak Base & Strong Acid Sample of Strong Base E.P < 7 Standard Acid is added Begin 10/3 26 Titration of Weak Acid with Strong Base ph 8 6 Legend: Acetic Acid Chloroacetic Acid 4 Formic Acid Vol NaOH Added, L 27 9

10 What is the ph when ml of M NaOH solution have been added to ml of M HCl solution? Because it is a strong acid-base reaction, the reaction will be: H + (aq)+oh (aq) H 2 O(l) The original number of moles of H + in the solution is: x 10-3 L x M HCl = moles Example The number of moles of OH - added is: x 10-3 L x M OH - = moles Which results in: = moles H + (aq) The total volume of solution is 0.048L L = 0.098L [H + ]= (.00020/.098L) = 2.0 x 10-3 ph= Exactly 100 ml of 0.10 M HNO 2 are titrated with a 0.10 M NaOH solution. What is the ph at the equivalence point? start (moles) HNO 2 (aq) + OH - (aq) NO - 2 (aq) + H2 O (l) end (moles) Final volume = 200 ml [NO - 2 ] = = 0.05 M NO - 2 (aq) +H2 + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) K b = [OH- ][HNO 2 ] = [NO 2- ] x +x x x x 2 = 2.2 x x 0.05 x 0.05 x 1.05 x 10-6 = [OH - ] x x poh = 5.98 ph = 14 poh = Acid-Base Indicators HIn (aq) [HIn] [In - ] [HIn] [In - ] H + (aq) + In - (aq) Color of acid (HIn) predominates Color of conjugate base (In - ) predominates 30 10

11 Solutions of Red Cabbage Extract ph 31 Choosing Indicators For Titrations will depend on the overall ph of the salt produced The titration curve of a strong acid with a strong base

12 Which indicator(s) would you use for a titration of HNO 2 with KOH? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, ph > 7 Use cresol red or phenolphthalein 34 Titration of a polyprotic acid: one equivalence point per acidic hydrogen Solutions of Red Cabbage Extract ph 36 12

13 hydrangea blossoms vary from pink to blue ph-dependent mobilization and uptake of soil aluminium into the plants 37 A new study by physicianscientists at NewYork- Presbyterian/Weill Cornell Medical Center used the device in patients with mild to moderate ulcerative colitis (UC), determining that they have significantly more acidic ph in their colons, compared with the average person a finding that may impact treatment strategy

14 Solubility Equilibria 41 Solubility Equilibria AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] K sp is the solubility product constant MgF 2 (s) Mg 2+ (aq) + 2F - (aq) K sp = [Mg 2+ ][F - ] 2 Ag 2 CO 3 (s) 2Ag + (aq) + CO 2-3 (aq) Ksp = [Ag + ] 2 [CO 2-3 ] Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 3-4 (aq) Ksp = [Ca 2+ ] 3 [PO 3-4 ] 2 Dissolution of an ionic solid in aqueous solution: Q < K sp Unsaturated solution No precipitate Q = K sp Saturated solution Q > K sp Supersaturated solution Precipitate will form

15 Molar solubility (mol/l) is the number of moles of solute dissolved in 1 L of a saturated solution. Solubility (g/l) is the number of grams of solute dissolved in 1 L of a saturated solution. 44 What is the solubility of silver chloride in g/l? AgCl (s) Ag + (aq) + Cl - (aq) Initial (M) Change (M) +s +s Equilibrium (M) s s [Ag + ] = 1.3 x 10-5 M [Cl - ] = 1.3 x 10-5 M K sp = 1.6 x K sp = [Ag + ][Cl - ] K sp = s 2 s = K sp s = 1.3 x 10-5 Solubility of AgCl = 1.3 x 10-5 mol AgCl 1 L soln g AgCl x = 1.9 x 10 1 mol AgCl -3 g/l

16 If 2.00 ml of M NaOH are added to 1.00 L of M CaCl 2, will a precipitate form? The ions present in solution are Na +, OH -, Ca 2+, Cl -. Only possible precipitate is Ca(OH) 2 (solubility rules). Is Q > K sp for Ca(OH) 2? [Ca 2+ ] = M [OH - = 4.0 x ] 0 M Q = [Ca 2+ ] 0 [OH - ] 0 2 = 0.10 x (4.0 x 10-4 ) 2 = 1.6 x 10-8 K sp = [Ca 2+ ][OH - ] 2 = 8.0 x 10-6 Q < K sp No precipitate will form 47 What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br - and Cl - at a concentration of 0.02 M? AgBr (s) Ag + (aq) + Br - (aq) K sp = 7.7 x [Ag + ] = AgCl (s) [Ag + ] = K sp = [Ag + ][Br - ] K sp 7.7 x 10 = -13 = 3.9 x 10 [Br - ] M Ag + (aq) + Cl - (aq) K sp = 1.6 x K sp = [Ag + ][Cl - ] K sp 1.6 x 10 = -10 = 8.0 x 10 [Cl - ] M 3.9 x M < [Ag + ] < 8.0 x 10-9 M AgCl AgBr 48 The Common Ion Effect and Solubility The presence of a common ion decreases the solubility of the salt. What is the molar solubility of AgBr in (a) pure water and (b) M NaBr? AgBr (s) K sp = 7.7 x s 2 = K sp s = 8.8 x 10-7 Ag + (aq) + Br - (aq) NaBr (s) Na + (aq) + Br - (aq) [Br - ] = M AgBr (s) Ag + (aq) + Br - (aq) [Ag + ] = s [Br - ] = s K sp = x s s = 7.7 x

17 ph and Solubility The presence of a common ion decreases the solubility. Insoluble bases dissolve in acidic solutions Insoluble acids dissolve in basic solutions Mg(OH) 2 (s) K sp = [Mg 2+ ][OH - ] 2 = 1.2 x K sp = (s)(2s) 2 = 4s 3 4s 3 = 1.2 x s = 1.4 x 10-4 M [OH - ] = 2s = 2.8 x 10-4 M poh = 3.55 ph = remove add Mg 2+ (aq) + 2OH - (aq) At ph less than Lower [OH - ] OH - (aq) + H + (aq) H 2 O (l) Increase solubility of Mg(OH) 2 At ph greater than Raise [OH - ] Decrease solubility of Mg(OH) 2 50 Complex Ion Equilibria and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Co 2+ (aq) + 4Cl - (aq) 2- CoCl 4 (aq) The formation constant or stability constant (K f ) is the equilibrium constant for the complex ion formation. Co(H 2 O) 6 2+ K f = 2- CoCl 4 2- [CoCl 4 ] [Co 2+ ][Cl - ] 4 HCl K f stability of complex 51 Effect of Complexation on Solubility AgNO Add 3 + NH NaCl 3 Ag(NH AgCl 3 )

18 53 54 Qualitative Analysis of Cations 55 5

19 Flame Test for Cations lithium sodium potassium copper 56 Chemistry In Action: How an Eggshell is Formed Ca 2+ (aq) + CO 3 2- (aq) CaCO 3 (s) carbonic CO 2 (g) + H 2 O (l) H 2 CO 3 (aq) anhydrase H 2 CO 3 (aq) HCO 3- (aq) H + (aq) + HCO 3- (aq) H + (aq) + CO 3 2- (aq) electron micrograph 57 6