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Chemistry Equilibrium

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Chemistry Equilibrium

  1. 1. Acid-Base Equilibria and Solubility Equilibria Chapter 16 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display. PowerPoint Lecture Presentation by J. David Robertson University of Missouri
  2. 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. Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). 16.2 The presence of a common ion suppresses the ionization of a weak acid or a weak base. CH 3 COONa ( s ) Na + ( aq ) + CH 3 COO - ( aq ) CH 3 COOH ( aq ) H + ( aq ) + CH 3 COO - ( aq ) common ion
  3. 3. Consider mixture of salt NaA and weak acid HA. p K a = -log K a Henderson-Hasselbalch equation 16.2 HA ( aq ) H + ( aq ) + A - ( aq ) NaA ( s ) Na + ( 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 = p K a + log [A - ] [HA] pH = p K a + log [conjugate base] [acid]
  4. 4. What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? 0.30 0.00 - x + x 0.30 - x 0.52 + x x 0.52 + x Common ion effect 0.30 – x  0.30 0.52 + x  0.52 HCOOH p K a = 3.77 = 4.01 16.2 Mixture of weak acid and conjugate base! HCOOH ( aq ) H + ( aq ) + HCOO - ( aq ) Initial ( M ) Change ( M ) Equilibrium ( M ) pH = p K a + log [HCOO - ] [HCOOH] pH = 3.77 + log [0.52] [0.30]
  5. 5. <ul><li>A buffer solution is a solution of: </li></ul><ul><li>A weak acid or a weak base and </li></ul><ul><li>The salt of the weak acid or weak base </li></ul><ul><li>Both must be present! </li></ul>A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. 16.3 Add strong acid Add strong base H + ( aq ) + CH 3 COO - ( aq ) CH 3 COOH ( aq ) OH - ( aq ) + CH 3 COOH ( aq ) CH 3 COO - ( aq ) + H 2 O ( l ) Consider an equal molar mixture of CH 3 COOH and CH 3 COONa
  6. 6. 16.3 HCl H + + Cl - HCl + CH 3 COO - CH 3 COOH + Cl -
  7. 7. 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 3 2- is a weak base and HCO 3 - is it conjugate acid buffer solution 16.3
  8. 8. = 9.20 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 0.050 M NaOH to 80.0 mL of the buffer solution? p K a = 9.25 = 9.17 start (moles) end (moles) 0.029 0.001 0.024 0.028 0.0 0.025 final volume = 80.0 mL + 20.0 mL = 100 mL 16.3 NH 4 + ( aq ) H + ( aq ) + NH 3 ( aq ) pH = p K a + log [NH 3 ] [NH 4 + ] pH = 9.25 + log [0.30] [0.36] NH 4 + ( aq ) + OH - ( aq ) H 2 O ( l ) + NH 3 ( aq ) pH = 9.25 + log [0.25] [0.28] [NH 4 + ] = 0.028 0.10 [NH 3 ] = 0.025 0.10
  9. 9. Chemistry In Action: Maintaining the pH of Blood 16.3
  10. 10. Titrations 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 ) 4.7
  11. 11. Strong Acid-Strong Base Titrations 16.4 NaOH ( aq ) + HCl ( aq ) H 2 O ( l ) + NaCl ( aq ) OH - ( aq ) + H + ( aq ) H 2 O ( l )
  12. 12. Weak Acid-Strong Base Titrations At equivalence point (pH > 7): 16.4 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 ) CH 3 COO - ( aq ) + H 2 O ( l ) OH - ( aq ) + CH 3 COOH ( aq )
  13. 13. Strong Acid-Weak Base Titrations At equivalence point (pH < 7): 16.4 HCl ( aq ) + NH 3 ( aq ) NH 4 Cl ( aq ) NH 4 + ( aq ) + H 2 O ( l ) NH 3 ( aq ) + H + ( aq ) H + ( aq ) + NH 3 ( aq ) NH 4 Cl ( aq )
  14. 14. 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) end (moles) 0.01 0.01 0.0 0.0 0.01 0.05 0.00 - x + x 0.05 - x 0.00 + x x x Final volume = 200 mL = 2.2 x 10 -11 0.05 – x  0.05 x  1.05 x 10 -6 = [OH - ] pOH = 5.98 pH = 14 – pOH = 8.02 HNO 2 ( aq ) + OH - ( aq ) NO 2 - ( aq ) + H 2 O ( l ) NO 2 - ( aq ) + H 2 O ( l ) OH - ( aq ) + HNO 2 ( aq ) Initial ( M ) Change ( M ) Equilibrium ( M ) [NO 2 - ] = 0.01 0.200 = 0.05 M K b = [OH - ][HNO 2 ] [NO 2 - ] = x 2 0.05- x
  15. 15. Acid-Base Indicators Color of acid (HIn) predominates Color of conjugate base (In - ) predominates 16.5 HIn ( aq ) H + ( aq ) + In - ( aq )  10 [HIn] [In - ]  10 [HIn] [In - ]
  16. 16. 16.5 pH
  17. 17. The titration curve of a strong acid with a strong base. 16.5
  18. 18. 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 16.5
  19. 19. Solubility Equilibria 16.6 K sp = [Ag + ][Cl - ] K sp is the solubility product constant K sp = [Mg 2+ ][F - ] 2 K sp = [Ag + ] 2 [CO 3 2 - ] K sp = [Ca 2+ ] 3 [PO 3 3 - ] 2 Dissolution of an ionic solid in aqueous solution: Q = K sp Saturated solution Q < K sp Unsaturated solution No precipitate Q > K sp Supersaturated solution Precipitate will form AgCl ( s ) Ag + ( aq ) + Cl - ( aq ) MgF 2 ( s ) Mg 2+ ( aq ) + 2F - ( aq ) Ag 2 CO 3 ( s ) 2Ag + ( aq ) + CO 3 2 - ( aq ) Ca 3 (PO 4 ) 2 ( s ) 3Ca 2+ ( aq ) + 2PO 4 3 - ( aq )
  20. 20. 16.6
  21. 21. 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. 16.6
  22. 22. What is the solubility of silver chloride in g/L ? K sp = [Ag + ][Cl - ] 0.00 + s 0.00 + s s s K sp = s 2 s = 1.3 x 10 -5 [Ag + ] = 1.3 x 10 -5 M [Cl - ] = 1.3 x 10 -5 M Solubility of AgCl = = 1.9 x 10 -3 g/L K sp = 1.6 x 10 -10 16.6 AgCl ( s ) Ag + ( aq ) + Cl - ( aq ) Initial ( M ) Change ( M ) Equilibrium ( M ) s = K sp  1.3 x 10 -5 mol AgCl 1 L soln 143.35 g AgCl 1 mol AgCl x
  23. 23. 16.6
  24. 24. If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl 2 , will a precipitate form? 16.6 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+ ] 0 = 0.100 M [OH - ] 0 = 4.0 x 10 -4 M K sp = [Ca 2+ ][OH - ] 2 = 8.0 x 10 -6 = 0.10 x (4.0 x 10 -4 ) 2 = 1.6 x 10 -8 Q < K sp No precipitate will form Q = [Ca 2+ ] 0 [OH - ] 0 2
  25. 25. 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 ? K sp = [Ag + ][Cl - ] K sp = 1.6 x 10 -10 16.7 K sp = 7.7 x 10 -13 K sp = [Ag + ][Br - ] 3.9 x 10 -11 M < [Ag + ] < 8.0 x 10 -9 M AgCl ( s ) Ag + ( aq ) + Cl - ( aq ) AgBr ( s ) Ag + ( aq ) + Br - ( aq ) [Ag + ] = K sp [Br - ] 7.7 x 10 -13 0.020 = = 3.9 x 10 -11 M [Ag + ] = K sp [Br - ] 1.6 x 10 -10 0.020 = = 8.0 x 10 -9 M
  26. 26. The Common Ion Effect and Solubility K sp = 7.7 x 10 -13 s 2 = K sp s = 8.8 x 10 -7 [Br - ] = 0.0010 M [Ag + ] = s [Br - ] = 0.0010 + s  0.0010 K sp = 0.0010 x s s = 7.7 x 10 -10 16.8 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) 0.0010 M NaBr? AgBr ( s ) Ag + ( aq ) + Br - ( aq ) NaBr ( s ) Na + ( aq ) + Br - ( aq ) AgBr ( s ) Ag + ( aq ) + Br - ( aq )
  27. 27. pH and Solubility K sp = [Mg 2+ ][OH - ] 2 = 1.2 x 10 -11 K sp = ( s )(2 s ) 2 = 4 s 3 4 s 3 = 1.2 x 10 -11 s = 1.4 x 10 -4 M [OH - ] = 2 s = 2.8 x 10 -4 M pOH = 3.55 pH = 10.45 At pH less than 10.45 Lower [OH - ] Increase solubility of Mg(OH) 2 At pH greater than 10.45 Raise [OH - ] Decrease solubility of Mg(OH) 2 16.9 <ul><li>The presence of a common ion decreases the solubility. </li></ul><ul><li>Insoluble bases dissolve in acidic solutions </li></ul><ul><li>Insoluble acids dissolve in basic solutions </li></ul>Mg(OH) 2 ( s ) Mg 2+ ( aq ) + 2OH - ( aq ) OH - ( aq ) + H + ( aq ) H 2 O ( l ) remove add
  28. 28. Complex Ion Equilibria and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. The formation constant or stability constant (K f ) is the equilibrium constant for the complex ion formation. 16.10 Co 2+ ( aq ) + 4Cl - ( aq ) CoCl 4 ( aq ) 2- K f = [CoCl 4 ] [Co 2+ ][Cl - ] 4 2- Co(H 2 O) 6 2+ CoCl 4 2- K f stability of complex
  29. 29. 16.10
  30. 30. 16.11
  31. 31. Qualitative Analysis of Cations 16.11
  32. 32. lithium sodium potassium copper 16.11 Flame Test for Cations
  33. 33. Chemistry In Action: How an Eggshell is Formed Ca 2+ ( aq ) + CO 3 2- ( aq ) CaCO 3 ( s ) H 2 CO 3 ( aq ) H + ( aq ) + HCO 3 - ( aq ) HCO 3 - ( aq ) H + ( aq ) + CO 3 2- ( aq ) CO 2 ( g ) + H 2 O ( l ) H 2 CO 3 ( aq ) carbonic anhydrase

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