CHM 112 Summer 2007 M. Prushan Acid-Base Equilibria and Solubility Equilibria Chapter 16

CHM 112 Summer 2007 M. Prushan 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) Na + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) common ion

CHM 112 Summer 2007 M. Prushan Consider mixture of salt NaA and weak acid HA. 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 = pK a + log [A - ] [HA] pK a = -log K a Henderson-Hasselbalch equation pH = pK a + log [conjugate base] [acid]

CHM 112 Summer 2007 M. Prushan What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x x x Common ion effect 0.30 – x  x  0.52 pH = pK a + log [HCOO - ] [HCOOH] HCOOH pK a = 3.77 pH = log [0.52] [0.30] = 4.01 Mixture of weak acid and conjugate base!

CHM 112 Summer 2007 M. Prushan 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. Add strong acid H + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) Add strong base 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

CHM 112 Summer 2007 M. Prushan Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) HF 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

CHM 112 Summer 2007 M. Prushan = 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 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 NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (moles) end (moles) pH = log [0.25] [0.28] [NH 4 + ] = final volume = 80.0 mL mL = 100 mL [NH 3 ] =

CHM 112 Summer 2007 M. Prushan Maintaining the pH of Blood

CHM 112 Summer 2007 M. Prushan 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)

CHM 112 Summer 2007 M. Prushan Strong Acid-Strong Base Titrations NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) OH - (aq) + H + (aq) H 2 O (l) 0.10 M NaOH added to 25 mL of 0.10 M HCl

CHM 112 Summer 2007 M. Prushan

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) CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) At equivalence point (pH > 7):

CHM 112 Summer 2007 M. Prushan

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 (aq) K sp = [Ag + ] 2 [CO ] Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO (aq) K sp = [Ca 2+ ] 3 [PO ] 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

CHM 112 Summer 2007 M. Prushan

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.

CHM 112 Summer 2007 M. Prushan What is the solubility of silver chloride in g/L ? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] Initial (M) Change (M) Equilibrium (M) s+s +s+s ss K sp = s 2 s = K sp  s = 1.3 x [Ag + ] = 1.3 x M [Cl - ] = 1.3 x M Solubility of AgCl = 1.3 x mol AgCl 1 L soln g AgCl 1 mol AgCl x = 1.9 x g/L K sp = 1.6 x

CHM 112 Summer 2007 M. Prushan

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+ ] 0 = M [OH - ] 0 = 4.0 x M K sp = [Ca 2+ ][OH - ] 2 = 8.0 x Q = [Ca 2+ ] 0 [OH - ] 0 2 = 0.10 x (4.0 x ) 2 = 1.6 x Q < K sp No precipitate will form

CHM 112 Summer 2007 M. Prushan 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) Ag + (aq) + Br - (aq) K sp = 7.7 x s 2 = K sp s = 8.8 x NaBr (s) Na + (aq) + Br - (aq) [Br - ] = M AgBr (s) Ag + (aq) + Br - (aq) [Ag + ] = s [Br - ] = s  K sp = x ss = 7.7 x

CHM 112 Summer 2007 M. Prushan 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) Mg 2+ (aq) + 2OH - (aq) 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 M [OH - ] = 2s = 2.8 x M pOH = 3.55 pH = At pH less than Lower [OH - ] OH - (aq) + H + (aq) H 2 O (l) remove Increase solubility of Mg(OH) 2 At pH greater than 10.45Raise [OH - ] add Decrease solubility of Mg(OH) 2