The common ion effect is the shift in

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The common ion effect is the shift in equilibrium caused by the addition of

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

Consider mixture of salt Na. A and weak acid HA. Na. A (s) Na+

Consider mixture of salt Na. A and weak acid HA. Na. A (s) Na+ (aq) + A- (aq) HA (aq) H+ (aq) + A- (aq) [H+] = Ka [HA] [A-] -log [H+] = -log Ka - log [HA] [A-] -] [A -log [H+] = -log Ka + log [HA] [A-] p. H = p. Ka + log [HA] [H+][A-] Ka = [HA] Henderson-Hasselbalch equation p. H = p. Ka + log p. Ka = -log Ka [conjugate base] [acid]

What is the p. H of a solution containing 0. 30 M HCOOH and

What is the p. H of a solution containing 0. 30 M HCOOH and 0. 52 M HCOOK? Ka = 1. 7 x 10 -4 Mixture of weak acid and conjugate base! HCOOH (aq) H+ (aq) + HCOO- (aq) Initial (M) Change (M) Equilibrium (M) 0. 30 0. 00 0. 52 -x +x +x 0. 30 - x x 0. 52 + x Common ion effect [HCOOH] = 0. 30 – x 0. 30 [HCOO-] p. H = p. Ka + log [HCOOH] [0. 52] = 4. 01 [0. 30] Ka = [x][0. 52+x] ÷ [0. 3 -x] = [x][0. 52] ÷ 0. 3 = 1. 7 x 10 -4 HCOOH p. Ka = ant log - 1. 7 x 10 -4 = 3. 77 [HCOO- ] = 0. 52 + x 0. 52 p. H = 3. 77 + log

A buffer solution is a solution of: 1. A weak acid or a weak

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 p. H upon the addition of small amounts of either acid or base.

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3/Na. HCO 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 32 - is a weak base and HCO 3 - is its conjugate acid buffer solution

Solubility Equilibria molar solubility = [S] Ag. Cl (s) Ag+ (aq) + Cl- (aq)

Solubility Equilibria molar solubility = [S] Ag. Cl (s) Ag+ (aq) + Cl- (aq) Ksp = [Ag+][Cl-] Ksp is the solubility product constant Mg. F 2 (s) Mg 2+ (aq) + 2 F- (aq) Ksp = [Mg 2+][F-]2 Ag 2 CO 3 (s) 2 Ag+ (aq) + CO 32 - (aq) Ksp = [Ag+]2[CO 32 -] Ca 3(PO 4)2 (s) 3 Ca 2+ (aq) + 2 PO 43 - (aq) Ksp = [Ca 2+]3[PO 43 -]2 Dissolution of an ionic solid in aqueous solution: Q < Ksp Unsaturated solution Q = Ksp Saturated solution Q > Ksp Supersaturated solution No precipitate Precipitate will form

What is the solubility of silver chloride in g/L ? (Ksp =1. 3 x

What is the solubility of silver chloride in g/L ? (Ksp =1. 3 x 10 -5) Initial (M) Ag. Cl (s) Ag+ (aq) + Cl- (aq) 0. 00 Change (M) Equilibrium (M) +s +s s s Ksp = 1. 6 x 10 -10 Ksp = [Ag+][Cl-] Ksp = s 2 s = K sp s = 1. 3 x 10 -5 [Ag+] = 1. 3 x 10 -5 M [Cl-] = 1. 3 x 10 -5 M 1. 3 x 10 -5 mol Ag. Cl 143. 35 g Ag. Cl Solubility of Ag. Cl = x = 1. 9 x 10 -3 g/L 1 L soln 1 mol Ag. Cl

(1) The molar solubility [S] of lead(II) iodate in water is 4. 0 10

(1) The molar solubility [S] of lead(II) iodate in water is 4. 0 10 -5 mol/L. Calculate Ksp for lead(II) iodate. A. 1. 6 10 -9 B. 6. 4 10 -14 C. 2. 6 10 -13 D. 4. 0 10 -5 E. 4. 0 10 -15 Pb. I 2 pb 2+(aq) + 2 I-(aq) s 2 s s = 4 x 10 -5 M Ksp = [s][2 s]2 = 4 s 3 = 4 x (4 x 10 -5)3 = 2. 6 x 10 -13 (2) Which of the following would decrease the Ksp for Pb. I 2? A. Lowering the p. H of the solution B. Adding a solution of Pb(NO 3)2 C. Adding a solution of KI D. None of these—the Ksp of a compound is constant at constant temperature. (3) In which one of the following solutions will acetic acid have the greatest percent ionization? A. 0. 1 M CH 3 COOH C. 0. 1 M CH 3 COOH plus 0. 1 M CH 3 COONa B. 0. 1 M CH 3 COOH dissolved in 1. 0 M HCl D. 0. 1 M CH 3 COOH plus 0. 2 M CH 3 COONa

(4) The Ksp for silver(I) phosphate is 1. 8 10 -18. Calculate the molar

(4) The Ksp for silver(I) phosphate is 1. 8 10 -18. Calculate the molar solubility of silver(I) phosphate. A. 1. 6 10 -5 M D. 7. 2 10 -1 M B. 2. 1 10 -5 M E. 1. 8 10 -1 M C. 3. 7 10 -5 M Ag 3 PO 4 3 Ag+(aq) + PO 43 -(aq) 3 S S Ksp = 1. 8 x 10 -18 = [3 s]3[s] = 27 s 4 S 4 = Ksp /27 = 1. 8 x 10 -18 / 27 = 6. 66667 x 10 -20 S = 4 6. 66667 x 10 -20 = 1. 6 x 10 -5 = [Ag+] = [Ag 3 PO 4] (5) Calculate the silver ion concentration in a saturated solution of silver(I) carbonate (Ksp = 8. 1 10 -12). A. 5. 0 10 -5 M B. 2. 5 10 -4 M C. 1. 3 10 -4 M D. 2. 0 10 -4 M Ag 2 CO 3 2 Ag 1+ + CO 32 - 2 s s Ksp = 8. 1 x 10 -12 = [2 s]2[s] = 4 s 3 = 8. 1 x 10 -12 /4 = 2. 03 x 10 -12 s = 3 2. 03 x 10 -12 = 1. 27 x 10 -4 [Ag+] = 2 s = 2 x 1. 27 x 10 -4 = 2. 53 x 10 -4 Molar. E. 8. 1 10 -4 M