1. 16.10

2. Complex Ion Equilibria and Solubility A complex ion can increase the solubility of a salt. Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) K f = [Ag(NH 3 ) 2 + ] [Ag + ][NH 3 ] 2 AgCl (s) Ag + (aq) + Cl - (aq) K sp = 1.6 x 10 -10 Consider What is we add NH 3 (Lewis base) ? Lewis acid Lewis base Complex ion = Formation constant = 1.7 x 10 7 Huge !!! Ag + is effectively removed from solution, allowing more AgCl to dissolve. AgCl (s) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) + Cl - (aq) Le Chatelier’s Principle

3. What is the molar solubility of AgCl in 0.10 M NH 3 ? Assumes = 5.0 x 10 -3 Ms << 0.010 Initial (M) Change (M) Equilibrium (M) 0.00 +s+s +s+s ss AgCl (s) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) + Cl - (aq)  -s (  - s) 0.10 - 2s 0.10 -2s AgCl (s) Ag + (aq) + Cl - (aq) K sp = 1.6 x 10 -10 Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) K f = 1.7 x 10 7 K = K sp K f = (1.6 x 10 -10 )(1.7 x 10 7 ) = 2.8 x 10 -3 K f = [Ag(NH 3 ) 2 + ] [Cl - ] [NH 3 ] 2 = 2.8 x 10 -3 = (s)(s) (0.10 -2s) 2 Check: (0.0050/0.10) x 100% = 5.0 % Molar solubility = s

4. What happens when we mix two solutions? Dilution Molarity = moles / L # moles = (conc.)(volume) = (Molarity)(L) mole 1 = C 1 V 1 = mole 2 = C 2 V 2 So, C 2 = C 1 V 1 / V 2 Reaction? same Dilute to new volume Q sp = K sp Saturated solution Q sp < K sp Unsaturated solution No precipitate Q sp > K sp Supersaturated solution Precipitate will form

5. A solution contains 0.10 M AgNO 3 (aq). When a 10.0 ml sample is mixed with 10.0 ml tap water (containing 1.0 x 10 -5 M Cl - ), will a white precipitate form? AgCl (s) Ag + (aq) + Cl - (aq) K sp = 1.6 x 10 -10 = [Ag + ] 0 [Cl - ] 0 Since Q sp > K sp, a precipitate will form Q sp = 0.10 M 10.0 ml 1.0 x 10 -5 M 10.0 ml 20.0 ml Q sp = 2.5 x 10 -7 > 1.6 x 10 -10 = K sp [Ag + ] 0 [Cl - ] 0

6. What pH is required to keep [Cu 2+ ] below 0.0010 M in a saturated solution of Cu(OH) 2 ? K sp,Cu(OH)2 = 2.2 x 10 -20 Cu(OH) 2 (s) Cu 2+ (aq) + 2 OH - (aq) K sp = 2.2 x 10 -20 = [Cu 2+ ] [OH - ] 2 K sp = 2.2 x 10 -20 = (0.0010 M) (c 2 ) How can we control the concentration of an ion in solution? At Equilibrium s 0.0010 M c c = 4.7 x 10 -9 = [OH - ] pH = 14 + log[OH - ] = 14 + (-8.33) = 5.67 => if pH is higher than 5.67, [OH - ] > 4.7 x 10 -9 M and [Cu 2+ ] will be less than 0.0010 M; at lower pH more Cu 2+ will be in solution

7. What concentration of NH 4 + is required to prevent the precipitation of Co(OH) 2 in a solution of 0.10 M Co(NO 3 ) 2 and 0.30 M NH 3 ?(Assume NH 4 + / NH 3 only acts as a buffer.) K sp,Co(OH)2 = 2.5 x 10 -10, K b,NH3 = 1.8 x 10 -5 Co(OH) 2 (s) Co 2+ (aq) + 2 OH - (aq) K sp = 2.5 x 10 -10 = [Co 2+ ] [OH - ] 2 = ( 0.10 M) [OH - ] 2 K b,NH3 = 1.8 x 10 -5 = [NH 4 + ][OH - ] = [NH 4 + ](5.0 x 10 -5 M) => Buffer will fix pH => fix [OH - ] => Find [NH 4 + ] that sets [OH - ] too low to cause Co(OH) 2 ppt. [OH - ] = 5.0 x 10 -5 or less to prevent ppt. NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) [NH 3 ] 0.30 M => [NH 4 + ] = 0.11 M

8. Calculate the concentration of the free metal ion (Hg 2+ ) in the presence of a ligand. Starting concentrations: [Hg 2+ ] 0 = 0.010 M [I - ] 0 = 0.78 M Hg 2+ + 4 I - HgI 4 2- K f = 1.0 x 10 +30 => Assume complete reaction with I - K f = 1.0 x 10 +30 = [HgI 4 2- ] [Hg 2+ ][I - ] 4 Large Hg 2+ + 4 I - HgI 4 2- start end 0.010 M0.78 M0.00 M 0.78 - 4(0.010) = 0.74 M 0.010 M But K f  8 Cannot be zero !!

9. => Assume x << 0.010 < 0.74 K f = 1.0 x 10 +30 = [HgI 4 2- ] = 0.10 - x [Hg 2+ ][I - ] 4 Hg 2+ + 4 I - HgI 4 2- At Equilibrium x 0.74 + 4x 0.10 -x (x)(0.74 + 4x) 4 (1.0 x 10 +30 )(0.74) 4 = 3.0 x 10 +31 = 1 0.010 x => x = 3.3 x 10 -32 M = [Hg 2+ ]

10. 16.11

11. Qualitative Analysis of Cations 16.11