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16.10. AgCl ( s ) Ag + ( aq ) + Cl - ( aq ). Complex Ion Equilibria and Solubility. A complex ion can increase the solubility of a salt. Consider. K sp = 1.6 x 10 -10. What is we add NH 3 (Lewis base) ?. Ag + ( aq ) + 2 NH 3 ( aq ) Ag(NH 3 ) 2 + ( aq ).
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AgCl (s) Ag+(aq) + Cl-(aq) Complex Ion Equilibria and Solubility A complex ion can increase the solubility of a salt. Consider Ksp= 1.6 x 10-10 What is we add NH3 (Lewis base) ? Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) Lewis acid Lewis base Complex ion [Ag(NH3)2+] = Formation constant = 1.7 x 107 Kf = [Ag+][NH3]2 Huge !!! Ag+ is effectively removed from solution, allowing more AgCl to dissolve. AgCl (s) + 2 NH3(aq) Ag(NH3)2+(aq) + Cl-(aq) Le Chatelier’s Principle
AgCl (s) Ag+(aq) + Cl-(aq) What is the molar solubility of AgCl in 0.10 M NH3? Ksp= 1.6 x 10-10 Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) Kf = 1.7 x 107 AgCl (s) + 2 NH3(aq) Ag(NH3)2+(aq) + Cl-(aq) 0.10 0.00 0.00 Initial (M) -s - 2s +s +s Change (M) ( - s) 0.10 -2s s s Equilibrium (M) Molar solubility = s K = Ksp Kf = (1.6 x 10-10)(1.7 x 107) = 2.8 x 10-3 [Ag(NH3)2+] [Cl-] (s)(s) = 2.8 x 10-3 = Kf = [NH3]2 (0.10 -2s)2 Assume s << 0.010 s = 5.0 x 10-3 M Check: (0.0050/0.10) x 100% = 5.0 %
What happens when we mix two solutions? Dilution Molarity = moles / L # moles = (conc.)(volume) = (Molarity)(L) mole1 = C1V1 = mole2 = C2V2 So, C2 =C1V1 / V2 Reaction? same Dilute to new volume No precipitate Unsaturated solution Qsp < Ksp Qsp = Ksp Saturated solution Precipitate will form Qsp > Ksp Supersaturated solution
AgCl (s) Ag+(aq) + Cl-(aq) A solution contains 0.10 M AgNO3 (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? Ksp= 1.6 x 10-10 = [Ag+] 0 [Cl-] 0 Qsp = 0.10 M 10.0 ml 1.0 x 10-5M 10.0 ml 20.0 ml 20.0 ml [Ag+] 0 [Cl-] 0 Qsp = 2.5 x 10-7 > 1.6 x 10-10 =Ksp Since Qsp > Ksp, a precipitate will form
How can we control the concentration of an ion in solution? What pH is required to keep [Cu2+] below 0.0010 M in a saturated solution of Cu(OH)2 ? Ksp,Cu(OH)2= 2.2 x 10-20 Cu(OH)2 (s) Cu2+(aq) + 2 OH-(aq) s 0.0010 M At Equilibrium c Ksp= 2.2 x 10-20 = [Cu2+] [OH-] 2 Ksp = 2.2 x 10-20 = (0.0010 M) (c2) 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 [Cu2+] will be less than 0.0010 M; at lower pH more Cu2+ will be in solution
What concentration of NH4+ is required to prevent the precipitation of Co(OH)2 in a solution of 0.10 M Co(NO3 )2 and 0.30 M NH3 ?(Assume NH4+ / NH3 only acts as a buffer.) Ksp,Co(OH)2= 2.5 x 10-10 , Kb,NH3= 1.8 x 10-5 => Buffer will fix pH => fix [OH-] => Find [NH4+] that sets [OH-] too low to cause Co(OH)2 ppt. Co(OH)2 (s) Co2+(aq) + 2 OH-(aq) Ksp= 2.5 x 10-10 = [Co2+] [OH-] 2 = (0.10 M) [OH-] 2 [OH-] = 5.0 x 10-5 or less to prevent ppt. NH3(aq) + H2O (l) NH4+(aq) + OH- (aq) Kb,NH3= 1.8 x 10-5 = [NH4+][OH-] = [NH4+](5.0 x 10-5 M) [NH3] 0.30 M => [NH4+] = 0.11 M
Calculate the concentration of the free metal ion (Hg2+) in the presence of a ligand. Starting concentrations: [Hg2+] 0 = 0.010 M [I-] 0 = 0.78 M Hg2+ + 4 I- HgI42- Kf= 1.0 x 10+30 Kf= 1.0 x 10+30 = [HgI42-] Large [Hg2+][I-]4 => Assume complete reaction with I- start 0.010 M 0.78 M 0.00 M Hg2+ + 4 I- HgI42- 0.78 - 4(0.010) = 0.74 M end 0.00 M 0.010 M Cannot be zero !! But Kf 8
Hg2+ + 4 I- HgI42- At Equilibrium x 0.10 -x 0.74 + 4x Kf= 1.0 x 10+30 = [HgI42-] = 0.10 - x [Hg2+][I-]4 (x)(0.74 + 4x) 4 => Assume x << 0.010 < 0.74 (1.0 x 10+30)(0.74)4 = 3.0 x 10+31 = 1 x 0.010 => x = 3.3 x 10-32 M = [Hg2+]
