Chapter 16 ContinuationAxBy ↔ xAy+ + yBx-Ksp= [Ay]x[Bx]-yAgI(s)↔ Ag+(aq) + I-(aq) Ksp= [Ag+][I-]Every insoluble solid will have molar solubility “s”AgI(s)↔ Ag+(aq) + I-(aq) s s Ksp= s2Cu(OH)2↔ Cu2+(aq) + 2OH-(aq) s 2s Ksp = s(2s)2 = 4s3Q<Ksp unsaturated solution ( no precipitate)Q >Ksp supersaturated solution (precipitate)Predicting Precipitation2 mL of 0.2 M of NaOH mixed with 1 L of 0.1 M CaCl2. Will a precipitate form?NaOH ↔ Na+ + OH-CaCl2↔ Ca2+ + 2Cl-Ca2+ + 2OH-1. What is the precipitate? Ca(OH)2Ksp= 8.0 x 10-62. Find Q!Q= [Ca2+][OH-]2You need to find the concentrations of both Ca2+ and OH-Moles of CaCl2= .1 M x 1L= .1 molesMoles of NaOH= .2 M x .002 L= .0004[CaCl2]=[Ca2+]= 0.1 moles / 1 L = .1 M Ca2+(usually you would add both the volumes of CaCl2 and NaOH but in this case it won’t make much of a difference since the change will be minimal)[OH-]= .0004 moles / 1 L= .0004 M OH-Q= [.1M][.0004]2 = 1.6 x 10-81.6 x 10-8 < Ksp= 8.6 x 10-6NO PRECIPITATION WILL OCCUR [Group 1 ions do not
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