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3)Assume the volume change is negligible. Ksp AgI = 8.3 10-17 , Ksp PbI2 = 7.9 10a) Write down the individual reactions between NaI and Ag(NO3) and PbI2b) What is/are the solid product/s if there is any would form?2. -9.
c) Which one of the solid product will precipitate first from the solution?
Answers1) (a) Fe(OH)2↔ Fe2++ 2 OH-(b)1.43×10-3gL×1 mol89.9 g=1.59×10-5molLFe(OH)2= 1.59x10-5 M = [Fe2+= 3.18x10-5 M = [OH-] [OH-] = 2 [Fe2+Ksp= [Fe2+][OH-]2= (1.59x10-5)(3.18x10-5)2= 1.61x10-14[H+] = 1.0 x 10-14/3.18 x 10-5= 3.14 x 10-10(c)pH = -log[H+] = 9.50ORpOH = -log[OH-] = -log(3.18x10-8) = 4.50pH = 14 - pOH = 9.50(d)50.0 mL of 3.00x10-3 M Fe2+diluted of 100.0 mL = 1.50x10-3 M Fe50.0 mL of 4.00x10-6 M OH-diluted of 100.0 mL = 2.00x10-6 M OHQ= [Fe2+][OH-]2= (1.50x10-3)(2.00x10-6)2= 6.00x10-15Precipitate will NOTform since Q< Ksp2) (a)i.Cu(OH)2Cu2++ 2 OH-ii. 1.72 x 10-6x 1 mol= 1.76 x 10-7mol/0.100 L 97.5 giii.Ksp= [Cu2+][OH-]2= [1.76x10-7][3.53x10-7]2= 2.20x10-20(b)i.Zn(OH)2x Zn2++ 2 OH-Ksp= [Zn2+][OH-]2pH 9.35 = pOH 4.65; [OH-] = 10-pOH[OH-] = 10-4.65= 2.24x10-5M[Zn2+] = solubility of Zn(OH)2in mol/[Zn2+] = Ksp/ [OH-]2= (7.7 x 10-17)/(2.24 x 10-5)2= 1.5 x 10-7ii.[Zn2+]init= 0.100Mx 50 mL 100 mL= 0.0500 [OH-]init= 0.300Mx 50 mL 100 mL= 0.150 X= conc. loss to get to equilibriumKsp= 7.7x10-17= (0.0500 - X)(0.150 - 2X)2 X = 0.05All the Zn2+ ions get precipitated.ORZn2+ + 2(OH)- Zn(OH)]]2+-LLMMM2
0.05M 0.15M initiallyAll the Zn2+ ions will be precipitated as the Zn2+ions is the limiting here. In that situation we have an equilibrium as Zn(OH)2Zn2+ + 2(OH)-and the Ksp expression should be followed.Ksp = 7.7 10-17= 4x3, x =[Zn2+] = 2.7 10-6M3)(a) Ksp = [Mg2+][F-]2= (1.21 x 10-3) (2 x 1.21 x 10-3)2= 7.09 x 10-9(b) Ksp = [Mg2+] ( 0.1)27.09 x 10-9 = [Mg2+] (0.1)2[Mg2+] = (7.09 x 10-9) / (0.01)= 7.09 x 10-7M(c) [Mg2+] = 3.00 x 10-3x0.1L /0.3l = 0.001M[F-] = 2.00 x 10-3x 0.2L / 0.3L = 0.00133MQ = Ion Product = [Mg2+] [F-]2= (1.00 x 10-3) (1.33 x 10-3)2= 1.77 x 10-9Since Q < Ksp, no precipitate will form.Note: conclusion must be consistent with Q value.d) Solubility of MgF2decreases with the increasing temperature, thus dissolution process is exothermic.MgF2→ Mg2+ + 2F-+ Q (or H)Reason:i) Increased temperature puts a stress on the system (LeChâtelier). The system will reduce the stress by shifting the equilibrium in the endothermic (left) direction.4) (a) i) AgCl(s) → Ag+(aq) + Cl-(aq)ii) 8.9 x 10-5g= 6.2 x 10-7(in 100 mL)143.32 g/mol(6.2 x 10-7mol/100 mL) (1,000 mL/ 1 L) = 6.2 x 10-6iii) Ksp= [Ag+][Cl-] = (6.2 x 10-6)2= 3.8 x 10-11
(b) Find Q and see if it is bigger than the Ksp. -6(c) Use common ion effect. You will know the [Cl-] and then solve for the [Pb2+] using the Kspequation.Ksp=1.6 x 10-5=[Pb2+][0.25M]2Therefore, [Pb2+]=2.56 x 10-4M5) (a)Ksp= 5.0 x 10-13= [Ag+][Br-](b) [Ag+]= [Br-] = (Ksp)1/2= 7.1 x 10-7M ( saturated) (c) Adding distilled water to a solution with solid ( it is already saturated)will not change the concentration. It will remain the same.The value of [Ag+] after addition of distilled water is equal to the value in part (b). The concentration of ions in solution in equilibrium with a solid does not depend on the volume of the solution.(d) moles AgBr = 5.0/188= 0.027 moles5.0 x 10-13= (0.027/V)( 0.027/V) = 7.3 x 10-4/V2V = 3.8 x 104L(e) moles Ag+= 0.010 x 1.5 x 10-4= 1.5 x 10-6moles Br-= 2.0 x 10-3L x 5.0 x 10-4M= 1.0 x 10-6total volume = 0.012 L[Br-] = 1.0 x 10-6/ 0.012 L=8.3 x 10-5M[Ag+] = 1.5 x 10-6/ 0.012 L=1.3 x 10-4MQ = 1.3 x 10-4x 8.3 x 10-5= 1.0 x 10-8Q > Ksp so precipitation should occur(f) Na +I- (aq) + AgBr (s) AgI (s) + Br- (aq) + Na+solid AgI - yellow colored, is precipitated. The solubility of AgBr is greater than the solubility of AgI. In presence of Br- and I- ions, the less soluble AgI got precipitated.