Solubility Homework Set Worked Out Solutions


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Unformatted text preview: 3(1.25 x 10‐2 M) +]=1.0 x 10‐6 M needed for Ag PO to ppt. [Ag 3 4 *Since the concentration of Ag+ needed for AgCl to precipitate is lower, it will precipitate first. 17 Dr. Fus CHEM 123 61. Will a precipitate form when 38 mL of 0.25 M Pb(NO3)2 and 12 mL of 0.35 M KBr are mixed? Why or why not? Pb2+(aq) + 2Br‐(aq) [Pb2+]= (0.25 mol/L)(0.038 L)=0.19 M (0.038 L + 0.012 L) [Br‐]=(0.35 mol/L)(0.012 L)=0.084 M (0.038 L + 0.012 L) Q=[Pb2+][Br‐]2=(0.19 M)(0.084 M)2=1.34 x 10‐3 Q>Ksp, therefore equilibrium will shift left and a precipitate forms. PbBr2(s) 62. A solution contains 0.10 M Mg(NO3)2 and 0.10 M Ca(NO3)2. If solid sodium oxalate, Na2C2O4, is added to the solution, what is [Ca2+] (M) when MgC2O4 begins to precipitate? (Assume no volume changes.) At 25 oC, Ksp of CaC2O4 is 2.3 x 10–9 and Ksp of MgC2O4 is 8.6 x 10–5. First, calculate the concentration of C2O42‐ when MgC2O4 begins to precipitate: Mg2+(aq) + C2O42‐(aq) MgC2O4(s) Q=[ Mg2+][ C2O42‐] 8.6 x 10‐5=(0.10 M)[ C2O42‐] 2‐]=8.6 x 10‐4 M needed for MgC O to ppt. [C2O4 24 Now, use this concentration of C2O42‐ in the Ksp expression for CaC2O4 to determine the concentration of Ca2+ at this point: Ca2+(aq) + C2O42‐(aq) CaC2O4(s) Ksp=[ Ca2+][ C2O42‐] 2.3 x 10‐9[ Ca2+](8.6 x 10‐4 M) [Ca2+]=2.7 x 10‐6 M 63. The Ksp for BaF2 is 1.0 10‐6. When 10 mL of 0.010 M NaF is mixed with 10 mL of 0.01 M BaNO3 will a precipitate form? Why or why not? BaF2(s) Ba2+(aq) + 2F‐(aq) 2+]= (0.01 mol/L)(0.010 L)=0.005 M [Ba (0.010 L + 0.010 L) [F‐]=(0.01 mol/L)(0.010 L)=0.005 M (0.010 L + 0.010 L) Q=[Ba2+][F‐]2=(0.005 M)(0.005 M)2=1.25 x 10‐7 Q<Ksp, therefore equilibrium will shift right and no precipitate forms. 18 Dr. Fus CHEM 123 64. A solution contains 0.005 M AsO43‐, 0.005 M I‐, and 0.005 M CO32‐. If AgNO3 is slowly added, in what order would the silver salts precipitate? Ksp = 1.0 x 10‐22 For Ag3AsO4, For AgI, Ksp = 8.3 x 10‐17 For Ag2CO3, Ksp = 8.1 x 10‐12 3Ag+(aq) + AsO43‐(aq) Ag3AsO4(s) Q=[Ag+]3[AsO43‐] 1.0 x 10‐22=[Ag+]3(0.005 M) [Ag+]=2.71 x 10‐7 M needed for Ag3AsO4 to ppt. Ag+(aq) + I‐(aq) Q=[Ag+][I‐] 8.3 x 10‐17=[Ag+](0.005 M) [Ag+]=1.66 x 10‐14 M needed for AgI to ppt. AgI(s) 2Ag+(aq) + CO32‐(aq) Ag2CO3(s) +]2[CO 2‐] Q=[Ag 3 8.1 x 10‐12=[Ag+]2(0.005 M) [Ag+]=4.02 x 10‐5 M needed for Ag2CO3 to ppt. *Lowest concentration will precipitate first: AgI precipitates first, then Ag3AsO4, then Ag2CO3. 65. Three beakers contain the following solutions: 1) 40.0 mL of 0.020 M Ca(NO3)2 2) 40.0 mL of 0.020 M Fe(NO3)2 3) 40.0 mL of 0.020 M Pb(NO3)2 If 10.0 mL of 0.050 M NaF is added to each beaker, in which beakers will a precipitate form? For CaF2, Ksp = 1.5 x 10‐10, FeF2 Ksp = 2.4 x 10‐6, PbF2 Ksp = 7.1 x 10‐7 MF2(s) M2+(aq) + 2F‐(aq) [M2+]= (0.020 mol/L)(0.040 L)=0.016 M (0.040 L + 0.010 L) [F‐]=(0.050 mol/L)(0.010 L)=0.010 M (0.040 L + 0.010 L) Q=[M2+][F‐]2=(0.016 M)(0.010 M)2=1.6 x 10‐6 Q=1.6 x 10‐6 for each equilibrium because the concentrations are the same. Compare th...
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This homework help was uploaded on 03/28/2014 for the course CHEM 123 taught by Professor Larosa during the Spring '06 term at Ohio State.

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