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Solubility Homework Set Worked Out Solutions

# 19 drfus chem123

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Unformatted text preview: or Ksp. Ksp can be thought of as a specific Q value that occurs at equilibrium. Q for given concentrations can be compared to Ksp to determine whether or not precipitation will occur: **If Q<Ksp, there are not enough products. The solution is unsaturated, and the reaction will shift to the right, so NO precipitate is formed. More solid can be dissolved. **If Q=Ksp, the reaction is at equilibrium. The solution is saturated. **If Q>Ksp, there are too many products. The solution is supersaturated, and the reaction will shift to the left, so a precipitate will form. 58. A solution contains Ca2+ at a concentration of 2.0 x 10‐4 M. If 40.0 mL of this solution is added to 25.0 mL of 5.0 x 10‐3 M NaF, will a precipitate form? If 40.0 mL of the Ca2+ solution is added to 25.0 mL of 5.0 x 10‐3 M Na3PO4 will a precipitate form? Ksp = 3.9 x 10‐11 For CaF2, Ksp = 2.0 x 10‐29 For Ca3(PO4)2, CaF2(s) Ca2+(aq) + 2F‐(aq) The concentrations of each ion must be calculated because the volumes change once they are added together: [Ca2+]= (2.0 x 10‐4 mol/L)(0.040 L)=1.23 x 10‐4 M (0.040 L + 0.0250 L) [F‐]=(5.0 x 10‐3 mol/L)(0.025 L)=1.92 x 10‐3 M (0.040 L + 0.0250 L) 2+][F‐]2=(1.23 x 10‐4 M)( 1.92 x 10‐3 M)2=4.55 x 10‐10 Q=[Ca Q>Ksp, therefore equilibrium will shift left and a precipitate forms. 16 Dr. Fus CHEM 123 Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43‐(aq) 2+]= (2.0 x 10‐4 mol/L)(0.040 L)=1.23 x 10‐4 M [Ca (0.040 L + 0.0250 L) [PO43‐]=(5.0 x 10‐3 mol/L)(0.025 L)=1.92 x 10‐3 M (0.040 L + 0.0250 L) Q=[Ca2+]3[ PO43‐]2=(1.23 x 10‐4 M)3( 1.92 x 10‐3 M)2=6.86 x 10‐18 Q>Ksp, therefore equilibrium will shift left and a precipitate forms. 59. Will a precipitate form when 4.5 mL of 0.025 M Pb(NO3)2 and 1.5 mL of 0.0065 M KF are mixed? How could you tell? PbF2(s) Pb2+(aq) + 2F‐(aq) [Pb2+]= (0.025 mol/L)(0.0045 L)=0.01875 M (0.0045 L + 0.0015 L) ‐]=(0.0065 mol/L)(0.0015 L)=0.001625 M [F (0.0045 L + 0.0015 L) Q=[Ca2+][F‐]2=(0.01875 M)(0.001625 M)2=4.95 x 10‐8 Q>Ksp, therefore equilibrium will shift left and a precipitate forms. 60. Which salt precipitates first and what is the minimum concentration of Ag+ necessary to cause this precipitation in a solution containing 7.5 mL of 0.025 M NaCl and 7.5 mL of 0.025 M Na3PO4? Ksp Ag3PO4 = 1.3 x 10‐20 Ksp AgCl = 1.8 x 10‐10 To determine which salt precipitates first, we need to look at Q, not Ksp, because precipitation occurs when Q>Ksp, which is not at equilibrium. We can, however, set Q equal to Ksp and solve for the minimum concentration of the ion required for precipitation to occur. Whichever concentration is lower, that solid will precipitate first: AgCl(s) Ag+(aq) + Cl‐(aq) [Cl‐]=(0.025 mol/L)(0.0075 L)=1.25 x 10‐2 M (0.0075 L + 0.0075 L) Q=[Ag+][Cl‐] 1.8 x 10‐10=[Ag+](1.25 x 10‐2 M) [Ag+]=1.4 x 10‐8 M needed for AgCl to ppt. Ag3PO4(s) 3Ag+(aq) + PO43‐(aq) [PO43‐]=(0.025 mol/L)(0.0075 L)=1.25 x 10‐2 M (0.0075 L + 0.0075 L) Q=[Ag+]3[PO43‐] 1.3 x 10‐20=[Ag+]...
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