Solubility Homework Set Worked Out Solutions

19 drfus chem123

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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+]...
View Full Document

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.

Ask a homework question - tutors are online