1A) Solid copper(II) nitrate is slowly added to 175 mL of a 0.0378 M potassium s

Question

1A) Solid copper(II) nitrate is slowly added to 175 mL of a 0.0378 M potassium sulfide solution. The concentration of copper(II) ion required to just initiate precipitation is ? M.

1B) Solid calcium nitrate is slowly added to 125 mL of a 0.0616 M sodium phosphate solution. The concentration of calciumion required to just initiate precipitation is ? M.

1C) Solid potassium sulfide is slowly added to 175 mL of a 0.178 M copper(II) nitrate solution until the concentration of sulfideion is 0.0501 M. The percent of copper(II) ion remaining in solution is ? %.

1D) Solid sodium phosphate is slowly added to 150 mL of a calcium chloride solution until the concentration of phosphate ion is 0.0101 M. The maximum amount of calcium ion remaining in solution is ? M.

Explanation / Answer

1A)

CuS(s) <-------> Cu2+ (aq) + S2-(aq)

Ksp= [Cu2+][S2-] = 6×10-37M2

  [K2S] = 0.0378M

therefore,

[S2-] = 0.0378M

substitung the S2 concentration

[Cu2+]×0.0378M = 6×10-37M2

     [Cu2+] = 1.59×10-35M

Therefore,

Concentraion of Cu2+ required to just start precipitate = 1.59×10-35M

1B)

Ca3(PO4)4 (s) <------> 3Ca2+(aq) + PO43-(aq)

Ksp = [Ca3+]3[PO4]3- = 1.3×10-26M5

[Na3PO4] = 0.0616M

[PO43-]= 0.0616M

substtituting the concentration of PO43-

[Ca2+]3×(0.0616M)2 = 1.3×10-26

[Ca2+]3 = 3.43×10-24M3

   [Ca2+] = 1.51×10-8M

Therefore,

Concentration of Ca2+ required just start to precipitate =1.51×10-8M

1C)

[Cu2+][S2-] = 6×10-37M2

[S2-]= 0.0501M

[Cu2+] × 0.0501M = 6×10-37M2

     [Cu2+] = 1.20×10-35M

at Initial,

[ Cu2+]= 0.178M

After addition

[Cu2+] =1.20×10-35M

% of Cu2+ ion reamaining = (1.20×10-35M/0.178M) ×100 = 6.74×10-33 %

1D)

Ksp of Ca3PO4 = 1.3×10-26M5

[Ca2+]3 × [PO43-]2 = 1.3×10-26M5

[PO43-] = 0.0101M

substituting the value of [PO43-]

[Ca2+]3 × (0.0101M) = 1.3×10-26M5

[Ca2+]3= 1.27×10-22M3

[Ca2+] = 5.03×10-8M

Therefore,

maximum amount of Ca2+ remains in solution=5.03×10-8M

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