1 liter of solution contains 0.020mol #Cd^(2+)# , 0.050mol #Cu^(+)# and 0.40mol KCN. Will CdS and #Cu_2S# precipitate if we add 0.0010mol #S^(2- )# to the solution?

Please explain what is happening in the solution before and after adding #S^(2-)# .
I have the calculus in my book, but dont understand what is happening in the solution exactly.

1 Answer
Sep 16, 2017

Answer:

WARNING! Long answer! A precipitate of #"CdS"# will form.

Explanation:

#bb("Before adding S"^"2-")#

The initial solution contains #"Cd"^"2", "Cu"^"+"#, and #"CN"^"-"#.

These will react to form the complex ions #"Cd(CN)"_4^"2-"; K_text(f) = 6.0 × 10^18# and #"Cu(CN)"_4^"3-"; K_text(f) = 2.0 × 10^30#.

We must calculate the concentrations of all species in solution.

Because the formation constants are so large, essentially all the #"Cd"^"2+"# and #"Cu"^"+"# will be converted to their complex ions.

Thus, we will have 0.020 mol/L #"Cd(CN)"_4^"2-"# and 0.050 mol/L #"Cu(CN)"_4^"3-"#.

#["CN"^"-"]# will decrease by 0.080 mol/L in forming the #"Cd"# complex and by 0.10 mol/L in forming the #"Cu"# complex.

At equilibrium, #["CN"^"-"] = "0.40 mol/L - 0.28 mol/L" = "0.12 mol/L"#

#color(white)(mmmmmmm)"Cd"^"2+" + "4CN"^"-" ⇌ "Cd(CN)"_4^"2-"#
#"I/mol·L"^"-1":color(white)(mll)0.020color(white)(mll)0.40color(white)(mmmml)0#
#"C/mol·L"^"-1":color(white)(m)"-0.020"color(white)(ml)"-0.28"color(white)(mml)"+0.020"#
#"E/mol·L"^"-1":color(white)(mml)xcolor(white)(mmll)0.12color(white)(mmml)0.020#

#K_text(f) = (["Cd"("CN")_4^"2-"])/(["Cd"^"2+"]["CN"^"-"]^4) = 0.020/(["Cd"^"2+"] × 0.12^4) = 6.0 × 10^18#

#["Cd"^"2+"] = 0.020/(0.12^4 × 6.0 × 10^18) = 1.61 × 10^"-17" color(white)(l)"mol/L"#

#color(white)(mmmmmmml)"Cu"^"+" + "4CN"^"-" ⇌ "Cu(CN)"_4^"3-"#
#"I/mol·L"^"-1":color(white)(mll)0.050color(white)(mll)0.40color(white)(mmmml)0#
#"C/mol·L"^"-1":color(white)(m)"-0.050"color(white)(ml)"-0.28"color(white)(mml)"+0.050"#
#"E/mol·L"^"-1":color(white)(mml)xcolor(white)(mmll)0.12color(white)(mmml)0.050#

#K_text(f) = (["Cu"("CN")_4^"3-"])/(["Cu"^"+"]["CN"^"-"]^4) = 0.050/(["Cu"^"+"] × 0.12^4) = 2.0 × 10^30#

#["Cu"^"2+"] = 0.050/(0.12^4 ×2.0 × 10^30) = 5.18 × 10^"-36" color(white)(l)"mol/L"#

#bb("After adding S"^"2-")#

Will we get a precipitate of #"CdS" (K_text(sp) = 1 × 10^"-27")#?

#color(white)(mmmmmm)"CdS"color(white)(m) ⇌color(white)(m)"Cd"^"2+" color(white)(m)+ color(white)(ml)"S"^"2-"#
#"I/mol·L"^"-1":color(white)(mmmmml)1.61 × 10^"-17"color(white)(m)0.0010#

#Q_text(sp) = ["Cd"^"2+"]["S"^"2-"] = 1.61 × 10^"-17" × 0.0010 = 1.61 × 10^"-20"#

#Q_text(sp) > K_text(sp)#, so a precipitate of #"CdS"# will form.

Will we get a precipitate of #"Cu"_2"S" (K_text(sp) = 2.0 × 10^"-47")#?

#color(white)(mmmmmm)"Cu"_2"S"color(white)(m) ⇌color(white)(m)"2Cu"^"+" color(white)(m)+ color(white)(ml)"S"^"2-"#
#"I/mol·L"^"-1":color(white)(mmmmmll)5.18 × 10^"-36"color(white)(mm)0.0010#

#Q_text(sp) = ["Cu"^"+"]^2["S"^"2-"] = (5.18 × 10^"-36")^2 × 0.0010 = 2.67 × 10^"-74"#

#Q_text(sp) < K_text(sp)#, so a precipitate of #"Cu"_2"S"# will not form.