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?

Question

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.

Wyatt Adkins · Accepted Answer

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

#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#. The concentrations of each species in solution must be determined. 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.

Naomi Aronson · Answer

undefined To determine whether CdS and Cu2S will precipitate upon adding 0.0010 mol of S2- to the solution, we need to calculate the ion product (Q) for each potential precipitate and compare it to their respective solubility products (Ksp). The ion product (Q) is the product of the concentrations of the ions in the solution at any given moment, while the solubility product (Ksp) is a constant value that represents the maximum product of the ion concentrations in a saturated solution of the precipitate at equilibrium. If Q > Ksp, precipitation will occur.

1. **Given:**
   - [Cd2+] = 0.020 mol/L
   - [Cu+] = 0.050 mol/L
   - Added [S2-] = 0.0010 mol in 1 L = 0.0010 mol/L
   - Ksp(CdS) ≈ 8 x 10^-27
   - Ksp(Cu2S) ≈ 6 x 10^-37

2. **Calculation of Q for CdS and Cu2S:**
   - Q for CdS = [Cd2+][S2-] = 0.020 mol/L * 0.0010 mol/L = 2 x 10^-5
   - Because Cu2S involves Cu+ in a 2:1 ratio to form Cu2S, the calculation for Q should ideally reflect this stoichiometry. However, for simplicity and to understand if there's any potential for precipitation, we'll consider the reaction with Cu+ directly as if it were forming CuS for an initial assessment: Q for CuS (approximation) = [Cu+][S2-] = 0.050 mol/L * 0.0010 mol/L = 5 x 10^-5

**Comparing Q to Ksp:**
- For CdS: Q = 2 x 10^-5 is much larger than Ksp = 8 x 10^-27. Therefore, CdS will precipitate.
- For Cu2S: Given the approximate Q calculated for Cu+ reacting with S2- (as if forming CuS for simplicity), which is much larger than Ksp for Cu2S, it suggests that in the correct stoichiometric balance, Cu2S could also precipitate due to the high relative concentration of Cu+ ions. However, the precise stoichiometric calculation for Cu2S would require considering the 2:1 copper to sulfur ratio, but given the very low Ksp for Cu2S, it's reasonable to expect that Cu2S would begin to precipitate as the concentration of available Cu+ is substantially above the threshold set by its Ksp.

**Conclusion:** Based on the comparison of Q to Ksp values, both CdS and Cu2S would precipitate upon the addition of 0.0010 mol of S2- to the solution.

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.

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.