A solution contains 2.0*10^-3 M Pb^2+ and 1.0*10^-4 M Cu^+. I^- is added to the

Question

A solution contains 2.0*10^-3 M Pb^2+ and 1.0*10^-4 M Cu^+. I^- is added to the solution little by little, will CuI (Ksp=1.0*10^-12) or PbI2 (Ksp=7.9*10^-9) precipitate first? Determine the concentration of I^- required to commence precipitation of each salt.

Explanation / Answer

Is there a typo?. I think it should be Cu? and not Cu²? A) The salts precipitate according to: Cu?(aq) + I?(aq) ? CuI(s) with Ksp = [Cu?]·[I?] and Pb²?(aq) + 2 I?(aq) ? PbI2(s) with Ksp = [Pb²?]·[I?]² Precipitation occurs when [I?] reaches a critical value, which is is given by Ksp and the metal ion concentration in solution. The salt for which has the lower critical [I?] will precipitate first. Compare the order of magnitude of the Ksp and you find that Copper iodide will precipitate first, To be sure about that it might be better to solve part B) and C) first and compare the values for [I?]. B) Ksp = [Pb²?]·[I?]² => [I?] = v( Ksp / [Pb?] ) = v( 1.4×10?8 / 2.0×10?³ = 2.65×10?³M C) Ksp = [Cu?]·[I?] => [I?] = Ksp / [Cu?] = 5.3×10?¹² / 1.0×10?4 = 5.3×10?8M D) After reaching concentration calculated in part C), further addition of iodide increases [I?] while [Cu?] decreases due to copper iodide precipitation. The ionic concentrations still satisfy the equilibrium condition. So plug in iodide concentration calculated in part B) and you have the copper ion concentration, when lead iodide starts to precipitate: Ksp = [Cu?]·[I?] => [Cu?] = Ksp / [I?] = 5.3×10?¹² / 2.65×10?³M = 2.0×10??M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.