Lead (IV) bromide is a sparingly soluble salt in aqueous solution and dissociate

Question

Lead (IV) bromide is a sparingly soluble salt in aqueous solution and dissociates to its respective ions in an equilibrium reaction. Write the solubility product (K_) expression for the lead (IV) bromide salt What is the concentration of lead ions if the solubility product of lead (IV) bromide salt is 3.5 times 10^5 and the concentration of bromide ions is 3mM? What is/are the effects of addition of calcium sulfate salt on the equilibrium constant of lead (IV) bromide? you must fully discuss your answer before you can receive full credit)

Explanation / Answer

PbBr4 <==> Pb^2+ + 4Br^-

Ksp = [Pb^2+] [Br-] ^4

--------------------------------------------

[Br-] = 3mM = 0.003 M

[Pb^2+] [Br-] ^4 =3.5 *10^-9

or, [Pb^2+] = 3.5 *10^-9/(0.003)^4 = 43.21 M

-----------------------------------------------------

CaSO4 reacts with PbBr4 as follows:

CaSO4 + Pb^2+ ==> PbSO4 + Ca^2+

As PBSO4 has a low value Ksp, it is easily precipitated out from the solution. As a result more and more PbBr4 dissociation equilibrium will shift towards right hand side.

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

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