The molar solubility of lead(II) iodate in water is 4.0 * 10^-5 mol/L. Calculate K_sp for lead(II) iodate.

Question

Quizplus · Accepted Answer

The balanced equation for the dissolution of lead (II) iodate is:
Pb(IO3)2(s) ⇌ Pb2+(aq) + 2 IO3-(aq)
The solubility product expression for Pb(IO3)2 is:
Ksp = [Pb2+][IO3-]^2

From the given molar solubility we know that [Pb2+] = 4.0×10^-5 M and [IO3-] = 8.0×10^-5 M.
Substituting this into the Ksp expression gives:

Ksp = (4.0×10^-5)(8.0×10^-5)^2 = 2.56×10^-12
Converting this to scientific notation gives:
Ksp = 2.6×10^-13 ≈ 2.6×10^-13
Therefore, the answer is choice C, 2.6×10^-13.

The Molar Solubility of Lead(II) Iodate in Water Is 4