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 solubility product constant, Ksp, expression for lead(II)iodate is:

Pb(IO3)2(s) ⇌ Pb2+(aq) + 2IO3-(aq)

The balanced equation shows that one mole of Pb(IO3)2 dissociates into one mole of lead ions and two moles of iodate ions. Thus, if we let x be the concentration of Pb2+ and 2x be the concentration of IO3-, then:

Ksp = [Pb2+][IO3-]^2

Substituting the molar solubility of Pb(IO3)2 into the above equation, we have:

Ksp = (4.0 × 10^-5)^1(2*4.0 × 10^-5)^2
     = 2.56 × 10^-14
     = 2.6 × 10^-13 (rounded to two significant figures)

Therefore, the correct answer is choice C.

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