Question

Consider the salt, lead(II) iodate, Pb(IO₃)₂.

Write the equilibrium reaction (with states), and the equilibrium-constant expression for the K_sp of this slightly soluble salt (including activity coefficients). Why doesn

Answer 1

Lead(II) iodate is a salt, and can be formed from an acid/base reaction:

2HIO3(aq) + Pb(OH)2(s) --> Pb(IO3)2(s) + 2H2O(l)

Perhaps what you mean is whether a solution of lead(II) iodate is acidic or basic. That is, do the ions hydrolyze to form H+ and make the solution acidic, or OH- and make the solution basic. Lead(II) iodate isn't particularly water soluble, so very little will dissolve in water. Since the salt is formed from a weak acid and weak base, the pH of the salt solution is not easy to predict without using the equilibrium constants. But since the Ka for iodic acid is 0.16, we can easily see that there is little tendency for IO3- to act as a base and form HIO3.

Pb2+ + 2H2O <==> Pb(OH)2(s) + 2H+ .... Keq is small

IO3^- + H2O <==> HIO3 + OH- ................ Keq is very, very small

Therefore, the H+ ion concentration will be greater than the OH- concentration, and the solution will be acidic.