Determine Ksp for Cu(IO3)2 by ICE Table. Molar solubility of Cu(IO3)2 is 2.7 x 10^-3

Answer 1

Cu(IO₃)₂(s) ==> Cu²⁺(aq) + 2IO₃^-(aq)

Ksp = [Cu²⁺][IO₃^-]²

[Cu²⁺] = 2.7x10^-3

[IO₃^-] = 2 x 2.7x10^-3 = 5.4x10^-3

Ksp = (2.7x10^-3)(5.4x10^-3)²

Ksp = 7.9x10^-8

Just noticed question mentions to use ICE table (why?)

Using ICE table (which isn't really necessary)

Cu(IO₃)₂(s) ==> Cu²⁺(aq) + 2IO₃^-(aq)

1......................0.................0.............Initial

-2.7x10^-3....+2.7x10^-3...+5.4x10^-3....Change

...................2.7x10^-3......5.4x10^-3.....Equilibrium

Answer 2

The solubility product constant, Ksp, is defined as the product of the ion concentrations raised to their stoichiometric coefficients in the equilibrium equation for a sparingly soluble salt.

The equation for the dissolution of copper(II) iodate (Cu(IO3)2) is:

Cu(IO3)2(s) ⇌ Cu²⁺(aq) + 2 IO₃⁻(aq)

Let's define x as the molar solubility of Cu(IO3)2. Then, the equilibrium concentrations of the ions in solution are:

[Cu²⁺] = x M [IO₃⁻] = 2x M

At equilibrium, the solubility product expression is:

Ksp = [Cu²⁺][IO₃⁻]² = x (2x)² = 4x³

We are given that the molar solubility of Cu(IO3)2 is 2.7 × 10^-3 M.

Therefore, x = 2.7 × 10^-3 M.

Substituting this value into the Ksp expression gives:

Ksp = 4x³ = 4(2.7 × 10^-3 M)³ = 6.6 × 10^-11

Therefore, the solubility product constant, Ksp, for Cu(IO3)2 is 6.6 × 10^-11.