Question

Calculate the solubility of copper (II) iodate in 0.16 M copper (II) nitrate. Ksp* is 7.4x10-8 M3. *You should know that the Ksp must refer to the copper iodate because all nitrate compounds are soluble and strong electrolytes!

Answer 1

Answer:

solubility in presence of 0.16M Cu(IO₃⁻)₂ = 3.4 x 10⁻⁴M*

Explanation:

        Cu(IO₃⁻)₂ ⇄           Cu⁺²          +   2(IO₃⁻)        

C(i)    ----------                0.16M                 0M

ΔC    ----------                   +x                   +2x

C(f)    ----------      0.16 + x ≅ 0.16M*        2x

Ksp = [Cu⁺²][IO₃⁻]²

7.4 x 10⁻⁸M³ = 0.16M(2x)² = 0.64x²

x = solubility in presence of 0.16M Cu(IO₃⁻)₂ = SqrRt(7.4x10⁻⁸M³/0.64M²)

= 3.4 x 10⁻⁴M*

*Note: This is consistent with the common ion effect in that a reduction in solubility is expected. The normal solubility of Cu(IO₃⁻)₂ in pure water at 25°C is ~2.7 x 10⁻³M.