16.1k views
1 vote
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!

User Danypata
by
6.0k points

1 Answer

2 votes

Answer:

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

Step-by-step 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.

User Janjust
by
5.4k points