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.
