Question

The Hg²⁺ ion forms complex ions with I^- as follows:

Hg2+ + I- HgI+	K1 = 1.0 × 108
HgI+ + I- HgI2	K2 = 1.0 × 105
HgI2 + I- HgI3-	K3 = 1.0 × 109
HgI3- + I- HgI42-	K4 = 1.0 × 108

A solution is prepared by dissolving 0.053 mol Hg(NO₃)₂ and 5.00 mol NaI in enough water to make 1.0 L of solution.

Calculate the equilibrium concentration of [HgI₄^2- ].

[HgI₄^2- ] = M

Calculate the equilibrium concentration of [I^-].

[I^-] = M

Calculate the equilibrium concentration of [Hg²⁺].

[Hg²⁺] = M

Answer 1

[HgI₄^2-] = 0.053 M

[I^-] = 4.8 M

[Hg²⁺] = 1.0 x 10^-30 M

Explanation

Overall equation : Hg²⁺ + 4 I^-$\rightleftharpoons$ HgI₄^2-

Overall K = (K₁) * (K₂) * (K₃) * (K₄)

Overall K = (1.0 x 10⁸) * (1.0 x 10⁵) * (1.0 x 10⁹) * (1.0 x 10⁸)

Overall K = 1.0 x 10³⁰

initial moles Hg²⁺ = 0.053 mol

initial moles I^- = 5.00 mol

moles HgI₄^2- formed = initial moles Hg²⁺

moles HgI₄^2- formed = 0.053 mol

moles I^- consumed = 4 * (initial moles Hg²⁺)

moles I^- consumed = 4 * (0.053 mol)

moles I^- consumed = 0.212 mol

moles I^- remaining = (initial moles I^-) - (moles I^- consumed)

moles I^- remaining = (5.00 mol) - (0.212 mol)

moles I^- remaining = 4.788 mol