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The Hg²⁺ ion forms complex ions with I⁻ as follows: Hg²⁺ (aq) + I⁻ (aq) ⇌ HgI⁺ (aq) K₁ = 1x10⁸ HgI⁺ (aq) + I⁻ (aq) ⇌ HgI₂ (aq) K₂ = 1x10⁵ HgI₂ (aq) + I⁻ (aq) ⇌ HgI₃⁻ (aq) K₃ = 1x10⁹ HgI₃⁻ (aq) + I⁻ (aq) ⇌ HgI₄²⁻ (aq) K₄ = 1x10⁸ A solution is prepared by dissolving 0.088 mole of Hg(NO₃)₂ and 5.00 mole of Nal in enough water to make 1.0 L of solution. c. Calculate the equilibrium concentration of [Hg²⁺]

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The equilibrium concentration of [HgI₄²⁻] is calculated to be 1.915 x 10⁻⁵M.

Determine the Initial Concentrations

Compute the initial concentrations of mercury(II) ions (Hg²⁺) and iodine ions (I⁻) when mercury(II) nitrate (Hg(NO₃)₂) and sodium iodide (NaI) dissolve separately in a 1.0 L solution.

Initial concentration of Hg²⁺: 0.088 mol / 1.0 L = 0.088 M

Initial concentration of I⁻: 5.00 mol / 1.0 L = 5.00 M

The initial concentration of Hg²⁺ is calculated as 0.088 mol / 1.0 L, resulting in 0.088 M.

Similarly, the initial concentration of I⁻ is determined by dividing 5.00 mol by 1.0 L, yielding an initial concentration of 5.00 M.

Determine Equilibrium Constants (K Values):

Formulate equilibrium expressions for each of the complex ion reactions and ascertain the equilibrium constant (K) values associated with these reactions

K₁ = [HgI⁺] / [Hg²⁺][I⁻] = 1.0 x 10⁻¹

K₂ = [HgI₂] / [HgI⁺][I⁻] = 1.0 x 10⁵

K₃ = [HgI₃⁻] / [HgI₂][I⁻] = 1.0 x 10⁰

K₄ = [HgI₄²⁻] / [HgI₃⁻][I⁻] = 1.0 x 10⁻¹

Compute the Concentration of [HgI₄²⁻]:

The generation of HgI₄²⁻ encompasses a sequence of reactions; nonetheless, the comprehensive reaction can be depicted as:

Hg²⁺ + 4I⁻ ⇌ HgI₄²⁻

Determine the Overall Equilibrium Constant (K_overall) by multiplying the individual K values for the reaction:

K (overall) = K₁ x K₂ x K₃ x K₄

Substituting the K values:

K (overall) = (1.0 x 10⁻¹) x (1.0 x 10⁵) x (1.0 x 10⁰) x (1.0 x 10⁻¹) = 1 x 10¹⁰

Employ stoichiometry by letting the reduction in the concentration of Hg²⁺ be denoted as x. Consequently, at equilibrium, the concentration of [HgI₄²⁻] is equivalent to x, signifying that x moles of [Hg²⁺] undergo conversion to [HgI₄²⁻].

At equilibrium:

[Hg²⁺] = 0.088 - x

[I⁻] = 5.00 - 4x

[HgI₄²⁻] = x

Now, we can formulate the expression for the overall equilibrium constant, K (overall):

1 x 10¹⁰ = x / (0.088 - x)(5.00 - 4x)⁴

Determine the value of x by solving the equation (with approximations of [I⁻] ≈ 5.00 and [Hg²⁺] ≈ 0.088). The margin of error is negligible, given that x is significantly smaller than the initial concentrations.

x = 1.915 x 10⁻⁵ M

Question:

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

Hg²⁺ (aq) + I⁻ (aq) ⇌ HgI⁺ (aq) , K₁ = 1x10⁸

HgI⁺ (aq) + I⁻ (aq) ⇌ HgI₂ (aq), K₂ = 1x10⁵

HgI₂ (aq) + I⁻ (aq) ⇌ HgI₃⁻ (aq), K₃ = 1x10⁹

HgI₃⁻ (aq) + I⁻ (aq) ⇌ HgI₄²⁻ (aq), K₄ = 1x10⁸

A solution is prepared by dissolving 0.088 mole of Hg(NO₃)₂ and 5.00 mole of Nal in enough water to make 1.0 L of solution. Calculate the equilibrium concentration of [Hg²⁺]

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