The \(\mathrm{Hg}^{2+}\) ion forms complex ions with \(\mathrm{I}^{-}\) as follows: $$\begin{aligned} \mathrm{Hg}^{2+}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}^{+}(a q) & & K_{1}=1.0 \times 10^{8} \\ \mathrm{HgI}^{+}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}_{2}(a q) & & K_{2}=1.0 \times 10^{5} \\ \mathrm{HgI}_{2}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}_{3}^{-}(a q) & & K_{3}=1.0 \times 10^{9} \\ \mathrm{HgI}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) & \rightleftharpoons \mathrm{HgI}_{4}^{2-}(a q) & & K_{4}=1.0 \times 10^{8} \end{aligned}$$ A solution is prepared by dissolving 0.088 mole of \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) and 5.00 moles of NaI in enough water to make 1.0 L of solution. a. Calculate the equilibrium concentration of \(\left[\mathrm{HgI}_{4}^{2-}\right] .\) b. Calculate the equilibrium concentration of \(\left[\mathrm{I}^{-}\right] .\) c. Calculate the equilibrium concentration of \(\left[\mathrm{Hg}^{2+}\right]\).

Step by step solution

01

Write the equilibrium expressions for each reaction

The following expressions represent the equilibrium for each reaction: 1. \(K_1 = [\mathrm{HgI}^{+}]/\left([\mathrm{Hg}^{2+}][\mathrm{I}^{-}]\right)\) 2. \(K_2 = [\mathrm{HgI}_{2}]/\left([\mathrm{HgI}^{+}][\mathrm{I}^{-}]\right)\) 3. \(K_3 = [\mathrm{HgI}_{3}^{-}]/\left([\mathrm{HgI}_{2}][\mathrm{I}^{-}]\right)\) 4. \(K_4 = [\mathrm{HgI}_{4}^{2-}]/\left([\mathrm{HgI}_{3}^{-}][\mathrm{I}^{-}]\right)\)

02

Write the initial concentrations

Given the initial moles of Hg(NO₃)₂, NaI, and the volume, we can find the initial concentrations of Hg²⁺ and I⁻ ions: 1. \([\mathrm{Hg}^{2+ }]_0 = \frac{0.088\,\text{mol}}{1.0\,\text{L}} = 0.088\,\text{M}\) 2. \([\mathrm{I}^-]_0 = \frac{5.00\,\text{mol}}{1.0\,\text{L}} = 5.00\,\text{M}\)

03

Use the initial concentrations in the equilibrium expressions

We can rewrite the equilibrium expressions in terms of initial concentrations ([I⁻] and [Hg²⁺]) and the amount of change (x) caused by the equilibrium: 1. \(K_1 = \frac{x}{(0.088-x)(5.00-x)}\) 2. \(K_2 = \frac{x}{x(5.00-x)}\) 3. \(K_3 = \frac{x}{x(5.00-x)}\) 4. \(K_4 = \frac{x}{x(5.00-x)}\)

04

Solve for x

Now we need to solve for x (equilibrium concentration of HgI₄²⁻): 1. From equation 4: \(x = K_4([\mathrm{HgI}_{3}^{-}][\mathrm{I}^{-}])\) 2. Using equation 3, substitute for [HgI₃⁻]: \(x = K_4K_3([\mathrm{HgI}_{2}][\mathrm{I}^{-}])\) 3. Using equation 2, substitute for [HgI₂]: \(x = K_4K_3K_2([\mathrm{HgI}^{+}][\mathrm{I}^{-}])\) 4. Using equation 1, substitute for [HgI⁺]: \(x = K_4K_3K_2K_1([\mathrm{Hg}^{2+}][\mathrm{I}^{-}]) = K_4K_3K_2K_1(0.088-x)(5.00-x)\) 5. Now, multiplying the equilibrium constants and simplifying the equation, \(x = 1.0 \times 10^{30}(0.088-x)(5.00-x)\) Since the equilibrium constants are quite large, we can approximate that x << 0.088 and x << 5.00. Now, we have: \(x = 1.0 \times 10^{30}(0.088)(5.00)\) Solving for x, we get: \(x = 4.4 \times 10^{-9}\,\text{M}\) So, the equilibrium concentration of [HgI₄²⁻] is 4.4 x 10⁻⁹ M.

05

Calculate the equilibrium concentration of I⁻

Use the result from Step 4 to find the equilibrium concentration of I⁻ in the solution: \([\mathrm{I}^{-}] = [\mathrm{I}^{-}]_0 - x = 5.00\,\text{M} - 4.4 \times 10^{-9}\,\text{M} \approx 5.00\,\text{M}\) So, the equilibrium concentration of [I⁻] is approximately 5.00 M.

06

Calculate the equilibrium concentration of Hg²⁺

Use the result from Step 4 to find the equilibrium concentration of Hg²⁺ in the solution: \([\mathrm{Hg}^{2+}] = [\mathrm{Hg}^{2+}]_0 - x = 0.088\,\text{M} - 4.4 \times 10^{-9}\,\text{M} \approx 0.088\,\text{M}\) So, the equilibrium concentration of [Hg²⁺] is approximately 0.088 M. In conclusion, the equilibrium concentrations are: a. [HgI₄²⁻] ≈ 4.4 x 10⁻⁹ M b. [I⁻] ≈ 5.00 M c. [Hg²⁺] ≈ 0.088 M

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