The \(\mathrm{Hg}^{2+}\) ion forms complex ions with \(\mathrm{I}^{-}\) as follows: \(\mathrm{Hg}^{2+}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}^{+}(a q) \quad K_{1}=1.0 \times 10^{\mathrm{s}}\) \(\mathrm{HgI}^{+}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{2}(a q) \quad K_{2}=1.0 \times 10^{5}\) \(\mathrm{HgI}_{2}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{3}-(a q) \quad K_{3}=1.0 \times 10^{\circ}\) \(\mathrm{HgI}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}{ }^{2-}(a q) \quad K_{4}=1.0 \times 10^{\mathrm{s}}\) A solution is prepared by dissolving \(0.088\) mole of \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) and \(5.00\) mole of Nal in enough water to make \(1.0 \mathrm{~L}\) 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 \(\left[\mathrm{Hg}^{2+}\right]\).

Short Answer

Expert verified
a. Equilibrium concentration of \(\left[\mathrm{HgI}_{4}{ }^{2-}\right] = 1.915 \times 10^{-5} \,\text{M}\) b. Equilibrium concentration of \(\left[\mathrm{I}^{-}\right] = 4.9992 \,\text{M}\) c. Equilibrium concentration of \(\left[\mathrm{Hg}^{2+}\right] = 0.08798 \,\text{M}\)

Step by step solution

01

Calculate Initial Concentrations

Find the initial concentrations of \(\mathrm{Hg}^{2+}\) and \(\mathrm{I}^{-}\) ions upon dissolution of \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) and NaI in 1.0 L solution, respectively. Initial concentration of \(\mathrm{Hg}^{2+}\): \(\frac{0.088\,\text{mol}}{1.0\,\text{L}} = 0.088\,\text{M}\) Initial concentration of \(\mathrm{I}^{-}\): \(\frac{5.00\,\text{mol}}{1.0\,\text{L}} = 5.00\,\text{M}\) ###Step 2: Determine K values###
02

Determine K values

Write equilibrium expressions for all the complex ion reactions and determine the K values for the reactions. \(K_1 = \frac{[\mathrm{HgI}^{+}]}{[\mathrm{Hg}^{2+}][\mathrm{I}^{-}]} = 1.0 \times 10^{\mathrm{s}}\) \(K_2 = \frac{[\mathrm{HgI}_{2}]}{[\mathrm{HgI}^{+}][\mathrm{I}^{-}]} = 1.0 \times 10^{5}\) \(K_3 = \frac{[\mathrm{HgI}_{3}^{-}]}{[\mathrm{HgI}_{2}][\mathrm{I}^{-}]} = 1.0 \times 10^{\circ}\) \(K_4 = \frac{[\mathrm{HgI}_{4}{ }^{2-}]}{[\mathrm{HgI}_{3}^{-}][\mathrm{I}^{-}]} = 1.0 \times 10^{\mathrm{s}}\) ###Step 3: Calculate the concentration of \(\left[\mathrm{HgI}_{4}{ }^{2-}\right]\)###
03

Calculate the concentration of \(\left[\mathrm{HgI}_{4}{ }^{2-}\right]\)

The formation of \(\mathrm{HgI}_{4}{ }^{2-}\) involves a series of reactions, but the overall reaction can be represented by: \(\mathrm{Hg}^{2+} + 4\mathrm{I}^{-} \rightleftharpoons \mathrm{HgI}_{4}{ }^{2-}\) The overall equilibrium constant for the reaction, \(K_{\text{overall}}\), can be found by multiplying the individual \(K\) values: \(K_{\text{overall}} = K_1 \times K_2 \times K_3 \times K_4\) Plug in the given \(K\) values: \(K_{\text{overall}} = (1.0 \times 10^{\mathrm{s}}) \times (1.0 \times 10^{5}) \times (1.0 \times 10^{0}) \times (1.0 \times 10^{\mathrm{s}}) = 1 \times 10^{10}\) Now use stoichiometry: Let the decrease in the concentration of \(\mathrm{Hg}^{2+}\) be x. Then, the concentration of \([\mathrm{HgI}_{4}{ }^{2-}]\) at equilibrium equals x, which is the same amount of \(x\) moles of \([\mathrm{Hg}^{2+}]\) gets converted to \([\mathrm{HgI}_{4}{ }^{2-}]\). At equilibrium: \([\mathrm{Hg}^{2+}] = 0.088 - x\) \([\mathrm{I}^{-}] = 5.00 - 4x\) \([\mathrm{HgI}_{4}{ }^{2-}] = x\) Now we can write the expression for \(K_{\text{overall}}\): \(1 \times 10^{10} = \frac{x}{(0.088 - x)(5.00 - 4x)^4}\) Solve for x (Approximate as \([I^{-}] \approx 5.00 - 4x \approx 5.00\) and \([Hg^{2+}] \approx 0.088 - x \approx 0.088\). This results in a very small error since x is much smaller than the initial concentrations): \(1 \times 10^{10} = \frac{x}{(0.088)(5.00)^4}\) \(x = 1.915 \times 10^{-5} \,\text{M}\) So the equilibrium concentration of \(\left[\mathrm{HgI}_{4}{ }^{2-}\right] = 1.915 \times 10^{-5} \,\text{M}\). ###Step 4: Calculate the concentration of \(\left[\mathrm{I}^{-}\right]\)###
04

Calculate the concentration of \(\left[\mathrm{I}^{-}\right]\)

Calculate the equilibrium concentration of \(\mathrm{I}^{-}\) ions based on the amount of \(\mathrm{I}^{-}\) ions that reacted to form \(\mathrm{HgI}_{4}{ }^{2-}\): \[\left[\mathrm{I}^{-}\right] = 5.0\,\text{M} - 4x = 5.0\,\text{M} - 4(1.915 \times 10^{-5}\,\text{M}) = 4.9992 \,\text{M} \] ###Step 5: Calculate the concentration of \(\left[\mathrm{Hg}^{2+}\right]\)###
05

Calculate the concentration of \(\left[\mathrm{Hg}^{2+}\right]\)

Calculate the equilibrium concentration of \(\mathrm{Hg}^{2+}\) ions based on the amount of \(\mathrm{Hg}^{2+}\) ions that reacted to form \(\mathrm{HgI}_{4}{ }^{2-}\): \[\left[\mathrm{Hg}^{2+}\right] = 0.088\,\text{M} - x = 0.088\,\text{M} - 1.915 \times 10^{-5}\,\text{M} = 0.08798 \,\text{M} \] #Final Answers# a. Equilibrium concentration of \(\left[\mathrm{HgI}_{4}{ }^{2-}\right] = 1.915 \times 10^{-5} \,\text{M}\) b. Equilibrium concentration of \(\left[\mathrm{I}^{-}\right] = 4.9992 \,\text{M}\) c. Equilibrium concentration of \(\left[\mathrm{Hg}^{2+}\right] = 0.08798 \,\text{M}\)

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Most popular questions from this chapter

A solution is prepared by mixing \(100.0 \mathrm{~mL}\) of \(1.0 \times 10^{-4} M\) \(\mathrm{Be}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{~mL}\) of \(8.0 \mathrm{M} \mathrm{NaF}\). $$ \begin{aligned} \mathrm{Be}^{2+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \mathrm{BeF}^{+}(a q) & & K_{1}=7.9 \times 10^{4} \\ \mathrm{BeF}^{+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \mathrm{BeF}_{2}(a q) & & K_{2}=5.8 \times 10^{3} \\ \mathrm{BeF}_{2}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \mathrm{BeF}_{3}-(a q) & & K_{3}=6.1 \times 10^{2} \\ \mathrm{BeF}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \mathrm{BeF}_{4}{ }^{2-}(a q) & & K_{4}=2.7 \times 10^{1} \end{aligned} $$ Calculate the equilibrium concentrations of \(\mathrm{F}^{-}, \mathrm{Be}^{2+}, \mathrm{BeF}^{+}\), \(\mathrm{BeF}_{2}, \mathrm{BeF}_{3}^{-}\), and \(\mathrm{BeF}_{4}{ }^{2-}\) in this solution.

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