Question

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

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

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###

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 \(\left[\mathrm{HgI}_{4}{ }^{2-}\right]\)###

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 \(\left[\mathrm{I}^{-}\right]\)###

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 \(\left[\mathrm{Hg}^{2+}\right]\)###

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 \(\left[\mathrm{Hg}^{2+}\right] = 0.08798 \,\text{M}\)

Key concepts

Complex Ion Formation

Complex ion formation is a key concept in understanding how certain ions interact within a solution to create compounds with specific structures and properties. It involves a central metal ion, like \( \text{Hg}^{2+} \), binding with several smaller ions or molecules, such as iodide ions (\( \text{I}^{-} \)), forming a complex. This process is crucial in many chemical reactions, particularly those involving transition metals.

In the exercise you've encountered, \( \text{Hg}^{2+} \) reacts with \( \text{I}^{-} \) ions through a series of steps to form various complex ions, such as \( \text{HgI}^{+} \), \( \text{HgI}_2 \), \( \text{HgI}_3^{-} \), and \( \text{HgI}_4^{2-} \). Each step represents a complex ion formation, where iodine ions are progressively added to the central mercury ion.

The complex ions formed are crucial in the determination of the equilibrium state of such reactions, as they define the final composition of the solution. Understanding this concept helps to predict the outcome of a reaction and manipulate the conditions to achieve a desired product.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of reactants and products in a reversible reaction at the point where their rates of formation are equal. This steady state allows us to calculate the amounts of each species present in the solution using the principles of chemical equilibrium.

In the context of the exercise, the equilibrium concentrations of \( \left[\text{HgI}_4^{2-}\right] \), \( \left[\text{I}^{-}\right] \), and \( \left[\text{Hg}^{2+}\right] \) are determined after the reaction has reached its equilibrium. Calculating these concentrations involves using the stoichiometry of each step of the complex ion formation and applying the equilibrium constant expression to find the unknown concentrations.

For example, the equilibrium concentration of \( \left[\text{HgI}_4^{2-}\right] \) is found by considering the initial amounts of \( \text{Hg}^{2+} \) and \( \text{I}^{-} \) and then solving for \( x \), which represents the amount of \( \text{Hg}^{2+} \) converted to \( \text{HgI}_4^{2-} \). By substituting this value back into the equilibrium expression, you can confirm the concentrations at equilibrium.

Stability Constant

Stability constants (\( K \)) are fundamental in determining how stable a complex ion is within a solution. They offer insight into the likelihood of a complex ion's formation by expressing the ratio of the concentration of the products to the concentration of the reactants, each raised to the power of their coefficients.

In the given problem, several stability constants are provided for each step of complex ion formation: \( K_1 \), \( K_2 \), \( K_3 \), and \( K_4 \). Each value indicates the stability of the respective complex ions \( \text{HgI}^{+} \), \( \text{HgI}_2 \), \( \text{HgI}_3^{-} \), and \( \text{HgI}_4^{2-} \).

The overall stability constant for the reaction sequence can be found by multiplying the individual stability constants. A high stability constant indicates that the complex ion formed is more stable, meaning it prefers to stay as a product rather than dissociating back into its reactants. This concept is crucial when predicting the final state of a reaction and understanding the behavior of complex ions under different conditions.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It provides a mathematical framework to determine how much of each substance is consumed or produced, based on balanced chemical equations.

In the exercise, stoichiometry plays a vital role in calculating the equilibrium concentrations after the formation of complex ions. Initially, the concentrations of \( \text{Hg}^{2+} \) and \( \text{I}^{-} \) are set up from the initial amounts given in the problem. As the reactions take place, the stoichiometric coefficients provide a basis for understanding how much of each reactant is converted into products through each step of the complex ion formation.

By using stoichiometry, one can know that for every mole of \( \text{Hg}^{2+} \), four moles of \( \text{I}^{-} \) are needed to ultimately form one mole of \( \text{HgI}_4^{2-} \). These proportions balance over the course of the reaction and are used to set up equations that describe how concentrations change, ultimately leading to the calculation of equilibrium concentrations.