Problem 67 - Chemistry An Atoms First Approach

Question

The overall formation constant for \(\mathrm{HgI}_{4}^{2-}\) is \(1.0 \times 10^{30}\) That is, $$ 1.0 \times 10^{30}=\frac{\left[\mathrm{HgI}_{4}^{2-}\right]}{\left[\mathrm{Hg}^{2+}\right]\left[\mathrm{I}^{-}\right]^{4}} $$ What is the concentration of \(\mathrm{Hg}^{2+}\) in \(500.0 \mathrm{mL}\) of a solution that was originally \(0.010\) \(M\) \(\mathrm{Hg}^{2+}\) and \(0.78\) \(M\) \(\mathrm{I}^{-} ?\) The reaction is $$\mathrm{Hg}^{2+}(a q)+4 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}^{2-}(a q)$$

Step-by-Step Solution

Answer

The concentration of Hg^2+ in 500.0 mL of the solution is approximately \(7.07 \times 10^{-3} M\).

Step 1: Initial concentration table

Set up a preliminary Ice Table: ``` [Hg^2+] [I^-] [HgI4^2-] Initial 0.010 M 0.78 M 0 ```

Step 2: Change in concentration

Since the reaction is Hg^2+(aq) + 4I^-(aq) <=> HgI4^2-(aq), let the change in concentration of Hg^2+ be x, and I^- be 4x. Thus, the concentration change of HgI4^2- will also be x. ``` [Hg^2+] [I^-] [HgI4^2-] Initial 0.010 M 0.78 M 0 Change -x -4x +x ```

Step 3: Equilibrium concentration

Equilibrium concentrations are as follows: ``` [Hg^2+] [I^-] [HgI4^2-] Equilibrium 0.010-x 0.78-4x x ```

Step 4: Formation constant equation

Write the equation for the formation constant of HgI4^2-, using the given number and equilibrium concentrations: \(1.0 \times 10^{30} = \frac{[HgI4^{2-}]}{[Hg^{2+}][I^-]^4}\) Substitute equilibrium concentrations from the table: \(1.0 \times 10^{30} = \frac{x}{(0.010-x)(0.78-4x)^4}\)

Step 5: Simplify and solve

Make the simplifying assumption that x will be much smaller than 0.010 and 0.78, so the expression becomes: \(1.0 \times 10^{30} = \frac{x}{(0.010)(0.78)^4}\) Solve for x: \(x = (0.010)(0.78)^4 (1.0 \times 10^{30})\) \(x \approx 2.93 \times 10^{-2}\)

Step 6: Calculate the concentration of Hg^2+

The concentration of Hg^2+ at equilibrium is (0.010-x). Substituting the value of x: [ Hg^2+ ] = (0.010 - 2.93 \times 10^{-2}) \approx 7.07 \times 10^{-3} M So, the concentration of Hg^2+ in 500.0 mL of the solution is approximately \(7.07 \times 10^{-3} M\).

Key Concepts

Formation Constant

The formation constant, often represented as \( K_f \), is a measure of the stability of a complex ion in solution. It essentially tells us how readily reactants form a complex ion at equilibrium. In our exercise, the formation constant for \( \mathrm{HgI}_4^{2-} \) is given as \( 1.0 \times 10^{30} \). This large value indicates that the formation of \( \mathrm{HgI}_4^{2-} \) is highly favored, meaning that almost all \( \mathrm{Hg}^{2+} \) and \( \mathrm{I}^- \) will form this complex in solution. The formation constant equation is set up as follows:

• Products are in the numerator.
• Reactants are in the denominator.
• Each reactant and product concentration is raised to the power of their stoichiometric coefficients in the reaction.

In our case, this is expressed in the equation: \[K_f = \frac{[\mathrm{HgI}_4^{2-}]}{[\mathrm{Hg}^{2+}][\mathrm{I}^-]^4}\] Given that \( K_f \) is extremely large, we can assume most of the iodine and mercury ions are present predominantly in the form of the complex ion at equilibrium.

ICE Table

The ICE table (Initial, Change, Equilibrium) is a useful tool for tracking the concentrations of species in equilibrium reactions. It helps us visualize how the concentrations change from their initial states to equilibrium. Here is how you construct an ICE table:

• Initial: List the initial concentrations of reactants and products before any reaction occurs. In our case, \( [\mathrm{Hg}^{2+}] \) is 0.010 M and \( [\mathrm{I}^-] \) is 0.78 M, while \([\mathrm{HgI}_4^{2-}]\) starts at 0 since it hasn't formed yet.
• Change: Define the changes using variables, often represented with \(x\). For the reaction \( \mathrm{Hg}^{2+} + 4\mathrm{I}^- \rightarrow \mathrm{HgI}_4^{2-} \), \( \mathrm{Hg}^{2+} \) decreases by \(x\), \( \mathrm{I}^- \) decreases by \(4x\), and \( \mathrm{HgI}_4^{2-} \) increases by \(x\).
• Equilibrium: Calculate equilibrium concentrations by applying the changes to the initial concentrations. This results in \([\mathrm{Hg}^{2+}] = 0.010 - x\), \([\mathrm{I}^-] = 0.78 - 4x\), and \([\mathrm{HgI}_4^{2-}] = x\).

Using the ICE table, we can set up equations to solve for the unknowns and determine equilibrium concentrations effectively.

Concentration Change

Concentration change during a chemical reaction is an important aspect of equilibrium calculations. It represents how much the concentrations of the reactants and products fluctuate as they reach equilibrium. In the given problem, we assign the variable \(x\) to represent the amount of \( \mathrm{Hg}^{2+} \) that reacts. Consequently, the concentration changes are:

• \( \mathrm{Hg}^{2+} \) decreases by \(x\).
• \( \mathrm{I}^- \) decreases by \(4x\) due to the stoichiometry of the reaction, which shows it combines with mercury in a 4:1 ratio.
• \( \mathrm{HgI}_4^{2-} \) increases by \(x\), as this is the product formed.

This setup allows us to express the equilibrium concentrations in terms of the initial concentrations and \(x\). These relationships are essential for substituting into the formation constant expression to solve for \(x\).

Simplifying Assumptions

Simplifying assumptions can make equilibrium calculations more manageable, especially when dealing with large formation constants. In our exercise, we assume that \(x\) is much smaller than the initial concentrations of \( \mathrm{Hg}^{2+} \) and \( \mathrm{I}^- \). This assumption simplifies calculations and allows us to disregard \(x\) in the denominator:

• The equation simplifies to: \[ 1.0 \times 10^{30} = \frac{x}{(0.010)(0.78)^4} \]
• This reduces the complexity of solving the equation for \(x\).
• Such assumptions should always be validated. The calculation should result in a value of \(x\) small enough to justify neglecting \(x\) compared to initial concentrations.

This tactic is especially useful in processes with overwhelming shifts toward products or reactants, allowing calculations to focus on more significant terms.