Problem 98 - Chemistry An Atoms First Approach

Question

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

Answer

In conclusion, the equilibrium concentrations are: a. \([\mathrm{HgI}_{4}^{2-}]\) ≈ 4.4 x 10⁻⁹ M b. \([\mathrm{I}^{-}]\) ≈ 5.00 M c. \([\mathrm{Hg}^{2+}]\) ≈ 0.088 M

Step 1: 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. \(K_4 = [\mathrm{HgI}_{4}^{2-}]/\left([\mathrm{HgI}_{3}^{-}][\mathrm{I}^{-}]\right)\)

Step 2: 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. \([\mathrm{I}^-]_0 = \frac{5.00\,\text{mol}}{1.0\,\text{L}} = 5.00\,\text{M}\)

Step 3: 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. \(K_4 = \frac{x}{x(5.00-x)}\)

Step 4: 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.

Step 5: 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.

Step 6: 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

Key Concepts

Equilibrium Expressions

When dealing with complex ion equilibrium, it's essential to understand how to write equilibrium expressions. These expressions help us determine the balance of reactants and products in a chemical reaction at equilibrium. For each reaction, the equilibrium constant (\(K\)) is calculated using the concentrations of the products and reactants. For example:

• For the reaction \(\mathrm{Hg}^{2+} + \mathrm{I}^{-} \rightleftharpoons \mathrm{HgI}^{+}\), the equilibrium expression is \(K_1 = [\mathrm{HgI}^{+}]/([\mathrm{Hg}^{2+}][\mathrm{I}^{-}])\).

This kind of expression shows the ratio of the concentration of the complex ion to the concentration of the ions forming it. Knowing how to construct these expressions is crucial for solving equilibrium problems efficiently.

Initial Concentrations

Initial concentrations refer to the amounts of reactants present before equilibrium is reached. In our example, we're given the moles of each substance and the volume of the solution to find these concentrations. By dividing the moles by the volume, we obtain:

• \([\mathrm{Hg}^{2+}]_0 = 0.088 \ \text{M}\)
• \([\mathrm{I}^{-}]_0 = 5.00 \ \text{M}\)

These initial concentrations set the stage for what changes occur during the establishment of equilibrium, as they are the starting values in the expressions used to solve for unknowns at equilibrium.

Equilibrium Constants

The equilibrium constant (\(K\)) determines the ratio of products to reactants at equilibrium. Large values, like \(K_1 = 1.0 \times 10^8\), indicate a strong tendency to form the product. Each unique reaction has its own \(K\) value, showcasing the stability of the formed complex ion.

In our problem, these constants are used to connect initial concentrations with the equilibrium concentrations in a series of linked reactions. They help predict how much of a substance will form, aiding in understanding how equilibrium shifts with different reaction conditions.

Approximation in Equilibrium Calculations

With large equilibrium constants, approximation is a helpful tool. When \(K\) is very high, equilibrium lies far to the right, indicating products form readily. This allows us to assume that "\(x\)", the change in concentration, is much smaller than the initial concentrations.

This simplifies calculations, as we can then approximate the change as negligible. For instance:

• If \(x = 4.4 \times 10^{-9} \ \text{M}\), this change is so minute compared to initial concentrations that it's often ignored for simplification.

Approximations thus transform complex equations into more manageable forms, letting us focus on understanding the core chemical balance without overly complicated math.