Problem 93 - Chemistry The Central Science

Question

Consider the reaction \(\mathrm{IO}_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}(a q) ; K_{c}=3.5 \times 10^{-2} .\) If you start with \(25.0 \mathrm{~mL}\) of a \(0.905 \mathrm{M}\) solution of \(\mathrm{NaIO}_{4}\), and then dilute it with water to \(500.0 \mathrm{~mL}\), what is the concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\) at equilibrium?

Step-by-Step Solution

Answer

The equilibrium concentration of \(\mathrm{H}_{4}\mathrm{IO}_{6}^{-}\) is approximately \(7.73 \times 10^{-4}\,\textnormal{M}\).

Step 1: 1. Calculating the initial concentration

To find the initial concentration of \(\mathrm{IO}_{4}^{-}\) after diluting the \(\mathrm{NaIO}_{4}\) solution, we will use the dilution formula: \[ C_1V_1 = C_2V_2 \] Where \(C_1\) is the initial concentration, \(V_1\) is the initial volume, \(C_2\) is the final concentration, and \(V_2\) is the final volume. In this case, we have: - \(C_1 = 0.905\,\textnormal{M}\) - \(V_1 = 25.0\,\textnormal{mL}\) - \(V_2 = 500.0\,\textnormal{mL}\) Solving for the final concentration \(C_2\), we get: \[ C_2 = \frac{C_1V_1}{V_2} = \frac{(0.905\,\textnormal{M})(25.0\,\textnormal{mL})}{500.0\,\textnormal{mL}} = 0.04525\,\textnormal{M} \] The initial concentration of \(\mathrm{IO}_{4}^{-}\) after dilution is \(0.04525\,\textnormal{M}\).

Step 2: 2. Writing the reaction quotient

The reaction quotient, \(Q_c\), is the ratio of the concentrations of products to the concentrations of reactants, each raised to the power of their stoichiometric coefficients. For the given reaction: \[ \mathrm{IO}_{4}^{-}(a q) + 2\,\mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{4}\mathrm{IO}_{6}^{-}(a q) \] The \(Q_c\) expression is: \[ Q_c = \frac{[\mathrm{H}_{4}\mathrm{IO}_{6}^{-}]}{[\mathrm{IO}_{4}^{-}]} \] Note that the concentration of water is omitted from the expression because it is a pure liquid.

Step 3: 3. Finding the equilibrium concentration

To solve for the equilibrium concentration of \(\mathrm{H}_{4}\mathrm{IO}_{6}^{-}\), we first define the changes in concentrations of \(\mathrm{IO}_{4}^{-}\) and \(\mathrm{H}_{4}\mathrm{IO}_{6}^{-}\) as \(x\). At equilibrium, the concentrations will be: - \([\mathrm{H}_{4}\mathrm{IO}_{6}^{-}]_{\textnormal{eq}} = x\) - \([\mathrm{IO}_{4}^{-}]_{\textnormal{eq}} = 0.04525 - x\) We can now replace these concentrations in our expression for \(K_c\), which is given as \(3.5 \times 10^{-2}\): \[ K_c = \frac{x}{0.04525 - x} \] Now, we can solve for x: \[ 3.5 \times 10^{-2} = \frac{x}{0.04525 - x} \] \[ x(0.04525 - x) = 3.5 \times 10^{-2} \] This equation is a quadratic equation, but since \(K_c\) is much smaller than \(1\), we can approximate it by assuming \(x \ll 0.04525\): \[ x \approx \frac{3.5 \times 10^{-2}}{0.04525} \] \[ x \approx 7.73 \times 10^{-4}\,\textnormal{M} \] Finally, the equilibrium concentration of \(\mathrm{H}_{4}\mathrm{IO}_{6}^{-}\) is approximately \(7.73 \times 10^{-4}\,\textnormal{M}\).

Key Concepts

Dilution

When you dilute a solution, you're essentially reducing the concentration of a solute by adding more solvent, typically water. This process doesn't change the amount of solute present, just its concentration. We use the dilution formula, \[ C_1V_1 = C_2V_2 \]which helps in calculating the new concentration after dilution. In this exercise, we started with a \(0.905\ ext{M}\) solution of \(\text{NaIO}_4\) and diluted it to \(500.0\ ext{mL}\), reducing its concentration to \(0.04525\ ext{M}\). A useful tip is to remember that the initial moles of solute remain unchanged during dilution.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of reactants and products that remain constant over time in a chemical equilibrium. In a reversible reaction like \(\mathrm{IO}_4^{-}+2 \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}_4\mathrm{IO}_6^{-}\), the forward and reverse reactions occur at the same rate. At equilibrium, the concentration of each species is determined by the equilibrium constant, \(K_c\). This exercise aimed to find the equilibrium concentration of \(\mathrm{H}_4\mathrm{IO}_6^{-}\), which is achieved by setting up an expression based on \(K_c\) using initial concentrations and changes due to reaction progress.

Reaction Quotient (Qc)

The reaction quotient \(Q_c\) helps determine the direction in which a reaction will proceed to reach equilibrium. It is calculated similarly to the equilibrium constant \(K_c\), but uses initial concentrations rather than equilibrium concentrations. For the given reaction:

• \[Q_c = \frac{[\mathrm{H}_4\mathrm{IO}_6^{-}]}{[\mathrm{IO}_4^{-}]}\]

The reaction quotient allows us to compare with \(K_c\):

• If \(Q_c < K_c\), the reaction moves forward.
• If \(Q_c > K_c\), the reaction reverses.
• When \(Q_c = K_c\), the system is at equilibrium.

In the context of the exercise, using \(Q_c\) helps establish starting conditions for reaching equilibrium.

Quadratic Approximation

Quadratic approximation is a technique used to simplify solving equations when a reaction involves small changes in concentration, which would otherwise require solving a quadratic equation. When \(K_c\) is small, it's safe to assume that the change in concentration \(x\) is much smaller than the initial concentration, letting us ignore \(x\) in the denominator. This simplification leads to a linear equation:

• \[x \approx \frac{K_c \times [\text{initial concentration of } \mathrm{IO}_4^{-}]}{1 + K_c}\]

In the exercise, we used this assumption to find the equilibrium concentration of \(\mathrm{H}_4\mathrm{IO}_6^{-}\) without solving a full quadratic equation, making the calculations much more straightforward.