Problem 95 - Chemistry The Central Science In Si Units

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 concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\) at equilibrium is approximately \(1.53 \times 10^{-3} \mathrm{M}\).

Step 1: Calculate Initial Concentration After Dilution

First, calculate the concentration of \(\mathrm{IO}_{4}^{-}\) after dilution. Initially, you have \(25.0\mathrm{~mL}\) of \(0.905\mathrm{M}\) \(\mathrm{NaIO}_{4}\), which when diluted to \(500.0\mathrm{~mL}\) results in:\[\text{Initial concentration} = \left( \frac{\text{initial volume} \times \text{initial concentration}}{\text{final volume}} \right) = \frac{25.0 \times 0.905}{500.0} = 0.04525 \mathrm{M}\]

Step 2: Write the Equilibrium Expression

For the given equilibrium reaction:\[ \mathrm{IO}_{4}^{-} + 2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{4} \mathrm{IO}_{6}^{-} \]The equilibrium constant expression is:\[ K_c = \frac{[\mathrm{H}_{4} \mathrm{IO}_{6}^{-}]}{[\mathrm{IO}_{4}^{-}]} = 3.5 \times 10^{-2} \]

Step 3: Set Up an ICE Table

Using an ICE table (Initial, Change, Equilibrium) for concentrations:\[\begin{array}{c|c|c|c} & [\mathrm{IO}_{4}^{-}] & [\mathrm{H}_{4} \mathrm{IO}_{6}^{-}] \hline\text{Initial (M)} & 0.04525 & 0 \\text{Change (M)} & -x & +x \\text{Equilibrium (M)} & 0.04525 - x & x \\end{array}\]

Step 4: Solve for \(x\) Using the Equilibrium Expression

Substitute the equilibrium concentrations into the expression for \(K_c\):\[3.5 \times 10^{-2} = \frac{x}{0.04525 - x}\]Solve for \(x\):\[x = 3.5 \times 10^{-2} \times (0.04525 - x)\]Expand and rearrange the equation:\[x = 1.58375 \times 10^{-3} - 3.5 \times 10^{-2}x\]Combine\(x\) terms:\[x + 3.5 \times 10^{-2}x = 1.58375 \times 10^{-3}\]\[(1 + 0.035)x = 1.58375 \times 10^{-3}\]\[1.035x = 1.58375 \times 10^{-3}\]\[x = \frac{1.58375 \times 10^{-3}}{1.035} = 1.53 \times 10^{-3} \mathrm{M}\]

Step 5: Interpret the Result

The concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\) at equilibrium is determined by the change in concentration \(x\). Therefore, the equilibrium concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\) is approximately \(1.53 \times 10^{-3} \mathrm{M}\).

Key Concepts

Equilibrium Constant (Kc)

Chemical Equilibrium refers to a state in a reversible reaction where the rates of the forward and reverse reactions are equal. This means that the concentrations of reactants and products remain constant over time. In this context, the Equilibrium Constant, denoted as \(K_c\), is an expression that relates the concentrations of the products and reactants at equilibrium for a given chemical reaction. For the reaction \(\mathrm{IO}_{4}^{-} + 2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{4} \mathrm{IO}_{6}^{-}\), the \(K_c\) is given as \(3.5 \times 10^{-2}\). This value informs us about the extent to which reactants are converted to products.

• A larger \(K_c\) (>1) indicates that the equilibrium position favors products.
• A smaller \(K_c\) (<1) suggests that the equilibrium position favors reactants.

For this reaction, since \(K_c\) is less than 1, reactants are favored. Hence, only a small amount of \(\mathrm{IO}_{4}^{-}\) converts to \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\) at equilibrium. Understanding \(K_c\) is crucial for predicting the outcomes of reactions and determining the concentrations of substances involved.

Dilution and Concentration Calculations

Dilution is the process of reducing the concentration of a solute in a solution, usually by adding more solvent. It's particularly important when preparing solutions for experiments to obtain the desired concentration. The formula used in dilution calculation is: \[\text{Initial Volume} \times \text{Initial Concentration} = \text{Final Volume} \times \text{Final Concentration} \]In our example, we started with \(25.0 \mathrm{~mL}\) of 0.905 M \(\mathrm{NaIO}_{4}\), and diluted this to a final volume of \(500.0 \mathrm{~mL}\). By plugging into the formula, \[\text{Final concentration} = \left( \frac{25.0 \times 0.905}{500.0} \right) = 0.04525 \mathrm{M}\]This process decreased the concentration of \(\mathrm{IO}_{4}^{-}\) in the solution. Key points about dilution:

• Dilution changes concentration, but the total number of moles stays the same.
• A greater dilution results in a lower concentration of solute.

It's a critical concept for accurately preparing chemical solutions, ensuring reactions proceed under controlled conditions.

ICE Table Method

The ICE Table is a systematic way to organize the changes in concentration of species in an equilibrium reaction. ICE stands for Initial, Change, and Equilibrium, which represent the concentration at different stages of the reaction. Using the ICE table helps simplify calculations related to equilibrium concentrations.For the reaction \( \mathrm{IO}_{4}^{-} + 2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{4} \mathrm{IO}_{6}^{-} \), the ICE table is set up by:- **Initial Concentrations**: Before any reaction occurs. \([\mathrm{IO}_{4}^{-}] = 0.04525 \mathrm{M}\), \([\mathrm{H}_{4} \mathrm{IO}_{6}^{-}] = 0 \mathrm{M}\)- **Change in Concentrations**: The change that occurs as the system moves to equilibrium. If \(x\) is the change, then \([\mathrm{IO}_{4}^{-}]\) decreases by \(x\) and \([\mathrm{H}_{4} \mathrm{IO}_{6}^{-}]\) increases by \(x\).- **Equilibrium Concentrations**: The concentrations at equilibrium are \([\mathrm{IO}_{4}^{-}] = 0.04525 - x\) and \([\mathrm{H}_{4} \mathrm{IO}_{6}^{-}] = x\).An ICE table simplifies complex reactions by organizing data in a clear way, allowing for more accurate calculations of equilibrium concentrations. It’s especially useful to predict not just what is present at equilibrium, but also how conditions affect the shift between reactants and products.