Problem 93 - Chemistry: The Central Science

Question

Consider the reaction \(\mathrm{IO}_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{4} \mathrm{IO}_{6}{\underline{\phantom{xx}}}^{-}(a q)\); \(K_{c}=3.5 \times 10^{-2}\). 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 \(\mathrm{H}_{4} \mathrm{IO}_{6}{\underline{\phantom{xx}}}^{-}\)at equilibrium?

Step-by-Step Solution

Answer

The equilibrium concentration of H₄IO₆⁻ is approximately \(1.52 \times 10^{-3}\) M.

Step 1: Calculate the initial concentration of IO₄^- after dilution.

To find the initial concentration of IO₄^- in the final solution after dilution, we can use the equation: C1V1 = C2V2 Where: C1 = initial concentration of IO₄^- V1 = initial volume of NaIO₄ solution C2 = final concentration of IO₄^- V2 = final volume of NaIO₄ solution Plug in the given values to find C2: (0.905 M)(25.0 mL) = C2(500.0 mL) C2 = (0.905 M * 25.0 mL) / 500.0 mL C2 = 0.04525 M The concentration of IO₄^- after dilution is 0.04525 M.

Step 2: Write down the balanced chemical equation and set up the ICE table.

Our balanced chemical equation is: IO₄^- (aq) + 2 H₂O (l) ⇌ H₄IO₆^- (aq) Set up the ICE table: | Species | Initial (M) | Change (M) | Equilibrium (M) | |---------|-------------|------------|-----------------| | IO₄^- | 0.04525 | -x | 0.04525 - x | | H₂O | | | | | H₄IO₆^- | 0 | +x | x | Here, x represents the change in concentration of IO₄^- and H₄IO₆^-. Notice that we have not included H₂O in the ICE table, as its concentration in the liquid form does not affect the reaction's equilibrium constant.

Step 3: Write the expression for Kc and plug in the equilibrium concentrations.

The expression for Kc is as follows: \[K_c = \frac{[H₄IO₆^-]}{[IO₄^-]}\] Given the equilibrium constant Kc = 3.5 × 10^(-2): \[3.5 \times 10^{-2} = \frac{x}{0.04525 - x}\]

Step 4: Solve for x, the equilibrium concentration of H₄IO₆^-.

We need to solve the equation for x: \[3.5 \times 10^{-2} = \frac{x}{0.04525 - x}\] Let us first multiply both sides by (0.04525 - x) to simplify the equation: \(x = 3.5 \times 10^{-2} (0.04525 - x)\) Now, let's expand and rearrange the equation: \(x = (3.5 \times 10^{-2})(0.04525) - (3.5 \times 10^{-2})x\) Combine x terms: \(x + (3.5 \times 10^{-2})x = (3.5 \times 10^{-2})(0.04525)\) Factor out x and solve: \(x(1 + 3.5 \times 10^{-2}) = (3.5 \times 10^{-2})(0.04525)\) Divide by (1 + 3.5 × 10^(-2)): \(x = \frac{(3.5 \times 10^{-2})(0.04525)}{1 + 3.5 \times 10^{-2}}\) Calculate the value of x: \(x \approx 1.52 \times 10^{-3}\) The equilibrium concentration of H₄IO₆^- is approximately 1.52 × 10^(-3) M.

Key Concepts

Equilibrium Constant

The equilibrium constant (\(K_c\)) is a ratio that provides a quantitative measure of the position of equilibrium for a reversible chemical reaction at a given temperature. For the reaction where reactants A and B form products C and D, the equilibrium constant expression is given as \( K_c = \frac{[C][D]}{[A][B]} \) where the concentrations are those at equilibrium, and the brackets denote molarity. The larger the value of the equilibrium constant, the more product there is relative to reactant at equilibrium. In our textbook exercise, the value of \( K_c \) for the reaction \( IO_4^- + 2 H_2O \rightleftharpoons H_4IO_6^- \) is \( 3.5 \times 10^{-2} \), which indicates that at equilibrium, the concentration of reactants is higher than that of the products.

ICE Table

The ICE table, which stands for Initial, Change, and Equilibrium, is a tool used to organize data and find the equilibrium concentrations of reactants and products in a chemical reaction. It's a systematic way to apply stoichiometry to reactions at equilibrium. In our problem, we created an ICE table for the reaction \( IO_4^- + 2 H_2O \rightleftharpoons H_4IO_6^- \), identifying the initial concentrations, the changes during the reaction (expressed as '-x' for reactants and '+x' for products), and the equilibrium concentrations. Since the solvent water is a pure liquid its concentration is not included in the ICE table as it remains essentially constant during the reaction.

Dilution Calculations

Dilution calculations are used to determine the new concentration of a solution after it has been diluted with additional solvent. The core principle used is that the amount of solute remains the same before and after dilution. The equation \( C_1V_1 = C_2V_2 \) encapsulates this idea, with \( C_1 \) and \( V_1 \) representing the initial concentration and volume, and \( C_2 \) and \( V_2 \) the final concentration and volume, respectively. In the context of our exercise, we used this equation to find the concentration of \( IO_4^- \) after diluting \( NaIO_4 \) solution from 25.0 mL to 500.0 mL.

Chemical Reaction Equilibrium

Chemical reaction equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. This dynamic state can be described by the equilibrium constant (\(K_c\)), which remains constant at a given temperature. An important point to note, as demonstrated by the textbook exercise, is that pure solids and liquids do not appear in the expression for the equilibrium constant, as their concentrations are unchanged.

Solving for Equilibrium

Solving for equilibrium involves finding the concentrations of all species at equilibrium. From the ICE table, we can express these concentrations in terms of a variable \( x \). Then, we use the equilibrium constant expression to write an equation in \( x \) and solve for it. As seen in the given problem, after setting up the expression for \( K_c \) and inserting our \( x \) values from the ICE table, we end up with a mathematical equation. Solving this equation gives us the value of \( x \) that corresponds to the concentration of \( H_4IO_6^- \) at equilibrium. It's sometimes necessary to solve a quadratic equation, but in our problem, we could simplify the expression to isolate \( x \) and find the answer.