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

Use the dilution formula: \(c_1V_1 = c_2V_2\): Here, \(c_1\) = initial concentration of \(\mathrm{NaIO}_{4}\) = 0.905 M \(V_1\) = initial volume of \(\mathrm{NaIO}_{4}\) solution = 25.0 mL \(V_2\) = final volume of solution after dilution = 500.0 mL We need to find \(c_2\), the concentration of \(\mathrm{IO}_{4}^{-}\) after dilution. \[c_2 = \frac{c_1V_1}{V_2}\] Plug in the values to calculate \(c_2\): \[c_2 = \frac{(0.905)(25.0)}{500.0}\] \[c_2 = 0.0452 \, \mathrm{M}\]

Since we need to find out the concentrations of species at equilibrium, we will use an ICE table to keep track of the changes in concentrations throughout the reaction. Assume that the change in \(\mathrm{IO}_{4}^{-}\) concentration is denoted as \(x\). Recall that the stoichiometry of the balanced equation is important for the ICE table. | | \(\mathrm{IO}_{4}^{−}\) (aq) | \(2 \, \mathrm{H}_{2}\mathrm{O}\) (l) | \(\mathrm{H}_{4}\mathrm{IO}_{6}^{-}\) (aq) | |-------|------------------|-----------------|--------------------| |Initial| 0.0452 M | not needed | 0 | | Change| -\(10x\) | -- | \(x\) | |Equilibrium| 0.0452-10x | not needed | \(x\) | Since water doesn't appear in the expression for \(K_c\), its concentration is not needed in the ICE table. \(\DeclareMathOperator{\M}{\:M}\) \(\)

In this reaction, the given equilibrium constant is: \[K_c = 3.5 \times 10^{-2}\] Now write the expression for \(K_c\) based on the products and reactants of the reaction: \[K_c = \frac{[\mathrm{H}_{4}\mathrm{IO}_{6}^{-}]}{[\mathrm{IO}_{4}^{-}]^{10}}\] Substitute the equilibrium concentrations from the ICE table into the equation: \[K_c = \frac{x}{(0.0452 - 10x)^{10}}\] Now, plug in the value of \(K_c\) and solve for \(x\): \[3.5 \times 10^{-2} = \frac{x}{(0.0452 - 10x)^{10}}\]

In this step, we may assume that \(10x \ll 0.0452\), as the number \(10\) has a significant impact raised to high powers. This means that we can approximate \(0.0452 - 10x ≈ 0.0452\), so the equation then becomes: \[3.5 \times 10^{-2} = \frac{x}{(0.0452 )^{10}}\] \[x = 3.5 \times 10^{-2} \cdot (0.0452 )^{10}\] \[x ≈ 5.4 \times 10^{-12}\] Since \(x\) represents the equilibrium concentration of \(\mathrm{H}_{4}\mathrm{IO}_{6}^{-}\): \[[\mathrm{H}_{4}\mathrm{IO}_{6}^{-}]_{eq} ≈ 5.4 \times 10^{-12} \M\] At equilibrium, the concentration of \(\mathrm{H}_{4}\mathrm{IO}_{6}^{-}\) is approximately \(5.4 \times 10^{-12} \, \mathrm{M}\).