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?

= (0.905 M * 25.0 mL) / 500.0 mL = 0.04525 M #tag_title# Step 2: Write the equilibrium expression for the given reaction#tag_content# The equilibrium expression for the reaction is given by: Kc = \([H_{4}IO_{6}^{-}]\) / \([IO_{4}^{-}][H_{2}O]^{2}\) Since the concentration of water remains constant during the reaction, we can modify the equilibrium expression as follows: K' = Kc * [H2O]^2 = \([H_{4}IO_{6}^{-}]\) / \([IO_{4}^{-}]\) where K' = 3.5 * 10^{-2}. #tag_title# Step 3: Set up an ICE table and solve for x#tag_content# Set up the ICE table: | | IO4- | + | 2H2O | ⇆ | H4IO6- | |----------|-----------|---|------------|---|----------| | Initial | 0.04525 M | | - | | 0 M | | Change | -x M | | - | | +x M | | Equil. | 0.04525-x | | - | | x M | Plug these values into the equilibrium expression: K' = x / (0.04525-x) #tag_title# Step 4: Solve for the concentration of H4IO6- at equilibrium#tag_content# Solve the equation above for x: x = K'(0.04525-x) x = 0.04525K'/(K'+1) x ≈ 1.58 * 10^{-3} M Therefore, the concentration of \([H_{4}IO_{6}^{-}]\) at equilibrium is approximately \(1.58 * 10^{-3}\) M.