A 50.0 -mL sample of \(0.00200\) \(M\) \(\mathrm{AgNO}_{3}\) is added to \(50.0 \mathrm{mL}\) of 0.0100 \(M \mathrm{NaIO}_{3} .\) What is the equilibrium concentration of \(\mathrm{Ag}^{+}\) in solution? \(\left(K_{\mathrm{sp}} \text { for } \mathrm{AgIO}_{3} \text { is } 3.0 \times 10^{-8} .\right)\)

We are given the following initial conditions: Sample volume of \(\mathrm{AgNO}_3\): 50.0 mL Molarity of \(\mathrm{AgNO}_3\): 0.00200 M Sample volume of \(\mathrm{NaIO}_3\): 50.0 mL Molarity of \(\mathrm{NaIO}_3\): 0.0100 M Calculate the initial moles of each reactant: Moles of \(\mathrm{AgNO}_3\) = Volume \(\times\) Molarity = \(50.0 \times 10^{-3} \mathrm{L} \) \(\times\) 0.00200 mol/L = 0.0001 mol Moles of \(\mathrm{NaIO}_3\) = Volume \(\times\) Molarity = \(50.0 \times 10^{-3} \mathrm{L} \) \(\times\) 0.0100 mol/L = 0.0005 mol

Write the balanced chemical equation for the reaction: \(\mathrm{AgNO}_3 + \mathrm{NAIO}_3 \rightarrow \mathrm{AgIO}_3 \downarrow + \mathrm{NaNO}_3\)

Set up an ICE (Initial, Change, Equilibrium) table to help keep track of the changes happening to the various species during the reaction: |\(\mathrm{AgNO}_3\) | \(\mathrm{NaIO}_3\) | \(\mathrm{AgIO}_3\) | \(\mathrm{NaNO}_3\) ---|---|---|---|--- Initial (M)| 0.00200| 0.0100| 0 | 0 Change (M)| -x | -x | +x | +x Equilibrium (M)| 0.00200-x | 0.0100-x | x | x As we can see in the table, we used x as a term to represent the changes in concentration during the reaction. The initial concentrations are calculated by dividing the moles found in step 1 by the total volume of the mixture: Total volume of the solution = 50.0 mL (from \(\mathrm{AgNO}_3\)) + 50.0 mL (from \(\mathrm{NaIO}_3\)) = 100 mL So, the initial concentrations are: \(\mathrm{AgNO}_3\): 0.0001 mol / 0.1 L = 0.00100 M \(\mathrm{NaIO}_3\): 0.0005 mol / 0.1 L = 0.00500 M

The solubility product constant \(K_{sp}\) expression for the reaction is: \(K_{sp} = [\mathrm{Ag}^{+}] [\mathrm{IO}_3^{-}]\) From the ICE table, we know that: \([\mathrm{Ag}^{+}] = x\) \([\mathrm{IO}_3^{-}] = 0.0100 - x\) Substituting these values into the equilibrium expression, we get: \(3.0 \times 10^{-8} = x(0.0100 - x)\)

Now we will solve the quadratic equation for x: \(3.0 \times 10^{-8} = x(0.0100 - x)\) To make the calculations easier, we can make an assumption that \(x << 0.0100\), so the equation becomes: \(3.0 \times 10^{-8} = 0.0100x\) Solve for x (equilibrium concentration of \(\mathrm{Ag}^{+}\)): \(x = \mathrm{Ag}^{+} = \frac{3.0 \times 10^{-8}}{0.0100} = 3.0 \times 10^{-6}\) Therefore, the equilibrium concentration of \(\mathrm{Ag}^{+}\) in the solution is approximately \(3.0 \times 10^{-6}\ \mathrm{M}\).