Question

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

Short answer

The equilibrium concentration of \(\mathrm{Ag}^{+}\) ions in the solution is \(4.0 \times 10^{-5} \mathrm{M}\).

Step-by-step solution

Determine the Initial Moles of Each Ion

Calculate the initial moles of \(\mathrm{AgNO}_{3}\) and \(\mathrm{NaIO}_{3}\) using the given volumes and concentrations. Moles of \(\mathrm{AgNO}_{3}\): \(0.00200 \mathrm{M} * 0.0500L = 1.00 \times 10^{-4} \mathrm{mol}\) Moles of \(\mathrm{NaIO}_{3}\): \(0.0100 \mathrm{M} * 0.0500L = 5.00 \times 10^{-4} \mathrm{mol}\)

Determine the Total Volume of the Solution

Add the volumes of the solutions to find the total volume: Total Volume: \(50.0 \mathrm{mL} + 50.0 \mathrm{mL} = 100.0 \mathrm{mL} = 0.100L\)

Set Up the Equilibrium Expression and ICE Table

Write the balanced equation for the formation of the precipitate, \(\mathrm{AgIO}_{3}\), from its ions, and write the equilibrium expression involving the solubility product constant, \(K_{\mathrm{sp}}\). Balanced Equation: \(\mathrm{Ag^{+} + IO^{-}_{3} \rightleftharpoons AgIO}_{3 \downarrow}\) Equilibrium Expression: \(K_{\mathrm{sp}} = \left[\mathrm{Ag}^{+}\right] \left[\mathrm{IO}_{3}^{-}\right]\) Next, set up an ICE table showing how the concentrations of the ions shift at equilibrium. | | Initial | Change | Equilibrium | | ----------------- |--------:|----------:|-----------:| | \(\mathrm{Ag}^{+}\) | \(1.0 \times 10^{-4}\mathrm{M}\) | \( -x\mathrm{M}\) | \((1.0 \times 10^{-4} - x)\mathrm{M}\) | | \(\mathrm{IO}_{3}^{-}\) | \(5.0 \times 10^{-4}\mathrm{M}\) | \( -x\mathrm{M}\) | \((5.0 \times 10^{-4} - x)\mathrm{M}\) |

Substitute the ICE Table Values into the Equilibrium Expression and Solve for x

Replace the ion concentrations in the equilibrium expression with the correspondings terms from the ICE table, and use the provided value of \(K_{\mathrm{sp}}\) to solve for x. \(3.0 \times 10^{-8} = (1.0 \times 10^{-4} - x)(5.0 \times 10^{-4} - x)\) Since \(K_{\mathrm{sp}}\) is small, we can assume that \(x\) is much smaller than the initial concentrations of both ions. Therefore, we can simplify the expression: \(3.0 \times 10^{-8} \approx (1.0 \times 10^{-4})(5.0 \times 10^{-4})\) Now, solve for x: \(x = \frac{3.0 \times 10^{-8}}{(1.0 \times 10^{-4})(5.0 \times 10^{-4})} = 6.0 \times 10^{-5} \mathrm{M}\)

Determine the Equilibrium Concentration of Ag+

The value of x represents the change in concentration of \(\mathrm{Ag}^{+}\) and \(\mathrm{IO}_{3}^{-}\) ions at equilibrium. To find the equilibrium concentration of \(\mathrm{Ag}^{+}\), subtract x from its initial concentration. Equilibrium Concentration of \(\mathrm{Ag}^{+}\): \((1.0 \times 10^{-4} - 6.0 \times 10^{-5}) \mathrm{M} = 4.0 \times 10^{-5} \mathrm{M}\) The equilibrium concentration of \(\mathrm{Ag}^{+}\) ions in the solution is \(4.0 \times 10^{-5} \mathrm{M}\).

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, often abbreviated as Ksp, is critical for understanding the solubility of ionic compounds in a solution. Ksp represents the maximum product of the concentrations of the ions in a solution at which the solution is at equilibrium with a solid phase of the compound. When the product of ion concentrations exceeds the Ksp value, the excess ions will form a precipitate, thus reducing their concentrations back to the equilibrium level.

For instance, in our exercise, the Ksp for silver iodate (AgIO3) is given as 3.0 x 10^-8. This value indicates the equilibrium point at which the solid silver iodate is in balance with its ions (Ag+ and IO3-) in the solution. If the ionic product of the concentrations of Ag+ and IO3- in the solution exceeds this Ksp, it signifies that more AgIO3 will precipitate until the ionic product matches the Ksp again.

ICE Table Method

The ICE table method—an acronym for Initial, Change, and Equilibrium—is a systematic approach to solve equilibrium problems in chemistry. This method relies on setting up a table to track the concentrations of reactants and products through the different stages of the reaction.

To implement the ICE method, we record the initial concentrations (I) of the reacting species, the changes (C) that occur as the system reaches equilibrium, and finally, the equilibrium concentrations (E). By applying the equilibrium expression for a reaction to the values obtained from the ICE table, we can solve for unknowns and find the equilibrium concentrations. In our problem, we use the ICE table to understand how the initial moles of Ag+ and IO3- ions change as the reaction proceeds to form AgIO3, allowing us to calculate the equilibrium concentration of Ag+.

Initial Moles of Reactants

Understanding the initial moles of reactants is essential for calculating equilibrium concentrations. To find these values, we multiply the concentration of each reactant by the volume of its solution. In our textbook example, we calculated the initial moles of AgNO3 and NaIO3 by using their given concentrations and the volume of the solutions in which they were dissolved.

These initial mole values serve as the starting point for the ICE table and are crucial for predicting how the reaction proceeds towards equilibrium. They give us insight into how much of each reactant is present before the reaction starts and enable us to track the progression of the reaction accurately.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, leading to no net change in the concentration of reactants and products. However, this does not mean that the reactants and products are at equal concentrations—rather, they have reached a stable ratio as defined by the equilibrium constant (K or Ksp for solubility equilibria).

In the context of our problem, the chemical equilibrium is approached as the solid AgIO3 dissolves and precipitates in the solution until the concentrations of Ag+ and IO3- ions no longer change. At this point, the system is in dynamic equilibrium and the concentrations of the ions in the solution are governed by the solubility product constant Ksp.