Question

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? \(\left(K_{\mathrm{sp}} \text { for } \mathrm{AgIO}_{3} \text { is } 3.0 \times 10^{-8} .\right)\)

Short answer

The equilibrium concentration of \(\mathrm{Ag}^{+}\) in the solution is approximately \(3.0 \times 10^{-6}\ \mathrm{M}\).

Step-by-step solution

Calculate initial moles of reactants

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

Identify the balanced reaction

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

Set up the ICE table

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

Set up the equilibrium expression

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: \(3.0 \times 10^{-8} = x(0.0100 - x)\)

Solve for 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 \(3.0 \times 10^{-6}\ \mathrm{M}\).

Key concepts

Ksp (Solubility Product Constant)

In chemistry, the **Ksp (Solubility Product Constant)** is pivotal when dealing with solutions and precipitates. It provides insights into the extent to which a compound can dissolve in a solution before it precipitates out.
When dealing with a reaction at equilibrium, the Ksp is the product of the molar concentrations of the ions involved, each raised to the power of its coefficient in the balanced chemical equation.
For instance, in the reaction where silver nitrate (\(\mathrm{AgNO}_3\)) reacts with sodium iodate (\(\mathrm{NaIO}_3\)), forming silver iodate (\(\mathrm{AgIO}_3\)), the Ksp expression is given as:
\[K_{sp} = [\mathrm{Ag}^+][\mathrm{IO}_3^-]\].
This expression signifies that the Ksp value, \(3.0 \times 10^{-8}\), is the product of the concentrations of silver and iodate ions at equilibrium. By comparing Ksp to the ion concentrations, you can determine whether a precipitate will form, remains at equilibrium, or if the ions will stay dissolved.

• **Ksp Expressions** are unique to each salt and depend heavily on temperature.
• In a supersaturated solution, ionic product exceeds Ksp leading to precipitation.
• In a saturated solution, ionic product equals Ksp, meaning no net change.

ICE Table

The **ICE Table** is a helpful tool used in equilibrium calculations to track the concentrations of species in a chemical reaction. ICE stands for Initial, Change, and Equilibrium, representing the three stages of a reaction's progression in terms of concentration.
Take for example the equilibrium scenario involving the reaction of \(\mathrm{AgNO}_3\) and \(\mathrm{NaIO}_3\). An ICE table helps visualize and compute how concentrations change as they reach equilibrium.
Initially, the concentrations of \(\mathrm{AgNO}_3\) and \(\mathrm{NaIO}_3\) in our 100 mL solution are \(0.00100\,\mathrm{M}\) and \(0.00500\,\mathrm{M}\) respectively, while those of the products are zero.

• **Initial Concentrations:** Derived directly from the provided or measured data.
• **Change in Concentrations:** Noted by the variable \(x\) in the table to represent the shift required to achieve equilibrium.
• **Equilibrium Concentrations:** These are calculated by adjusting the initial concentrations by the changes observed, typically represented as \(c - x\), \(0 + x\), etc.

Using an ICE table provides a structured method to set up the problem and simplify the mathematics of equilibrium calculations. This approach reduces potential errors and aids in visualizing the process.

Equilibrium Concentration

**Equilibrium Concentration** refers to the molar concentration of reactants and products in a chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the reactants and products' concentrations remain constant over time.
In the context of our example, we're particularly interested in the equilibrium concentration of \(\mathrm{Ag}^+\). The task is to calculate this via the Ksp value and the ICE table data. The equilibrium concentration aids in predicting the behavior of reactions, especially in determining if conditions favor the formation of products or reactants.
The general steps to find equilibrium concentration include:

• Utilizing initial concentrations and changes provided by the ICE table.
• Substituting these values into the **Ksp expression** and solving for the unknown.
• Making assumptions for simplification, such as neglecting \(x\) when small relative to other concentrations.

Ultimately, the computed equilibrium concentration of \(\mathrm{Ag}^+\) is \(3.0 \times 10^{-6}\,\mathrm{M}\). Understanding this concept is crucial because it reveals how concentrations stabilize, whether a system is in dynamic balance, and dictates the extent to which a compound can dissolve.