Question

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(7.2 \times 10^{-2}-M \mathrm{KIO}_{3}\) solution is \(6.0 \times 10^{-9} \mathrm{mol} / \mathrm{L} .\) Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\).

Short answer

The \(K_{sp}\) value for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) is approximately \(1.558 \times 10^{-18}\).

Step-by-step solution

Write the balanced dissolution reaction

First, we need to write the balanced dissolution reaction for the \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\): \[\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{IO}_{3}^{-}(aq)\]

Set up the expression for \(K_{sp}\)

Next, we will set up the expression for the solubility product constant, \(K_{sp}\), using the balanced dissolution reaction. For a reaction with the general form \(AxBy(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)\), the \(K_{sp}\) expression is: \[K_{sp} = [A^{y+}]^{x}[B^{x-}]^{y}\] For \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\), the \(K_{sp}\) expression will be: \[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{IO}_{3}^-]^2\]

Calculate the molar concentrations with the given solubility

We are given that the solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) equals \(6.0 \times 10^{-9}\mathrm{mol}/\mathrm{L}\) in the \(7.2 \times 10^{-2} \mathrm{M}\) \(\mathrm{KIO}_{3}\) solution. Let the molar concentration of \(\mathrm{Pb}^{2+}\) be \(s = 6.0 \times 10^{-9}\mathrm{mol}/\mathrm{L}\). Since the dissolution reaction produces two moles of \(\mathrm{IO}_{3}^-\) ions for each mole of \(\mathrm{Pb}^{2+}\), the increase in molar concentration of \(\mathrm{IO}_{3}^-\) due to \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}\) dissolution will be \(2s\). Considering the given concentration of \(\mathrm{KIO}_{3}\), the total concentration of \(\mathrm{IO}_{3}^-\) ions equals \(7.2\times 10^{-2}+ 2s\).

Substitute the values into the \(K_{sp}\) expression

Now we will use the values from Step 3 and substitute them into the \(K_{sp}\) expression: \[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{IO}_{3}^-]^2 = s \times (7.2\times10^{-2} + 2s)^2\]

Calculate the \(K_{sp}\) value

Finally, we will use the solubility value of \(s = 6.0 \times 10^{-9}\mathrm{mol}/\mathrm{L}\) to find the \(K_{sp}\) value: \[K_{sp} = (6.0 \times 10^{-9})(7.2 \times 10^{-2} + 2(6.0 \times 10^{-9}))^2\] You can use a calculator to find the final value: \[K_{sp} \approx 1.558 \times 10^{-18}\] The \(K_{sp}\) value for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) is approximately \(1.558 \times 10^{-18}\).

Key concepts

Chemical Solubility

Understanding chemical solubility is crucial for students delving into the world of solution chemistry. Solubility essentially refers to the ability of a substance, known as the solute, to dissolve in a solvent and form a homogeneous solution at a particular temperature and pressure. Solubility can be expressed quantitatively as a concentration, indicating how much solute dissolves in a certain volume of solvent.

The solubility of a compound is determined by a variety of factors, including temperature, pressure, and the chemical nature of the solute and solvent. For instance, the introduction of a common ion may lead to a decrease in solubility due to a principle known as the common ion effect. Notably, in the provided exercise, the solubility of lead(II) iodate is affected by the presence of potassium iodate in solution, illustrating this concept.

Dissolution Reaction

A dissolution reaction involves the process where a solid, liquid, or gaseous solute mixes with a solvent, resulting in a uniform dispersion at the molecular or ionic level. For ionic solids, the dissolution is typically characterized by the dissociation of the compounds into its constituent ions.

The equation representing the dissolution of lead(II) iodate in water is an excellent example, where the solid dissociates into lead(II) ions and iodate ions in the aqueous solution. This process is reversible, and when the system has reached equilibrium, the rate at which the solid dissolves equals the rate at which it re-forms, leading to a dynamic equilibrium.

Molar Concentration

Molar concentration, also known as molarity, is a measure of the concentration of a solute in a solution. It is expressed in moles per liter (mol/L) and is calculated by dividing the amount of solute (in moles) by the volume of solution (in liters).

In our example involving lead(II) iodate, we have a given solubility, which tells us the amount of solute that can dissolve in a specific volume of solvent. By knowing this value, we can calculate the molar concentrations of the ions produced during the dissolution reaction. Recognizing the stoichiometry of the reaction is key, as it influences the resulting ion concentrations from a known amount of solute.

Ksp Calculation

The Ksp, or solubility product constant, quantifies the solubility of a compound under equilibrium conditions. It is derived from the concentrations of the ions at equilibrium. The Ksp expression is unique for each sparingly soluble compound and is based on the balanced chemical equation of its dissolution.

To calculate Ksp, we must substitute the equilibrium concentrations of the ions into the Ksp expression. In situations where the concentrations are affected by other factors, such as common ions from other dissolved substances, these must be taken into account. The exercise we examined reveals how the initial ion concentration from potassium iodate influences the dissolution of lead(II) iodate and hence, its Ksp.