Question

Use the given molar solubilities in pure water to calculate \(K_{\text { sp }}\) for each compound. a. \(\mathrm{BaCrO}_{4}\) ; molar solubility \(=1.08 \times 10^{-5} \mathrm{M}\) b. \(\mathrm{Ag}_{2} \mathrm{SO}_{3} ;\) molar solubility \(=1.55 \times 10^{-5} \mathrm{M}\) c. \(\mathrm{Pd}(\mathrm{SCN})_{2} ;\) molar solubility \(=2.22 \times 10^{-8} \mathrm{M}\)

Short answer

\(K_{\text{sp}}\) for BaCrO4 is \(1.17 \times 10^{-10}\), for Ag2SO3 is \(1.49 \times 10^{-14}\), and for Pd(SCN)2 is \(4.36 \times 10^{-23}\).

Step-by-step solution

Write the Dissociation Equations

Write the balanced chemical equations for the dissociation of the compounds in water. For (a) \(\mathrm{BaCrO}_4(s) \rightarrow \mathrm{Ba}^{2+}(aq) + \mathrm{CrO}_4^{2-}(aq)\), (b) \(\mathrm{Ag}_2\mathrm{SO}_3(s) \rightarrow 2\mathrm{Ag}^{+}(aq) + \mathrm{SO}_3^{2-}(aq)\), and (c) \(\mathrm{Pd(SCN)}_2(s) \rightarrow \mathrm{Pd}^{2+}(aq) + 2\mathrm{SCN}^-(aq)\).

Write the Expression for \(K_{\text{sp}}\)

Express the solubility product constant \(K_{\text{sp}}\) for each compound based on the stoichiometry of the dissociation. For (a) \(K_{\text{sp}} = [\mathrm{Ba}^{2+}][\mathrm{CrO}_4^{2-}]\), (b) \(K_{\text{sp}} = [\mathrm{Ag}^{+}]^2[\mathrm{SO}_3^{2-}]\), and (c) \(K_{\text{sp}} = [\mathrm{Pd}^{2+}][\mathrm{SCN}^-]^2\).

Calculate the \(K_{\text{sp}}\) for BaCrO4

Plug in the molar solubility into the \(K_{\text{sp}}\) expression for \(\mathrm{BaCrO}_4\): \(K_{\text{sp}} = (1.08 \times 10^{-5})(1.08 \times 10^{-5}) = 1.17 \times 10^{-10}\).

Calculate the \(K_{\text{sp}}\) for Ag2SO3

Find the concentrations of ions from the molar solubility of \(\mathrm{Ag}_2\mathrm{SO}_3\): \(K_{\text{sp}} = (2 \times 1.55 \times 10^{-5})^2(1.55 \times 10^{-5}) = (3.10 \times 10^{-5})^2(1.55 \times 10^{-5}) = 1.49 \times 10^{-14}\).

Calculate the \(K_{\text{sp}}\) for Pd(SCN)2

Use the molar solubility to calculate the \(K_{\text{sp}}\) for \(\mathrm{Pd(SCN)}_2\): \(K_{\text{sp}} = (2.22 \times 10^{-8})(2 \times 2.22 \times 10^{-8})^2 = (2.22 \times 10^{-8})(4.44 \times 10^{-8})^2 = 4.36 \times 10^{-23}\).

Key concepts

Molar Solubility

Molar solubility is a measure of how much of a substance can dissolve in a given volume of solvent to reach saturation under specified conditions, such as temperature and pressure. It is expressed in moles per liter (M), indicating the number of moles of solute that can dissolve in one liter of solvent to form a saturated solution.

In the context of the exercise, the molar solubility values provided are key to finding the solubility product constants (\(K_{sp}\)) for various compounds. These constants help us understand at what concentration a compound will start precipitating out of the solution when it reaches its saturation point.

Understanding molar solubility is crucial for predicting solubility patterns in different environments which can be particularly important in fields like pharmaceuticals, where the solubility of drugs affects their efficacy and safety.

Dissociation Equations in Chemistry

Dissociation equations in chemistry describe how ionic compounds separate into their constituent ions when dissolved in water. This process is crucial for understanding reactions in solution and is represented by balanced chemical equations.

In the provided solution, step 1 involves writing the dissociation equations for each ionic compound. For instance, \(\mathrm{BaCrO}_4(s) \rightarrow \mathrm{Ba}^{2+}(aq) + \mathrm{CrO}_4^{2-}(aq)\) illustrates that barium chromate dissociates into barium ions and chromate ions in water. The stoichiometry of these equations is essential for deriving the correct solubility product constant expressions in step 2, as it determines the relationship between the concentrations of the dissolved ions.

Understanding how to write and balance dissociation equations is a foundational skill in chemistry that enables students to predict the products of reactions and calculate equilibrium constants accurately.

Chemistry Equilibrium

Chemistry equilibrium refers to a state where the concentrations of reactants and products in a reversible chemical reaction remain constant over time, as the rates of the forward and reverse reactions are equal. Equilibrium constants, such as the solubility product constant (\(K_{sp}\)), give us insight into the position of equilibrium and the extent to which a reaction proceeds.

The solubility product constant is a special type of equilibrium constant that applies to saturated solutions of ionic compounds. It is a numerical value that represents the product of the concentrations of the ions of a compound raised to the power of their coefficients from the balanced equation when equilibrium is established between the solid and its dissolved ions.

In the exercise, once the dissociation equations are written and the expressions for the solubility product constant are determined, students can use the solubility values to calculate the respective \(K_{sp}\) for each compound by plugging them into the equilibrium expression and evaluating. This practice highlights the direct connection between solubility, dissociation, and equilibrium in chemistry.