Problem 50 - Chemistry The Science in Context

Question

Algae in the genus Closterium contain structures built from barium sulfate (barite). Calculate the solubility in moles per liter of \(\mathrm{BaSO}_{4}\) in water at \(25^{\circ} \mathrm{C}\) given that \(K_{\mathrm{sp}}=1.08 \times 10^{-10}\).

Step-by-Step Solution

Answer

Answer: The solubility of BaSO4 in water at 25°C is 3.29 x 10^-6 mol L^-1.

Step 1: Write down the dissolution and equilibrium expressions of BaSO4

The dissolution reaction of barium sulfate (BaSO4) can be written as: \(\mathrm{BaSO}_{4}(s) \leftrightarrows \mathrm{Ba}^{2+}(aq) + \mathrm{SO}_{4}^{2-}(aq)\) The equilibrium expression for this reaction is given by the solubility product constant (Ksp): \(K_{\text{sp}} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]\)

Step 2: Define the solubility of BaSO4

Let the solubility of BaSO4 in water be represented by the variable "s". Then the concentration of Ba2+ and SO42- ions in the solution will both be "s".

Step 3: Substitute solubility values in the equilibrium expression

Now we substitute the solubility values into the equilibrium expression: \(K_{\text{sp}} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = (s)(s) = s^2\)

Step 4: Solve for solubility "s"

We are given the value of \(K_{\text{sp}} = 1.08 \times 10^{-10}\). Use the equilibrium expression from Step 3 to solve for solubility (s): \(1.08 \times 10^{-10} = s^2\) Taking the square root of both sides: \(s = \sqrt{1.08 \times 10^{-10}}\) Calculating the value for "s": \(s = 3.29 \times 10^{-6}\)

Step 5: Express the solubility in moles per liter

The solubility (s) in moles per liter of BaSO4 in water at 25°C is: \(s = 3.29 \times 10^{-6}\; \mathrm{mol\; L^{-1}}\)

Key Concepts

Barium Sulfate

Barium sulfate, represented by the chemical formula \( \text{BaSO}_4 \), is a white crystalline solid commonly found in nature as the mineral barite. It has a variety of practical applications, such as in radiology where it is used as a contrast agent in medical imaging due to its opacity to X-rays. This compound is sparingly soluble in water, meaning it dissolves very little in water. When barium sulfate is added to water, it partially dissociates into barium ions \( \text{Ba}^{2+} \) and sulfate ions \( \text{SO}_4^{2-} \). However, because it is not very soluble, only a small amount dissociates at equilibrium. In chemical terms, this equilibrium can be expressed with an equilibrium expression, which is key to understanding how its solubility behavior is quantified.

Ksp

The term \( K_{\text{sp}} \) stands for the solubility product constant, a valuable concept in chemistry that helps us understand the solubility of sparingly soluble compounds like barium sulfate. The \( K_{\text{sp}} \) is unique to each compound and depends on temperature. This constant reflects the level at which the solid solute can dissolve in the solvent to reach a state of equilibrium.Barium sulfate has a very low \( K_{\text{sp}} \) value, which means that it does not dissolve well in water. In this particular exercise, the provided \( K_{\text{sp}} \) of \( 1.08 \times 10^{-10} \) indicates just how limited the soluble ions in the solution are at equilibrium. Because the \( K_{\text{sp}} \) is a constant value for a given temperature, it is used as a reference to calculate the maximum amount of substance that can dissolve in water before the system reaches saturation.

Equilibrium Expression

The equilibrium expression is an equation derived from the dissolution of a compound at equilibrium. For barium sulfate, this expression is represented as:\[ K_{\text{sp}} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]This equation is essential in chemistry as it provides a mathematical way to express the concentration levels of ions when a compound is dissolved to its maximum solubility.In our exercise, both the barium and sulfate ions produced by the dissolution of barium sulfate have the same concentration, represented by "s" for solubility. Hence, when substituting solubility values, the equilibrium expression becomes:\[ K_{\text{sp}} = s^2 \]This expression indicates that the solubility can be calculated by taking the square root of the \( K_{\text{sp}} \). This is crucial for determining how many moles of barium sulfate can dissolve in a liter of water, providing insight into its solubility characteristics.

Solubility Product Constant

The solubility product constant, \( K_{\text{sp}} \), specifically applies to sparingly soluble salts. This constant is manipulated using stoichiometric relations to find the solubility of the salt in question. By using the \( K_{\text{sp}} \), we can understand how much of a substance can feasibly dissolve before the solution becomes saturated, and additional substance settles as a solid.For barium sulfate, using the equation \( K_{\text{sp}} = s^2 \) and a given \( K_{\text{sp}} = 1.08 \times 10^{-10} \), we can calculate the solubility, "s", by solving:\[ s = \sqrt{1.08 \times 10^{-10}} \]Executing this calculation yields a solubility of approximately \( 3.29 \times 10^{-6} \text{ mol/L} \). This calculated solubility in moles per liter tells us how much barium sulfate can dissociate into ions and remain in solution, a fundamental property utilized in various scientific and industrial applications where precise measured solubility is important.