Problem 50 - Chemistry: An Atoms-Focused Approach

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 barium sulfate (BaSO4) in water at \(25^{\circ} \mathrm{C}\) is approximately \(3.29 \times 10^{-6}\) moles per liter.

Step 1: Write the balanced chemical equation.

First, we need to write the balanced chemical equation for the dissolution of barium sulfate in water: \[ \mathrm{BaSO_4 \rightleftharpoons Ba^{2+} + SO_4^{2-}} \]

Step 2: Set up the solubility equation.

To calculate the solubility of BaSO4, we need to set up an equation based on the Ksp expression. For the dissolution of BaSO4, the Ksp expression is: \[ K_{sp} = [\mathrm{Ba^{2+}}] [\mathrm{SO_4^{2-}}] \] Let the solubility of BaSO4 in moles per liter be S. Since there is a 1:1 stoichiometric ratio between BaSO4 and the ions produced, at equilibrium, the molar concentrations of \(\mathrm{Ba^{2+}}\) and \(\mathrm{SO_4^{2-}}\) will both be "S". Now, we can substitute these values in the Ksp expression: \[ K_{sp} = S^2 \]

Step 3: Solve for S using the given Ksp value.

Now, we can solve for the solubility "S". Substitute the given value of Ksp into the equation: \[ S^2 = 1.08 \times 10^{-10} \] Take the square root of both sides to solve for S: \[ S = \sqrt{1.08 \times 10^{-10}} \]

Step 4: Calculate the solubility.

Finally, use a calculator to compute the value of S (solubility): \[ S \approx 3.29 \times 10^{-6} \] Thus, the solubility of BaSO4 in water at \(25^{\circ} \mathrm{C}\) is approximately \(3.29 \times 10^{-6}\) moles per liter.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating concept in chemistry that describes the state at which a chemical reaction reaches a balance between its forward and reverse processes. Imagine it like a seesaw with children on both sides. At equilibrium, no single child is going up or down—things remain balanced. In the context of our problem, the dissolution of barium sulfate ( \( \mathrm{BaSO_4} \)) into barium ions and sulfate ions is a process that reaches equilibrium.

• When the forward rate of dissolution (producing \( \mathrm{Ba^{2+}} \) and \( \mathrm{SO_4^{2-}} \)) equals the reverse rate (recombining to form \( \mathrm{BaSO_4} \)), equilibrium is achieved.
• No net change is observed in the concentration of reactants and products at this point.

Understanding this balanced state is essential for solving problems related to solubility, like calculating the solubility product constant (\( K_{sp} \)).

Ksp Calculation

The solubility product constant, or \( K_{sp} \), is a unique equilibrium constant representing the degree to which a compound can dissolve in a solution. For barium sulfate (\( \mathrm{BaSO_4} \)), we explore the equation:

• \[ K_{sp} = [\mathrm{Ba^{2+}}][\mathrm{SO_4^{2-}}] \]
• Each ion, \( \mathrm{Ba^{2+}} \) and \( \mathrm{SO_4^{2-}} \), originates directly from one molecule of dissociated barium sulfate due to their 1:1 stoichiometric relationship.
• At equilibrium, the concentration of both ions will be \( S \) moles per liter. Thus, \( K_{sp} = S^2 \).

Calculating the \( K_{sp} \) for an ionic compound allows you to determine its solubility, which is crucial in predicting whether a precipitate will form in a solution.

Saturated Solution

A saturated solution is one where no additional solute can dissolve at a given temperature and pressure. In our scenario, the barium sulfate (\( \mathrm{BaSO_4} \)) reaches a point where it dissolves until the solution is fully saturated.

• At saturation, the solution holds the maximum concentration of dissolved ions such as \( \mathrm{Ba^{2+}} \) and \( \mathrm{SO_4^{2-}} \).
• Any further addition of \( \mathrm{BaSO_4} \) will only restore the dynamic balance with solid undissolved barium sulfate at the bottom.
• The solubility value indicates how much \( \mathrm{BaSO_4} \) the water can contain at equilibrium—here, \( 3.29 \times 10^{-6} \) moles per liter at \( 25^\circ \mathrm{C} \).

Understanding saturation is vital as it directly impacts whether more solute can dissolve or if excess will precipitate.

Stoichiometry

Stoichiometry is like a mathematical recipe used in chemistry that specifies the proportions of substances. It helps us understand the quantitative relationships in chemical reactions. In our problem involving barium sulfate (\( \mathrm{BaSO_4} \)), stoichiometry plays a crucial role.

• The solubility reaction shows a 1:1:1 ratio between \( \mathrm{BaSO_4} \), \( \mathrm{Ba^{2+}} \), and \( \mathrm{SO_4^{2-}} \) ions.
• This ratio means that for every molecule of \( \mathrm{BaSO_4} \) that dissolves, one \( \mathrm{Ba^{2+}} \) ion and one \( \mathrm{SO_4^{2-}} \) ion enter the solution.
• Use stoichiometry to determine concentrations of ions, essential for calculating \( K_{sp} \) and understanding solution behavior in chemical processes.

By applying stoichiometric principles, you can accurately predict the outcomes of mixing chemicals, making it a vital skill in chemistry.