Question

Consider the reaction $$\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)$$ At \(25^{\circ} \mathrm{C}\), the equilibrium constant is \(K_{c}=2.4 \times 10^{-5}\) for this reaction. (a) If excess \(\operatorname{CaSO}_{4}(s)\) is mixed with water at \(25^{\circ} \mathrm{C}\) to produce a saturated solution of \(\mathrm{CaSO}_{4},\) what are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) ? (b) If the resulting solution has a volume of \(1.4 \mathrm{~L},\) what is the minimum mass of \(\operatorname{CaSO}_{4}(s)\) needed to achieve equilibrium?

Short answer

The equilibrium concentrations of both \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) are \(4.9\times 10^{-3}\, \mathrm{M}\), and the minimum mass of \(\mathrm{CaSO}_{4}(s)\) needed to achieve equilibrium in a 1.4 L solution is 0.93 g.

Step-by-step solution

Write the expression for the equilibrium constant#

We need to write the expression for the equilibrium constant \(K_c\) for the given reaction: \(\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\) According to the reaction, the formation of 1 mole of \(\mathrm{Ca}^{2+}\) ion is accompanied by the formation of 1 mole of \(\mathrm{SO}_{4}^{2-}\) ion. Therefore, we can simplify the expression for \(K_c\) as follows: \[ K_c = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}] \]

Calculate the equilibrium concentrations of the ions#

Let the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) be \(x\). According to the reaction, their concentrations are equal. We then have: \(K_c = x^2\) Given \(K_c = 2.4\times 10^{-5}\), we can now solve for \(x\): \(x^2 = 2.4\times 10^{-5}\) \(x = \sqrt{2.4\times 10^{-5}} = 4.9\times 10^{-3}\, \mathrm{M}\) Thus, the equilibrium concentrations of both \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) are \(4.9\times 10^{-3}\, \mathrm{M}\).

Calculate the minimum mass of CaSO4 needed to achieve equilibrium#

Given that the solution has a volume of 1.4 L, we can calculate the moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) at equilibrium: moles of \(\mathrm{Ca}^{2+}\) (and \(\mathrm{SO}_{4}^{2-}\)) = Molar concentration × Volume moles of \(\mathrm{Ca}^{2+}\) (and \(\mathrm{SO}_{4}^{2-}\)) = \(4.9\times 10^{-3}\, \mathrm{M}\times 1.4\, \mathrm{L} = 6.86\times 10^{-3}\, \mathrm{moles}\) Now, we can calculate the minimum mass of \(\mathrm{CaSO}_{4}\) needed using its molar mass. The molar mass of \(\mathrm{CaSO}_{4}\) can be found by adding the individual molar masses of \(\mathrm{Ca}\), \(\mathrm{S}\), and \(\mathrm{O}\): Molar mass of $\mathrm{CaSO}_{4} = 40.08 + 32.06 + 4(16.00) = 136.14\, \mathrm{g/mol}\) Finally, we can calculate the minimum mass by multiplying the moles of \(\mathrm{CaSO}_{4}\) needed by its molar mass: Minimum mass of \(\mathrm{CaSO}_{4}\) = moles × molar mass Minimum mass of \(\mathrm{CaSO}_{4}\) = \(6.86\times 10^{-3}\, \mathrm{moles} \times 136.14\, \mathrm{g/mol} = 0.93\, \mathrm{g}\) Therefore, the minimum mass of \(\mathrm{CaSO}_{4}(s)\) needed to achieve equilibrium in a 1.4 L solution is 0.93 g.

Key concepts

Equilibrium Constant

In the world of chemistry, the equilibrium constant, denoted as \( K_c \), plays a vital role in understanding how chemical reactions reach a point of balance. When a chemical reaction reaches equilibrium, the rate at which the reactants convert to products is equal to the rate at which products convert back into reactants. This is where the concept of \( K_c \) comes into play.

For a simple reaction such as \( \mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq)+\mathrm{SO}_{4}^{2-}(aq) \), the equilibrium constant expression can be written based solely on the concentrations of the products, as solids like \( \mathrm{CaSO}_{4}(s) \) do not appear in the \( K_c \) expression. Hence, we have:
\[ K_c = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}] \]

This expression tells us that at equilibrium, the product of the concentrations of \( \mathrm{Ca}^{2+} \) and \( \mathrm{SO}_{4}^{2-} \) is constant at a given temperature. The equilibrium constant is a crucial indicator of the extent to which the reactants are converted to products at equilibrium. In our case, with \( K_c = 2.4 \times 10^{-5} \), it suggests that the concentrations of ions in the solution are relatively small.

Saturated Solution

A saturated solution occurs when the maximum amount of a solute has been dissolved in a solvent at a particular temperature, beyond which no more solute will dissolve. In the context of chemical equilibrium, saturation plays an important role.

For the reaction \( \mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq)+\mathrm{SO}_{4}^{2-}(aq) \), a saturated solution is achieved when there's an equilibrium between the undissolved \( \mathrm{CaSO}_{4} \) and the dissolved ions \( \mathrm{Ca}^{2+} \) and \( \mathrm{SO}_{4}^{2-} \). At this point, the solution contains the maximum concentration of dissolved ions that corresponds to the equilibrium constant \( K_c \) for the reaction.

When you mix excess \( \mathrm{CaSO}_{4}(s) \) with water, the solid will continue to dissolve until the solution becomes saturated, and then stop. This saturation is represented by the constant concentrations of \( \mathrm{Ca}^{2+} \) and \( \mathrm{SO}_{4}^{2-} \) ions at equilibrium. The balance between the dissolved ions and the undissolved solid determines the dynamic state of saturation.

Calcium Sulfate Solubility

Calcium sulfate, \( \mathrm{CaSO}_{4} \), is a compound that demonstrates limited solubility in water, meaning that only a small amount will dissolve before the solution becomes saturated. Understanding its solubility requires exploring the chemical equilibrium of its dissolution.

In the reaction \( \mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq)+\mathrm{SO}_{4}^{2-}(aq) \), solubility is represented by the concentrations of the dissolved ions. As calculated, both \( \mathrm{Ca}^{2+} \) and \( \mathrm{SO}_{4}^{2-} \) ions reach a concentration of \( 4.9 \times 10^{-3} \, \mathrm{M} \) when the solution is saturated at \( 25^{\circ} \mathrm{C} \).

The solubility of \( \mathrm{CaSO}_{4} \) is important for various applications, including its use in construction materials like plaster and its role in soil treatments. In aqueous systems, the solubility will dictate how much calcium sulfate can remain dissolved before precipitating as a solid. This balance ensures that reactions stay at their equilibrium concentrations under constant conditions.