Question

What are \(\left[\mathrm{Sr}^{2+}\right]\) and \(\left[\mathrm{SO}_{4}^{2-}\right]\) in a saturated solution of \(\mathrm{SrSO}_{4}(\mathrm{~s}) ?\) The \(K_{\mathrm{sp}}\) of \(\mathrm{SrSO}_{4}(\mathrm{~s})\) is \(3.8 \times\) \(10^{-4}\)

Short answer

[Sr^{2+}] and [SO_4^{2-}] are both 0.0195 M in a saturated solution of SrSO_4.

Step-by-step solution

Understanding the Problem

We need to find the concentrations of \( \mathrm{Sr}^{2+} \) and \( \mathrm{SO}_{4}^{2-} \) in a saturated solution for \( \mathrm{SrSO}_4 \). The given soluble product constant \( K_{\mathrm{sp}} \) determines the extent to which \( \mathrm{SrSO}_4 \) dissolves.

Write the Dissociation Equation

In aqueous solution, \( \mathrm{SrSO}_4 \) dissociates into its ions: \[ \mathrm{SrSO}_4 (s) \rightleftharpoons \mathrm{Sr}^{2+} (aq) + \mathrm{SO}_4^{2-} (aq) \] Each molecule of \( \mathrm{SrSO}_4 \) produces one \( \mathrm{Sr}^{2+} \) ion and one \( \mathrm{SO}_4^{2-} \) ion.

Apply the Solubility Product Expression

The solubility product \( K_{\mathrm{sp}} \) is given by: \[ K_{\mathrm{sp}} = [\mathrm{Sr}^{2+}][\mathrm{SO}_4^{2-}] \] We know \( K_{\mathrm{sp}} = 3.8 \times 10^{-4} \). Both ion concentrations equal \( s \) in a saturated solution.

Set Up the Equation

Since the dissociation of \( \mathrm{SrSO}_4 \) yields equal concentrations of \( \mathrm{Sr}^{2+} \) and \( \mathrm{SO}_4^{2-} \), let \( [\mathrm{Sr}^{2+}] = s \) and \( [\mathrm{SO}_4^{2-}] = s \). So, \[ K_{\mathrm{sp}} = s^2 \] Thus, we solve: \[ s^2 = 3.8 \times 10^{-4} \]

Solve for s

We take the square root of both sides to get the concentration \( s \): \[ s = \sqrt{3.8 \times 10^{-4}} \] Calculating gives us: \[ s \approx 0.0195 \]

Final Concentrations

Since \( s = 0.0195 \), the concentrations of \( \mathrm{Sr}^{2+} \) and \( \mathrm{SO}_4^{2-} \) in a saturated solution are:\[ [\mathrm{Sr}^{2+}] = 0.0195 \; \text{M} \] \[ [\mathrm{SO}_4^{2-}] = 0.0195 \; \text{M} \]

Key concepts

Solubility Product Constant

The solubility product constant, often symbolized as \( K_{\text{sp}} \), is a crucial term in understanding how substances dissolve in water. It's especially important when dealing with sparingly soluble compounds like \( \mathrm{SrSO}_4 \). Essentially, it tells us the maximum number of ions a compound's solution can contain before it begins to precipitate out, meaning the compound cannot dissolve any more at that temperature.
For example, here's how it works: if you know the \( K_{\text{sp}} \) of a compound, you can find out the concentrations of its ions when it is saturated. For \( \mathrm{SrSO}_4 \), the \( K_{\text{sp}} \) is \( 3.8 \times 10^{-4} \), which means when the reaction equilibrium is reached, the product of the ionic concentrations equals this number.

• Determining \( K_{\text{sp}} \) for a salt helps predict solubility without needing extra calculations every time.
• It also aids in comparing which compounds are more soluble in water.

Dissociation Equation

When a solid dissolves in water, it separates into its constituent ions, a process known as dissociation. This is described by a dissociation equation, showing the breakup of the compound. For \( \mathrm{SrSO}_4 \), the dissociation in water can be written as: \[ \mathrm{SrSO}_4 (\text{s}) \rightleftharpoons \mathrm{Sr}^{2+} (\text{aq}) + \mathrm{SO}_4^{2-} (\text{aq}) \] This equation illustrates that each formula unit of \( \mathrm{SrSO}_4 \) will provide one \( \mathrm{Sr}^{2+} \) ion and one \( \mathrm{SO}_4^{2-} \) ion when dissolved.

• The double arrow in the equation signifies a dynamic equilibrium, indicating that the reaction can go in both forward and backward directions.
• The state symbols '(s)' for solid and '(aq)' for aqueous shows the physical states of the reactants and products.

This dissociation equation is vital because it helps us understand the stoichiometry, or the relative quantities of ions, and determines how we set up the expressions involving \( K_{\text{sp}} \).

Ion Concentration

Ion concentration refers to the amount of an ion dissolved in a solution, generally expressed in moles per liter (M). In a saturated solution, these concentrations are at equilibrium where the solution is fully saturated with ions. We focus on this because it tells us if the solution can or cannot dissolve more solute.
Using the example of \( \mathrm{SrSO}_4 \), in a saturated solution, the concentrations of \( [\mathrm{Sr}^{2+}] \) and \( [\mathrm{SO}_4^{2-}] \) determine the limit of solubility for the solution.

• These concentrations are directly related to the \( K_{\text{sp}} \) value, as found through the expression \( K_{\text{sp}} = [\mathrm{Sr}^{2+}] [\mathrm{SO}_4^{2-}] \).
• If the concentrations of ions in solution exceed their equilibrium values, precipitation forms, creating solid particles that settle out of solution.

Calculation of Solubility

Solubility calculation revolves around determining how much solute can dissolve in a solvent to form a saturated solution. This is directly related to the solubility product constant, \( K_{\text{sp}} \). To find solubility, consider the following: if \( [\mathrm{Sr}^{2+}] \) = \( s \) and \( [\mathrm{SO}_4^{2-}] \) = \( s \) in a saturated solution of \( \mathrm{SrSO}_4 \), according to the \( K_{\text{sp}} \) expression \( K_{\text{sp}} = s^2 \).
By plugging the given \( K_{\text{sp}} \) value \( 3.8 \times 10^{-4} \) into the equation, you can solve \( s^2 = 3.8 \times 10^{-4} \) and take the square root, yielding \( s \approx 0.0195 \) M. This means in the solution, each ion's concentration is \( 0.0195 \) M.

• This figure represents the maximum concentration of dissolved ions the solution can have without forming solid \( \mathrm{SrSO}_4 \).
• Understanding this calculation helps in practical applications such as predicting how compounds behave in different environments.

Through these steps, we can assess and manipulate the solubility of different substances.