Question

The value of \(K_{s p}\) for \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) is \(2.1 \times 10^{-20}\). The \(\mathrm{AsO}_{4}{ }^{3-}\) ion is derived from the weak acid \(\mathrm{H}_{3} \mathrm{AsO}_{4}\left(\mathrm{p} K_{a 1}=\right.\) \(2.22 ; \mathrm{pK}_{a 2}=6.98 ; \mathrm{pK}_{a 3}=11.50 \mathrm{~J}\) (a) Calculate the molar solubility of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water. (b) Calculate the \(\mathrm{pH}\) of a saturated solution of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water.

Short answer

The molar solubility is approximately \(8.5 \times 10^{-5} \text{ M}\), and the pH is about 12.7.

Step-by-step solution

Understanding Ionic Dissolution

For the compound \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\), it dissociates in water as follows: \[\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2} \rightleftharpoons 3\mathrm{Mg}^{2+} + 2\mathrm{AsO}_{4}^{3-}\]Let \(s\) be the molar solubility of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\). Then, \([\mathrm{Mg}^{2+}] = 3s\) and \([\mathrm{AsO}_{4}^{3-}] = 2s\).

Writing the Solubility Product Expression

The solubility product \(K_{sp}\) for the dissolution is given by:\[K_{sp} = [\mathrm{Mg}^{2+}]^3 [\mathrm{AsO}_{4}^{3-}]^2\]Substituting the expressions of \([\mathrm{Mg}^{2+}]\) and \([\mathrm{AsO}_{4}^{3-}]\), we get:\[K_{sp} = (3s)^3(2s)^2 = 108s^5\]We know \(K_{sp} = 2.1 \times 10^{-20}\).

Solving for Molar Solubility

Set the solubility product expression equal to the given \(K_{sp}\):\[108s^5 = 2.1 \times 10^{-20}\]Solve for \(s\):\[s^5 = \frac{2.1 \times 10^{-20}}{108}\]\[s = \left( \frac{2.1 \times 10^{-20}}{108} \right)^{1/5}\] After calculation, find \(s\) to be approximately \(8.5 \times 10^{-5} \text{ M}\).

Calculate pH of the Solution

In a saturated solution, \([\mathrm{AsO}_{4}^{3-}] = 2s\). Since \([\mathrm{AsO}_{4}^{3-}]\) comes from \(\mathrm{H}_{3}\mathrm{AsO}_{4}\) dissociation, it affects the \([\mathrm{H}^{+}]\) concentration.The equilibrium for \(\mathrm{AsO}_{4}^{3-}\) and \(\mathrm{H}^{+}\) can be estimated via the conjugate acid-base relation. Due to negligible influence at extremely low concentrations, assume \([\mathrm{H}^{+}] \approx \sqrt{(2 \times 10^{-9})}[\mathrm{AsO}_{4}^{3-}] \). Calculate:\[\mathrm{pH} \approx -\log(\sqrt{2 \times 8.5 \times 10^{-9}})\], \[\mathrm{pH} \approx 12.7\].

Key concepts

Solubility Product Constant

When we talk about the solubility product constant, often represented as \(K_{sp}\), we are essentially discussing the tendency of a compound to dissolve in water. It is a constant at a given temperature for a specific dissolution reaction. The \(K_{sp}\) helps predict whether a precipitate will form in solution.
Imagine you have a sparingly soluble compound, like \(\mathrm{Mg}_3(\mathrm{AsO}_4)_2\). When it dissolves, it reaches an equilibrium with its ions in solution, and the solubility product expression can be derived from this equilibrium.
For \(\mathrm{Mg}_3(\mathrm{AsO}_4)_2\), it dissociates as follows:
\[\mathrm{Mg}_3(\mathrm{AsO}_4)_2 \rightleftharpoons 3\mathrm{Mg}^{2+} + 2\mathrm{AsO}_4^{3-}\].
The \(K_{sp}\) expression becomes: \[K_{sp} = [\mathrm{Mg}^{2+}]^3 [\mathrm{AsO}_4^{3-}]^2\].
This equation defines the point where the concentrations of ions in a solution will not change further unless the solution conditions are altered, like temperature. It's important to note that \(K_{sp}\) does not change for a given compound and temperature, making it a cornerstone in predicting solubility.

Dissolution Reactions

Dissolution reactions play a critical role in understanding how and why compounds dissolve in solvents. For any dissolution reaction to occur, we require a chemical compound to disperse into its constituent ions or molecules in a solvent when equilibrium is achieved.
Consider \(\mathrm{Mg}_3(\mathrm{AsO}_4)_2\), a sparingly soluble compound that, when placed in water, will dissociate into three \(\mathrm{Mg}^{2+}\) ions and two \(\mathrm{AsO}_4^{3-}\) ions. This process is represented by the reaction:
\[\mathrm{Mg}_3(\mathrm{AsO}_4)_2 \rightleftharpoons 3\mathrm{Mg}^{2+} + 2\mathrm{AsO}_4^{3-}\].
Here's how we calculate the molar solubility, \(s\), which is the number of moles that dissolve per liter:

• At equilibrium, \([\mathrm{Mg}^{2+}] = 3s\) and \([\mathrm{AsO}_4^{3-}] = 2s\).
• The solubility product expression (from \(K_{sp}\)) becomes \(108s^5\).

By knowing the \(K_{sp}\), we can solve for \(s\) as shown in the problem steps. Solving \(108s^5 = 2.1 \times 10^{-20}\) gives us the molar solubility of about \(8.5 \times 10^{-5}\) M. Understanding dissolution at this level helps predict reactions in natural and industrial processes.

pH Calculation

The pH of a solution is a measure of its acidity or alkalinity. It is calculated as the negative logarithm of the hydrogen ion concentration, \([\mathrm{H}^+]\). For dissolution reactions, especially those involving weak acids or bases, the pH indicates how these ions affect the solution.
In the context of \(\mathrm{Mg}_3(\mathrm{AsO}_4)_2\)'s dissolution, the \(\mathrm{AsO}_4^{3-}\) ion can potentially affect the solution's pH. It's derived from \(\mathrm{H}_3\mathrm{AsO}_4\), a weak acid, which can influence \([\mathrm{H}^+]\) levels.
To estimate the pH of a saturated solution, use the relationship:
\[\mathrm{pH} \approx -\log(\sqrt{2s})\] where \(s\) reflects \([\mathrm{AsO}_4^{3-}]\).
For \(s = 8.5 \times 10^{-5}\) M, calculate \([\mathrm{H}^+]\) and simplify as done in step 4 to retrieve a pH around 12.7. This indicates a solution that's quite basic, due to the weak acidic behavior of \(\mathrm{H}_3\mathrm{AsO}_4\). Understanding pH is crucial for analyzing chemical environments, ensuring safety, and optimizing reactions.