Problem 35 - Analytical Chemistry

Question

The solubility product of \(\mathrm{Bi}_{2} \mathrm{~S}_{3}\) is \(1 \times 10^{-97}\) and that of \(\mathrm{HgS}\) is \(4 \times 10^{-53} .\) Which is the least soluble?

Step-by-Step Solution

Answer

\(\mathrm{Bi}_2 \mathrm{S}_3\) is the least soluble.

Step 1: Define Solubility Product

The solubility product \(K_{sp}\) of a salt is the product of the molar concentrations of its ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation for the salt's dissolution. These are given values for \(\mathrm{Bi}_2 \mathrm{S}_3\) as \(1 \times 10^{-97}\) and \(\mathrm{HgS}\) as \(4 \times 10^{-53}\).

Step 2: Write Dissolution Equations

The dissolution of \(\mathrm{Bi}_2 \mathrm{S}_3\) in water can be described by:\[\text{Bi}_2\text{S}_3 (s) \rightleftharpoons 2\text{Bi}^{3+} (aq) + 3\text{S}^{2-} (aq)\]The dissolution of \(\mathrm{HgS}\) in water is:\[\text{HgS} (s) \rightleftharpoons \text{Hg}^{2+} (aq) + \text{S}^{2-} (aq)\]

Step 3: Express Solubility Products in Terms of Ion Concentrations

For \(\mathrm{Bi}_2 \mathrm{S}_3\):\[K_{sp} = [\text{Bi}^{3+}]^2 [\text{S}^{2-}]^3\]For \(\mathrm{HgS}\):\[K_{sp} = [\text{Hg}^{2+}] [\text{S}^{2-}]\]

Step 4: Assume Molar Solubility \(s\)

Assume the molar solubility of \(\mathrm{Bi}_2 \mathrm{S}_3\) is \(s\):\[[\text{Bi}^{3+}] = 2s, \quad [\text{S}^{2-}] = 3s\]For \(\mathrm{HgS}\) assume the solubility is \(s\):\[[\text{Hg}^{2+}] = s, \quad [\text{S}^{2-}] = s\]

Step 5: Calculate Solubility Expression

For \(\mathrm{Bi}_2 \mathrm{S}_3\):\[K_{sp} = (2s)^2 (3s)^3 = 4s^2 \times 27s^3 = 108s^5 = 1 \times 10^{-97}\]For \(\mathrm{HgS}\):\[K_{sp} = s \times s = s^2 = 4 \times 10^{-53}\]

Step 6: Solve for \(s\) - Molar Solubility

For \(\mathrm{Bi}_2 \mathrm{S}_3\):\[108s^5 = 1 \times 10^{-97}\]\[s^5 = \frac{1 \times 10^{-97}}{108}\]\[s = \left(\frac{1 \times 10^{-97}}{108}\right)^{1/5}\]For \(\mathrm{HgS}\):\[s^2 = 4 \times 10^{-53}\]\[s = \sqrt{4 \times 10^{-53}}\]

Step 7: Compare Solubilities

Calculate numeric values:For \(\mathrm{Bi}_2 \mathrm{S}_3\), solve \[s \approx \left(9.26 \times 10^{-100}\right)^{1/5} \approx 3.61 \times 10^{-20}\]For \(\mathrm{HgS}\):\[s \approx 2 \times 10^{-26.5} \approx 2 \times 10^{-27}\]Comparing these, \(\mathrm{Bi}_2 \mathrm{S}_3\) has a smaller solubility.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental principle in understanding how reactions reach a state of balance. In the context of solubility, it describes the point at which the rate of the solid dissolving into solution equals the rate at which the ions recombine to form the solid. This dynamic equilibrium state is represented by the equilibrium constant known as the solubility product constant, or \( K_{sp} \).

• Reversed and forward reactions occur at equal rates when a system is at equilibrium.
• For dissolution, this involves the solute ions separating and joining back to form the solid.

The equilibrium expressions for \( \mathrm{Bi}_2 \mathrm{S}_3 \) and \( \mathrm{HgS} \) show how the ions balance in solution, with coefficients indicating the number of each ion involved. For example, \( \mathrm{Bi}_2 \mathrm{S}_3 \) dissociates into two \( \mathrm{Bi}^{3+} \) ions and three \( \mathrm{S}^{2-} \) ions. This equilibrium state allows us to set up equations using the \( K_{sp} \) values to find the concentration of ions at saturation.

Molar Solubility

Molar solubility is the number of moles of solute that can dissolve per liter of solution before the solution becomes saturated. This concept is crucial because it enables us to relate the solubility product \( K_{sp} \) to the actual concentration of ions in the solution. For both \( \mathrm{Bi}_2 \mathrm{S}_3 \) and \( \mathrm{HgS} \), understanding molar solubility helps us determine which compound is less soluble.

• It is commonly represented by \( s \), where \( s \) denotes the molarity of the solute in a saturated solution.
• The calculations often involve complex stoichiometric ratios, particularly when multiple ions are released per formula unit of solute.

For instance, to find the molar solubility of \( \mathrm{Bi}_2 \mathrm{S}_3 \), we calculate \( s \) using the equation \( 108s^5 = 1 \times 10^{-97} \). Solving for \( s \) reveals the concentration of ions at equilibrium, showing that \( \mathrm{Bi}_2 \mathrm{S}_3 \) is the least soluble due to its lower \( s \) value compared to \( \mathrm{HgS} \).

Inorganic Chemistry

Inorganic chemistry deals with the properties and behaviors of inorganic compounds. Solubility product and solubility are key topics within this field, particularly when it comes to understanding salts like \( \mathrm{Bi}_2 \mathrm{S}_3 \) and \( \mathrm{HgS} \). These topics highlight the role of ionic compounds in various chemical reactions and processes.

• Focuses on non-organic substances, including metals, minerals, and organometallic compounds.
• Important for predicting how substances will interact in a solution.

In the exercise, solving the solubility challenge required applying inorganic chemistry principles to calculate and compare the solubility of two metal sulfides. By examining their solubility products, students learn how different compounds behave in water, emphasizing the predictive power of \( K_{sp} \) within inorganic systems.