Question

Mercuric sulfide, \(\mathrm{HgS},\) is one of the least soluble salts known, with \(K_{\mathrm{sp}}=1.6 \times 10^{-54}\) at 25 ?. Calculate the solubility of \(\mathrm{HgS}\) in moles per liter and in grams per liter.

Short answer

The solubility of \(\mathrm{HgS}\) at 25°C is approximately \(1.26 \times 10^{-27}\, \mathrm{mol/L}\) or \(2.93 \times 10^{-25}\, \mathrm{g/L}\).

Step-by-step solution

Write the solubility product expression

Write the solubility product (\(K_\mathrm{sp}\)) expression for the dissociation of mercuric sulfide (\(\mathrm{HgS}\)). \(K_\mathrm{sp} = [\mathrm{Hg^{2+}}][\mathrm{S^{2-}}]\)

Substitute the solubility for the concentrations

Since the molar ratios of \(\mathrm{Hg^{2+}}\) and \(\mathrm{S^{2-}}\) are 1:1, we can substitute the solubility "s" for the concentrations of the ions. \(K_\mathrm{sp} = [s][s] = s^2\)

Solve for solubility s (in moles per liter)

Given the \(K_\mathrm{sp}\) value of \(1.6 \times 10^{-54}\), solve for the solubility "\(s\)" in moles per liter: \(s^2 = 1.6 \times 10^{-54}\) Take the square root of both sides to find the value of s: \(s = \sqrt{1.6 \times 10^{-54}}\) \(s \approx 1.26 \times 10^{-27} \, \mathrm{mol/L}\)

Convert solubility to grams per liter

We now have the solubility in moles per liter. Convert this to grams per liter by using the molar mass of \(\mathrm{HgS}\). Molar mass of \(\mathrm{HgS} = 200.6\, (\mathrm{Hg}) + 32.1\, (\mathrm{S}) = 232.7\, \mathrm{g/mol}\) Multiply the solubility in moles per liter by the molar mass of \(\mathrm{HgS}\): \(s_\mathrm{g/L} = (1.26 \times 10^{-27} \, \mathrm{mol/L}) (232.7 \, \mathrm{g/mol})\) \(s_\mathrm{g/L} \approx 2.93 \times 10^{-25}\, \mathrm{g/L}\) So, the solubility of \(\mathrm{HgS}\) at 25°C is approximately \(1.26 \times 10^{-27}\, \mathrm{mol/L}\) or \(2.93 \times 10^{-25}\, \mathrm{g/L}\).

Key concepts

Ksp (Solubility Product Constant)

The solubility product constant, often represented as \(K_{sp}\), plays a crucial role in predicting the solubility of sparingly soluble compounds in water. It is a specific type of equilibrium constant that applies to the dissolution of ionic compounds into their constituent ions in a saturated solution.
For example, when a salt like mercuric sulfide (\(\text{HgS}\)) dissolves in water, it dissociates into mercuric ions (\(\text{Hg}^{2+}\)) and sulfide ions (\(\text{S}^{2-}\)). The equilibrium expression for this dissociation is given as:
\[ K_{sp} = [\text{Hg}^{2+}][\text{S}^{2-}]. \]
Here, \(K_{sp}\) is the product of the molar concentrations of the constituent ions, each raised to the power of their stoichiometric coefficients in the balanced equation. This constant helps understand how much of the salt can dissolve in water before reaching its equilibrium.
A low \(K_{sp}\) value indicates that very little solute will dissolve, making the compound typically insoluble or sparingly soluble.

• For instance, mercuric sulfide has a very low \(K_{sp}\) of \(1.6 \times 10^{-54}\), reflecting its extremely low solubility.
• This implies that when equilibrium is reached, the concentrations of mercuric and sulfide ions in solution are tiny.

Mercuric Sulfide

Mercuric sulfide, with the chemical formula \(\text{HgS}\), is renowned for being one of the least soluble compounds known to science. It exists in two crystalline forms: cinnabar (red) and metacinnabar (black). In nature, it's primarily found as cinnabar, the red mercury ore.
The minimal solubility of mercuric sulfide is a result of its solid crystalline structure and strong ionic bonds. When placed in water, it dissociates slightly to provide very small amounts of mercuric and sulfide ions:

• These ions contribute to its minute solubility in solution.
• Given its \(K_{sp}\) value, it is clear that only an infinitesimal amount of solid dissolves.

Understanding the solubility of mercuric sulfide is crucial not just for chemical calculations, but also for environmental and industrial contexts.
Since it is so insoluble, the release of mercury into the environment is limited if in the form of \(\text{HgS}\).

Chemical Equilibrium

Chemical equilibrium is a state of balance in a chemical reaction, where the rate of the forward reaction equals the rate of the reverse reaction. This balance leads to constant concentrations of reactants and products in a closed system.
In the context of solubility and the dissolution of sparingly soluble salts like mercuric sulfide, equilibrium is established between the undissolved solid and its ions in solution.

• At this state, there is no net change in the amount of dissolved ions over time.
• The solubility product constant \(K_{sp}\) helps define this equilibrium for the dissolution process.

Consider the dissolution of \(\text{HgS}\):
\[\text{HgS(s)} \rightleftharpoons \text{Hg}^{2+}\text{(aq)} + \text{S}^{2-}\text{(aq)}\] At equilibrium, the concentrations of \(\text{Hg}^{2+}\) and \(\text{S}^{2-}\) remain constant.

• The saturation point of the solution is reached, meaning it cannot dissolve more solute unless conditions change.
• This principle guides the calculation of solutions' concentration limits, essential in determining solubility.

Chemical equilibrium is not static; it is dynamic. Particles continue to exchange between reactants and products, but quantities remain consistent in a stable equilibrated system.