Question

Mercuric sulfide is an extremely insoluble compound, having \(K_{\mathrm{sp}}=3 \times 10^{-53}\) near room temperature. Calculate the solubility of \(\mathrm{HgS}\) in mol/L at this temperature.

Short answer

The solubility of Mercuric sulfide (HgS) in water at room temperature is approximately \(1.73 \times 10^{-27}\) mol/L.

Step-by-step solution

Write the balanced chemical equation for the dissociation of HgS in water

The dissociation of Mercuric sulfide (HgS) in water can be written as follows: \[ HgS(s) \rightleftharpoons Hg^{2+}(aq) + S^{2-}(aq) \]

Set up an expression for the solubility product constant (Ksp)

Based on the dissociation equation, we can write an expression for Ksp as follows: \[ K_{sp} = [Hg^{2+}] [S^{2-}] \]

Define the solubility and the concentrations of the ions

Let the solubility of HgS be represented by "x" mol/L. Since both ions are produced in a 1:1 ratio upon the dissociation of HgS, the concentrations of the ions in the saturated solution will be: \[ [Hg^{2+}] = x \] and \[ [S^{2-}] = x \]

Substitute the ion concentrations in the Ksp expression and solve for x

Now, we can substitute the values of the ion concentrations in terms of x into the Ksp expression: \[ K_{sp} = x \cdot x = x^2 \] Given that the Ksp value for HgS is 3 × 10⁻⁵³, we can write the equation as: \[ 3 \times 10^{-53} = x^2 \] To solve for x (the solubility of HgS in mol/L), we can take the square root of both sides: \[ x = \sqrt{3 \times 10^{-53}} \] Now, we can calculate the value of x using a calculator: \[ x \approx 1.73 \times 10^{-27} \, mol/L \]

Report the solubility of HgS

The solubility of Mercuric sulfide (HgS) in water at room temperature is approximately \(1.73 \times 10^{-27}\) mol/L.

Key concepts

Ksp Calculation

The solubility product constant, or Ksp, is crucial for understanding how much of a barely soluble compound will dissolve in water. When we deal with Ksp, we're looking at the equilibrium between a solid and its ions in a saturated solution. This concept is especially important for compounds like mercuric sulfide (HgS), which are only slightly soluble.

Calculating the Ksp involves a few steps. First, we write the chemical equation for the substance as it dissociates in water. For HgS, the equation is:

• \[ HgS(s) \rightleftharpoons Hg^{2+}(aq) + S^{2-}(aq) \]

In this balanced equation, the solid HgS breaks down into mercury and sulfide ions.

Next, express the Ksp as the product of the concentrations of the ions:

• \[ K_{sp} = [Hg^{2+}] [S^{2-}] \]

Our next task is substituting these concentrations into the Ksp expression to find the solubility, which is done by representing the solubility as 'x'. Therefore, for mercuric sulfide, Ksp becomes:

• \[ x^2 = K_{sp} \]

Finally, solving for 'x' provides the molar solubility of HgS.

Chemical Equilibrium

Chemical equilibrium is a state where reactants and products are formed at a constant rate, creating a stable mixture. For insoluble compounds, this involves the solid phase and its dissolved ions.

In the case of HgS, the chemical equilibrium can be illustrated by its dissociation:

• \[ HgS(s) \rightleftharpoons Hg^{2+}(aq) + S^{2-}(aq) \]

This reversible reaction shows how the solid mercuric sulfide reaches equilibrium with its ionized form in a solution. At equilibrium, the rate at which HgS dissolves equals the rate at which ions recombine to form solid HgS.

The constant concentration means that calculating Ksp relies heavily on this principle of equilibrium. Whenever dealing with such weakly soluble salts, this equilibrium state helps predict how changes in conditions (like temperature or the presence of other ions) might alter solubility.

Insoluble Compounds

In chemistry, insoluble compounds refer to those substances that dissolve only slightly in a solvent. Mercuric sulfide (HgS) is an excellent example as it barely dissolves in water. Its very low solubility is reflected by the extremely small Ksp value: \(3 \times 10^{-53}\).

These compounds still establish an equilibrium between the dissolved ions and the undissolved solid. Despite their low solubility, understanding and calculating how they dissolve is crucial, especially in contexts like environmental science or industrial processes.

Recognizing insoluble compounds involves checking solubility rules and considering how they interact under different conditions. Most importantly, such compounds remind us that even small amounts of ions can significantly impact chemical reactions and equilibrium dynamics.