Question

What mass of \(\operatorname{ZnS}\left(K_{\text {sp }}=2.5 \times 10^{-22}\right.\) ) will dissolve in \(300.0 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} ?\) Ignore the basic properties of \(\mathrm{S}^{2-}\).

Short answer

Approximately \(1.46 \times 10^{-8}\) g of ZnS will dissolve in 300.0 mL of 0.050 M Zn(NO₃)₂ solution.

Step-by-step solution

Write the balanced dissolution equation for ZnS

\(\mathrm{ZnS_{(s)}} \leftrightarrows \mathrm{Zn^{2+}_{(aq)}} + \mathrm{S^{2-}_{(aq)}}\)

Find equilibrium concentration of S²⁻ ions using Ksp

The Ksp expression is given by: \[K_{sp} = [\mathrm{Zn^{2+}}][\mathrm{S^{2-}}]\] Plugging in the given value for Ksp and the concentrations at equilibrium, we get: \[2.5 \times 10^{-22} = (x)(x)\] \[x^2 = 2.5 \times 10^{-22}\] \[x = \sqrt{2.5 \times 10^{-22}} = 5 \times 10^{-12} \mathrm{M}\] So, the equilibrium concentration of S²⁻ ions is \(5 \times 10^{-12}\) M.

Calculate the total concentration of Zn²⁺ ions

The total concentration of Zn²⁺ ions comes from the dissolution of both \(\mathrm{ZnS}\) and \(\mathrm{Zn(NO_3)_2}\). The concentration of Zn²⁺ ions from the dissolution of \(\mathrm{ZnS}\) is equal to the concentration of S²⁻ ions. Therefore, the total concentration of Zn²⁺ ions is: \[\mathrm{Total\, Zn^{2+}\, concentration} = [\mathrm{Zn^{2+}}]_{\mathrm{from\, ZnS}} + [\mathrm{Zn^{2+}}]_{\mathrm{from\, Zn(NO_3)_2}\] \[= 5 \times 10^{-12} \mathrm{M} + 0.050 \mathrm{M}\]

Determine the concentration of dissolved ZnS

The concentrations of Zn²⁺ ions and S²⁻ ions at equilibrium are the same. The concentration of dissolved ZnS is equal to the concentration of S²⁻ ions, which we calculated earlier: \[[\mathrm{ZnS}]= 5 \times 10^{-12} \mathrm{M}\]

Convert the concentration of ZnS to mass

Now we will find the mass of ZnS that will dissolve in 300.0 mL of solution: Molar mass of ZnS = 65.38 g/mol (Zn) + 32.07 g/mol (S) = 97.45 g/mol Mass of ZnS = concentration x volume x molar mass \[m = (5 \times 10^{-12}\, \mathrm{M})(300 \times 10^{-3}\, \mathrm{L})(97.45\, \mathrm{g/mol})\] \[m = 1.46 \times 10^{-8}\, \mathrm{g}\] So, approximately \(1.46 \times 10^{-8}\) g of ZnS will dissolve in 300.0 mL of 0.050 M Zn(NO₃)₂ solution.

Key concepts

Equilibrium Concentration

In chemistry, equilibrium concentration is a vital concept, particularly when dealing with solubility and dissolution reactions. Let's dive into what it implies using Zinc Sulfide (ZnS) as our example. At equilibrium, the concentration of ions in a solution no longer changes, meaning the rate of the forward reaction matches the rate of the backward reaction.
This is represented in the dissolution of ZnS, where at equilibrium, you have ions of Zn²⁺ and S²⁻ suspended in solution.
When the question involves finding equilibrium concentration, you need to work with the solubility product constant, known as the Ksp.

For ZnS, the Ksp is given by the equation:

• \[ K_{sp} = [\mathrm{Zn^{2+}}][\mathrm{S^{2-}}] \]

This setup represents the equilibrium concentrations of the ions. Solving for the concentration means incorporating the given Ksp value. Let's say, for this case, Ksp for ZnS is \(2.5 \times 10^{-22}\).
By setting up the equation, you find that both ion concentrations in ZnS are the square root of this Ksp value, which yields \(5 \times 10^{-12}\, M\).

This value reflects the minimal concentrations of ions capable of existing in equilibrium without precipitating out of the solution.

Precipitation Reaction

Precipitation reactions occur when two soluble salts in solutions combine to form an insoluble compound, known as a precipitate.
This is crucial in a variety of chemical applications, including purification and qualitative analysis of ions.
For ZnS, though the solubility is low, precipitation relevance stems from its Ksp value: a smaller Ksp indicates greater potential for forming a precipitate.

If ZnS is mixed in a solution already containing Zn²⁺ ions (as with the initial concentration of Zn(NO₃)₂), the solution reaches a point where it cannot hold more ions in the dissolved state.
At that stage, additional ZnS will remain a solid and not dissolve, demonstrating the precipitation habit.
The calculation of exactly how much ZnS can dissolve—in this case, \(1.46 \times 10^{-8}\, g\) in 300.0 mL solution—aligns with understanding the delicate balance of ion concentrations where precipitation occurs.

Ionic Equilibrium

Ionic equilibrium in a solution involves ions being present in a state where their concentrations are stable over time, hence the system is in a steady state.
Ionic equilibrium governs reactions such as the dissolution of salts like ZnS within a solution containing additional ions like Zn²⁺ from Zn(NO₃)₂.

In essence, achieving ionic equilibrium signifies that the concentrations of the various ions no longer change as they continuously engage in the processes of dissolving and precipitating at the same rates.
This stability allows for precise calculations of ion concentrations, as seen with:

• \[ [\mathrm{S^{2-}}] = 5 \times 10^{-12} \, M \]

The total concentration of Zn²⁺ ions combines both contributions from the original and dissolved states:

• \[ \mathrm{[Zn^{2+}]_{Total} = [Zn^{2+}]_{from\, ZnS} + [Zn^{2+}]_{from\, Zn(NO_3)_2} = 0.050 \, M + 5 \times 10^{-12} \, M } \]

By recognizing these values, one can predict the outcome of possible further additions or temperature changes affecting the solvability or precipitating behavior.