Question

What is the maximum possible concentration of \(\mathrm{Ni}^{2+}\) ion in water at \(25^{\circ} \mathrm{C}\) that is saturated with \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{S}\) and maintained at \(\mathrm{pH} 3.0\) with HCl?What is the maximum possible concentration of \(\mathrm{Ni}^{2+}\) ion in water at \(25^{\circ} \mathrm{C}\) that is saturated with \(0.10\) \(M\) \(\mathrm{H}_{2} \mathrm{S}\) and maintained at \(\mathrm{pH} 3.0\) with \(\mathrm{HCl}\)?

Short answer

The maximum possible concentration of Ni²⁺ ion in water at 25°C that is saturated with 0.10 M H₂S and maintained at pH 3.0 with HCl is 3.3 x 10⁻⁴ M.

Step-by-step solution

Finding the [H⁺] concentration *at equilibrium*

Given that the pH of the solution is 3.0, we can calculate the [H⁺] concentration using the pH formula: pH = -log10[H⁺] 3.0 = -log10[H⁺] Solving for [H⁺]: [H⁺] = 10⁻³M

Calculating the [HS⁻] concentration at equilibrium

Using the Ka1 value and the equilibrium expression for H₂S: Ka1 = [H⁺][HS⁻] / [H₂S] Rearranging to solve for [HS⁻]: [HS⁻] = (Ka1 * [H₂S]) / [H⁺] Plugging in the given values for [H₂S] = 0.10M and [H⁺] = 10⁻³M: [HS⁻] = (9.1 x 10⁻⁸ * 0.10) / 10⁻³ [HS⁻] = 9.1 x 10⁻⁶M

Calculating the [S²⁻] concentration at equilibrium

Using the Ka2 value and the equilibrium expression for HS⁻: Ka2 = [H⁺][S²⁻] / [HS⁻] Rearranging to solve for [S²⁻]: [S²⁻] = (Ka2 * [HS⁻]) / [H⁺] Plugging in the calculated [HS⁻] = 9.1 x 10⁻⁶M and [H⁺] = 10⁻³M: [S²⁻] = (1.0 x 10⁻¹³ * 9.1 x 10⁻⁶) / 10⁻³ [S²⁻] = 9.1 x 10⁻¹⁰M

Finding the maximum possible concentration of Ni²⁺

Now, we use the Ksp value and the equilibrium expression for NiS: Ksp = [Ni²⁺][S²⁻] Rearranging to solve for [Ni²⁺]: [Ni²⁺] = Ksp / [S²⁻] Plugging in the calculated [S²⁻] = 9.1 x 10⁻¹⁰M and Ksp = 3.0 x 10⁻¹³: [Ni²⁺] = (3.0 x 10⁻¹³) / (9.1 x 10⁻¹⁰) [Ni²⁺] = 3.3 x 10⁻⁴M So, the maximum possible concentration of Ni²⁺ ion in water at 25°C that is saturated with 0.10 M H₂S and maintained at pH 3.0 with HCl is 3.3 x 10⁻⁴ M.

Key concepts

Ni²⁺ Ion Concentration

Ni²⁺ ion concentration in a solution can be influenced by various factors, such as the presence of other ions and the pH level of the solution.
The exercise challenges us to find the maximum concentration of nickel ions ( Ni²⁺ ) in a saturated environment where hydrogen sulfide ( H₂S ) is present.
In such scenarios, understanding the solubility equilibrium becomes crucial.
Sulfide ions ( S²⁻ ) , which result from dissolved H₂S in the water, play an important role in determining the nickel ion concentration.

By calculating the concentrations of HS⁻ and S²⁻ ions through their equilibrium expressions involving their respective dissociation constants, we get the necessary details to find the Ni²⁺ concentration using the solubility product constant ( K_{sp} ) .

• This involves calculating the S²⁻ ion concentration, which is vital due to its direct relation with Ni²⁺ in the solubility product ( Ksp ) expression of NiS compound.
• A higher S²⁻ concentration would imply a lower maximum Ni²⁺ that can exist in water before precipitation occurs, as a result of the equilibrium constraints detailed by K_{sp} .

This principle of equilibria allows chemists to predict and control the ion concentrations in a solution, preventing undesirable precipitations.

pH Calculation

Calculating the pH of a solution is a fundamental skill in chemistry, reflecting the acidic or basic nature of the solution.
In this exercise, the solution is maintained at a pH of 3.0, indicating a relatively acidic solution.
The pH is calculated using the formula defined by pH = -log_{10}[ H⁺] .

Since pH is easily found or given, it is useful to derive the H⁺ concentration from it whenever necessary.

• For example, a pH of 3.0 gives us H⁺ concentration as 10^{-3} M, depicting a considerable amount of acidic character.
• It's important to understand how pH influences solubility rates and equilibria by altering ion concentrations in the solution, particularly for complex equilibria like the one in this example with nickel and sulfide.

Being able to manipulate pH provides control over chemical reactions and equilibria in a solution, an essential aspect in both academic and industrial chemistry applications.

Ksp Value

The solubility product constant, Ksp, defines the point at which a solute will precipitate from a solution, making it a critical aspect when discussing chemical equilibria.
For this problem, the Ksp value of NiS represents the equilibrium between dissolved Ni²⁺ and S²⁻ ions in the saturated solution.

• Given as 3.0 × 10^{-13}, this value shows how sparingly soluble NiS is in the environment.
• Higher values of Ksp would indicate greater solubility and thus, higher concentrations of Ni²⁺ in the solution are possible without triggering precipitation.

The calculation involves solving the expression Ksp = [Ni²⁺][S²⁻] for Ni²⁺ by substituting known Ksp and calculated S²⁻ values.
This allows determining the maximum permissible Ni²⁺ in solution, providing insights into the solubility limits imposed by the system's conditions. By mastering the use of Ksp values, you can predict how compounds will behave in different solutions, a useful tool in many fields of science and engineering.