Question

Approximately \(0.14 \mathrm{~g}\) of nickel(II) hydroxide, \(\mathrm{Ni}(\mathrm{OH})_{2}(s),\) dissolves per liter of water at 20 ' \(\mathrm{C}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.

Short answer

The solubility product constant, \(K_{sp}\), for nickel(II) hydroxide, Ni(OH)\(_{2}\), at 20°C is approximately 1.33 x 10\(^{-8}\).

Step-by-step solution

Convert mass of Ni(OH)2 to moles

We are given that approximately 0.14g of Ni(OH)2 dissolves in 1L of the water. First, we need to find the number of moles of Ni(OH)2. Given that, Molar mass of Ni(OH)2 = 58.69 (Ni) + 2(16.00 + 1.01) = 93.71 g/mol Mass of Ni(OH)2 = 0.14g To find moles, use the formula: Moles of Ni(OH)2 = mass / molar mass Moles of Ni(OH)2 = 0.14g / 93.71 g/mol = 1.493 x 10^-3 mol/L

Find the concentrations of Ni²⁺ and OH⁻ at equilibrium

Since the dissociation of Ni(OH)2 results in a 1:1 ratio of Ni²⁺ and Ni(OH)2, the concentration of Ni²⁺ at equilibrium is equal to the moles of Ni(OH)₂ per liter of the water. The concentration of OH⁻ ions will be twice that of Ni²⁺ ions due to the 1:2 ratio in the balanced dissociation equation. [Ni²⁺] = 1.493 x 10^-3 mol/L [OH⁻] = 2 x [Ni²⁺] = 2 x 1.493 x 10^-3 mol/L = 2.986 x 10^-3 mol/L

Calculate Ksp using the ion concentrations

Now that we have the concentrations of both ions at equilibrium, we can use the Ksp expression to find the solubility product constant: Ksp = [Ni²⁺][OH⁻]² Ksp = (1.493 x 10^-3)(2.986 x 10^-3)² Ksp = (1.493 x 10^-3)(8.916 x 10^-6) Ksp = 1.33 x 10^-8 So, the Ksp for Ni(OH)2 at 20°C is approximately 1.33 x 10^-8.

Key concepts

Nickel(II) Hydroxide

Nickel(II) Hydroxide, with the chemical formula \(\mathrm{Ni(OH)_2}\), is an inorganic compound that consists of nickel cations and hydroxide anions. It is a green solid and finds its applications in various fields like battery materials.

In the context of solubility, \(\mathrm{Ni(OH)_2}\) is sparingly soluble in water. This limited solubility means that not all of the compound will dissolve in a given volume of water. Instead, an equilibrium is established between the undissolved solid and the ions in solution. The point at which this equilibrium is reached is crucial for calculating the solubility product constant \(K_{\text{sp}}\).

Understanding the properties of \(\mathrm{Ni(OH)_2}\) and its behaviour in aqueous solutions is essential for working through problems involving its dissolution.

Molar Mass Calculation

Calculating the molar mass of a compound is vital for converting between mass and moles, which is frequent in chemistry problems. To determine the molar mass of \(\mathrm{Ni(OH)_2}\), we must consider the atomic masses of each element in the compound.

Here's how to do it:

• Find the atomic mass of nickel (Ni), which is approximately 58.69 g/mol.
• Find the atomic mass of oxygen (O), approximately 16.00 g/mol, and hydrogen (H), approximately 1.01 g/mol.
• The \(\mathrm{OH}^-\) ion consists of one oxygen and one hydrogen, so its mass is \(16.00 + 1.01 = 17.01\) g/mol.
• Since there are two \(\mathrm{OH}^-\) ions in nickel(II) hydroxide, their total mass is \(2 \times 17.01 = 34.02\) g/mol.
• Add the mass of nickel: \(58.69 + 34.02 = 92.71\) g/mol.

The molar mass of \(\mathrm{Ni(OH)_2}\) thus calculated is 92.71 g/mol. Knowing this value allows you to convert any given mass of \(\mathrm{Ni(OH)_2}\) into moles, essential for subsequent calculations.

Concentration Calculation

In chemistry, calculating the concentration of a substance in a solution is a foundational skill. Given the amount of \(\mathrm{Ni(OH)_2}\) that dissolves in water, we can determine the concentrations of the resulting ions at equilibrium.

Here's a step-by-step explanation:

• First, convert the mass of \(\mathrm{Ni(OH)_2}\) (0.14 g) to moles using its molar mass (92.71 g/mol):
  \[\text{Moles of } \mathrm{Ni(OH)_2} = \frac{0.14}{92.71} = 1.493 \times 10^{-3} \text{ mol/L} \]
• When \(\mathrm{Ni(OH)_2}\) dissolves, it dissociates into \(\mathrm{Ni^{2+}}\) and \(\mathrm{OH^-}\) ions.
• The dissociation reaction of \(\mathrm{Ni(OH)_2}\) gives one \(\mathrm{Ni^{2+}}\) and two \(\mathrm{OH^-}\) ions per formula unit.
• Thus,
  \[\text{[Ni}^{2+}\text{]} = 1.493 \times 10^{-3} \text{ mol/L}\]
  \[\text{[OH}^-\text{]} = 2 \times 1.493 \times 10^{-3} = 2.986 \times 10^{-3} \text{ mol/L}\]

Knowing these concentrations allows us to calculate the solubility product constant, \(K_{\text{sp}}\), which gives significant insight into a compound's solubility characteristics.

Dissolution Equilibrium

Dissolution equilibrium refers to the state where the dissolution of a compound in a solvent is balanced by its precipitation out of the solution. For compounds like \(\mathrm{Ni(OH)_2}\), which only partially dissolve, understanding this equilibrium helps us understand their behavior in solutions.

The dissolution of \(\mathrm{Ni(OH)_2}\) can be represented by the equation:\[\mathrm{Ni(OH)_2(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^-(aq)}\]In this equation, the solid \(\mathrm{Ni(OH)_2}\) dissociates into nickel ions \((\mathrm{Ni^{2+}})\) and hydroxide ions \((\mathrm{OH^-})\).

• At equilibrium, the rate of dissolution equals the rate of precipitation.
• Solubility product constant, \(K_{\text{sp}}\), helps quantify this equilibrium.
• \(K_{\text{sp}}\) is calculated using:
   \[K_{\text{sp}} = [\mathrm{Ni^{2+}}][\mathrm{OH^-}]^2\]
• This equation highlights that the product of the ion concentrations raised to the power of their coefficients in the equation remains constant at equilibrium.

This understanding of dissolution equilibrium is crucial when predicting how much of a substance will dissolve in a particular environment and under certain conditions.