Question

Calculate the molar solubility of \(\mathrm{Ni}(\mathrm{OH})_{2}\) when buffered at \(\mathrm{pH}\) (a) 8.0, (b) \(10.0\), (c) \(12.0\).

Short answer

The molar solubility of \(\mathrm{Ni}(\mathrm{OH})_{2}\) at different pH values are: (a) \(5.48 \times 10^{-4}\) M for pH = 8.0 (b) \(5.48 \times 10^{-8}\) M for pH = 10.0 (c) \(5.48 \times 10^{-12}\) M for pH = 12.0

Step-by-step solution

Write the balanced equation

The balanced equilibrium reaction of \(\mathrm{Ni}(\mathrm{OH})_{2}\) dissolution is: \( \mathrm{Ni}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Ni}^{2+}(aq) + 2\mathrm{OH}^-(aq) \) Step 2: Calculate the concentration of hydroxide ions from the given pH values

Calculate the hydroxide concentration

Using the formula \( \mathrm{pOH} = 14 - \mathrm{pH}\), we can calculate the pOH values. To find the concentration of hydroxide ions, we will use the formula: \( [\mathrm{OH}^-] = 10^{-\mathrm{pOH}} \) We will do this for all three given pH values (8.0, 10.0, and 12.0). Step 3: Write the expression for the solubility product constant, \(K_{sp}\)

Write \(K_{sp}\) expression

Since the balanced equation is: \( \mathrm{Ni}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Ni}^{2+}(aq) + 2\mathrm{OH}^-(aq) \) The \(K_{sp}\) expression is: \( K_{sp} = [\mathrm{Ni}^{2+}][\mathrm{OH}^-]^2 \) Step 4: Calculate the molar solubility of \(\mathrm{Ni}(\mathrm{OH})_{2}\) at different pH values

Calculate molar solubility

We will substitute the calculated \([\mathrm{OH}^-]\) for each pH value into the \(K_{sp}\) expression. For \(\mathrm{Ni}(\mathrm{OH})_{2}\), the \(K_{sp}\) value is 5.48 x 10^{-16}. (a) For pH = 8.0: \( [\mathrm{OH}^-] = 10^{-\mathrm{pOH}} = 10^{-6} \) \( K_{sp} = [\mathrm{Ni}^{2+}][\mathrm{OH}^-]^2 => [\mathrm{Ni}^{2+}] = \dfrac{K_{sp}}{[\mathrm{OH}^-]^2} = \dfrac{5.48 \times 10^{-16}}{(10^{-6})^2} = 5.48 \times 10^{-4} M \) (b) For pH = 10.0: \( [\mathrm{OH}^-] = 10^{-4} \) \( [\mathrm{Ni}^{2+}] = \dfrac{5.48 \times 10^{-16}}{(10^{-4})^2} = 5.48 \times 10^{-8} M \) (c) For pH = 12.0: \( [\mathrm{OH}^-] = 10^{-2} \) \( [\mathrm{Ni}^{2+}] = \dfrac{5.48 \times 10^{-16}}{(10^{-2})^2} = 5.48 \times 10^{-12} M \) The molar solubility of \(\mathrm{Ni}(\mathrm{OH})_{2}\) at different pH values are: (a) 5.48 x 10^{-4} M for pH = 8.0 (b) 5.48 x 10^{-8} M for pH = 10.0 (c) 5.48 x 10^{-12} M for pH = 12.0

Key concepts

Solubility Product Constant

The Solubility Product Constant, often denoted as \( K_{sp} \), is a measure of the solubility of a compound under equilibrium conditions. It describes the extent to which a solid can dissolve in water. Each compound that dissolves partially or fully in water will reach a balance between the dissolved ions and the undissolved solid.

For a given ionic compound \( A_xB_y \), which dissociates into \( x \) cations \( A^{y+} \) and \( y \) anions \( B^{x-} \), the solubility product expression can be written as:

\[ K_{sp} = [A^{y+}]^x[B^{x-}]^y \]

In the case of nickel(II) hydroxide, \( \mathrm{Ni(OH)_2} \), the balance reaction is:

\( \mathrm{Ni(OH)_2 (s) \rightleftharpoons Ni^{2+} (aq) + 2OH^- (aq)} \)

The corresponding \( K_{sp} \) expression is:

\[ K_{sp} = [\mathrm{Ni^{2+}}][\mathrm{OH^-}]^2 \]

This equation implies that the concentration of \( \mathrm{Ni^{2+}} \) and the concentration of \( \mathrm{OH^-} \) are interdependent. Changes in the concentration of \( \mathrm{OH^-} \) can significantly affect the solubility of \( \mathrm{Ni(OH)_2} \).
Understanding the solubility product constant is essential in predicting whether precipitation will occur under specific conditions.

pH Calculation

Calculating pH and its related parameter, pOH, is fundamental in understanding how acidic or basic a solution is. The pH of a solution is a measure of the hydrogen ion concentration \( [H^+] \), whereas pOH is related to the hydroxide ion concentration \( [OH^-] \).

These values are connected through their relationship with water's autoionization constant \( K_w \), which at room temperature is \( 1.0 \times 10^{-14} \). Thus, the formula:

\[ \text{pH} + \text{pOH} = 14 \]

Using this, the hydroxide ion concentration can be approached from a given pH.
For example, if the pH is 8.0:

• Determine the pOH: \( \text{pOH} = 14 - \text{pH} = 14 - 8 = 6 \).
• Calculate \( [\mathrm{OH^-}] \) from the pOH: \( [\mathrm{OH^-}] = 10^{-\mathrm{pOH}} = 10^{-6} \).

This approach sets the stage to determine the degree of dissociation for substances like \( \mathrm{Ni(OH)_2} \) within specific pH ranges.

Equilibrium Reactions

Equilibrium reactions represent a state in which the rates of the forward and reverse reactions are equal, maintaining a constant concentration of reactants and products over time. When dealing with solubility, understanding equilibrium helps clarify how a compound dissolves and the conditions under which it remains saturated.

In solubility contexts, upon dissolving \( \mathrm{Ni(OH)_2} \) in water, equilibrium is established:

\( \mathrm{Ni(OH)_2 (s) \rightleftharpoons Ni^{2+} (aq) + 2OH^- (aq)} \)

At this point, the rate at which \( \mathrm{Ni(OH)_2} \) dissolves balances with the rate at which \( \mathrm{Ni^{2+}} \) and \( \mathrm{OH^-} \) combine to form \( \mathrm{Ni(OH)_2} \), thereby stabilizing the concentrations involved in the \( K_{sp} \) expression.

• Adding \( \mathrm{OH^-} \) to the solution shifts the equilibrium left, decreasing solubility, according to Le Chatelier's principle.
• Changing the pH affects \( [\mathrm{OH^-}] \), hence shifting the equilibrium position.

Equilibrium reactions and their responses to changes in conditions, like pH, are vital in predicting and controlling chemical processes in both laboratory and industrial settings.