Question

The \(\mathrm{pH}\) of a saturated solution of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is 9.8 . Determine the \(K_{\mathrm{sp}}\) for \(\mathrm{Mg}(\mathrm{OH})_{2}\)

Short answer

\(K_{\mathrm{sp}} = [Mg^{2+}][OH^-]^2 = (10^{-4.2}/2)(2 \cdot 10^{-4.2})^2 = 1.6 \times 10^{-11}\)

Step-by-step solution

Understanding pH and pOH

The \(\mathrm{pH}\) of a solution is a measure of its acidity or basicity. It is related to the concentration of hydrogen ions \(\left[ H^+ \right]\) in the solution. Since the given \(\mathrm{pH}\) is 9.8, we can find the pOH by the relation \(\mathrm{pOH} = 14 - \mathrm{pH}\). This will give us the \(\mathrm{pOH}\) of the solution.

Calculate pOH

Subtracting the \(\mathrm{pH}\) from 14 gives us the \(\mathrm{pOH}\): \(\mathrm{pOH} = 14 - 9.8 = 4.2\).

Calculate \(\left[ OH^- \right]\)

Using the pOH, calculate the concentration of hydroxide ions \(\left[ OH^- \right]\) in the solution by the formula \(\left[ OH^- \right] = 10^{-\mathrm{pOH}}\).

Determine Hydroxide Ion Concentration

Substitute the value of pOH to get \(\left[ OH^- \right] = 10^{-4.2}\).

Write the Dissociation Expression of \(Mg(OH)_2\)

The dissociation of \(Mg(OH)_2\) in water can be represented as: \([Mg(OH)_2 (s) \leftrightarrows Mg^{2+} (aq) + 2 OH^- (aq)]\).

Set up the Expression for \(K_{sp}\)

The solubility product constant, \(K_{sp}\), for \(Mg(OH)_2\) is given by \(K_{sp}=[Mg^{2+}][OH^-]^2\).

Find the Concentration of \(Mg^{2+}\)

Since every mole of \(Mg(OH)_2\) produces one mole of \(Mg^{2+}\) and two moles of \(OH^-\), if \(s\) is the solubility of \(Mg(OH)_2\), then \(\left[ Mg^{2+} \right] = s\) and \(\left[ OH^- \right] = 2s\). But from Step 4, we know \(\left[ OH^- \right]\), hence \(s = \frac{\left[ OH^- \right]}{2}\).

Calculate the Value of \(s\)

Substitute \(\left[ OH^- \right]\) into the expression for \(s\) to find its value.

Calculate \(K_{sp}\)

Substitute the values of \(\left[ Mg^{2+} \right]\) and \(\left[ OH^- \right]\) into the expression for \(K_{sp}\) to calculate its value.

Solve for \(K_{sp}\)

Finally, plug in the value of \(s\) into the \(K_{sp}\) expression to obtain the solubility product for \(Mg(OH)_2\).

Key concepts

pH and pOH Calculations

Understanding the relationship between pH and pOH is crucial when working with solutions in chemistry. pH is the logarithmic scale used to specify the acidity or basicity of an aqueous solution. It is inversely related to the concentration of hydrogen ions \( [H^+] \), with lower pH values indicating higher acidity. Conversely, pOH measures the basicity of a solution and relates to the hydroxide ion concentration \( [OH^-] \).

The pH scale ranges from 0 to 14 in pure water, where a pH of 7 is considered neutral. A pH less than 7 indicates an acidic solution, while a pH greater than 7 indicates a basic solution. The relationship between pH and pOH is simple but powerful: together they always add up to 14 at 25°C due to the water dissociation constant \(K_w\). Thus, knowing one immediately allows the calculation of the other using the formula \( pOH = 14 - pH \).

Additionally, the inverse logarithmic scale means that each whole number change in pH represents a tenfold change in hydrogen ion concentration. This helps explain how a seemingly small change in pH can translate to a substantial change in the chemistry of a solution.

Hydroxide Ion Concentration

Hydroxide ion concentration is a direct measure of the basicity of a solution and plays a key role in understanding solubility and reaction dynamics in chemistry. The concentration of hydroxide ions \( [OH^-] \) in a solution can be quickly calculated if the pOH is known. The formula \( [OH^-] = 10^{-pOH} \) reveals that as the pOH decreases, the hydroxide ion concentration increases.

This is because the pOH is also a logarithmic measure, similar to pH, but instead of reflecting hydrogen ion concentration, it represents hydroxide ion concentration. For a basic solution, a lower pOH corresponds to a higher concentration of \( [OH^-] \), indicating a stronger base. Knowing \( [OH^-] \) is essential when working with solubility product constants \( K_{sp} \) in precipitate formation, and in titration calculations where neutralization reactions take place. Typically, hydroxide ion concentration plays a pivotal role in determining the solubility of compounds in water, as well as the direction and extent of chemical equilibrium in aqueous reactions.

Dissociation of Mg(OH)2

When magnesium hydroxide \( Mg(OH)_2 \) is dissolved in water, it dissociates to form magnesium ions \( Mg^{2+} \) and hydroxide ions \( OH^- \). This dissociation can be described by the equilibrium equation: \[ Mg(OH)_2 (s) \leftrightarrows Mg^{2+} (aq) + 2 OH^- (aq) \].

The extent of this dissociation is governed by the solubility product constant \( K_{sp} \), which for \( Mg(OH)_2 \) is expressed as \( K_{sp} = [Mg^{2+}][OH^-]^2 \). \( K_{sp} \) is a temperature-dependent value that indicates the maximum amount of \( Mg(OH)_2 \) that can dissolve in water to reach saturation without forming a precipitate. If the product of the ionic concentrations exceeds \( K_{sp} \) under given conditions, additional \( Mg(OH)_2 \) will not dissolve and instead will precipitate out of solution.

In solving solubility problems, it's important to remember that the stoichiometry of the dissociation reaction directly affects the concentrations of the ions produced. For each mole of \( Mg(OH)_2 \) that dissolves, it yields one mole of \( Mg^{2+} \) and two moles of \( OH^- \) ions. Therefore, if the solubility of \( Mg(OH)_2 \) is represented as \( s \), we have \( [Mg^{2+}] = s \) and \( [OH^-] = 2s \). The solubility product must always be considered in light of the molar relationships established by the chemical equation representing dissolution.