Problem 131 - Chemistry: An Atoms-Focused Approach

Question

Which has a higher \(\mathrm{pH}\), a saturated solution of magnesium hydroxide \(\left(K_{\mathrm{sp}}=5.6 \times 10^{-12}\right)\) or a saturated solution of calcium hydroxide \(\left(K_{\mathrm{sp}}=4.7 \times 10^{-6}\right) ?\)

Step-by-Step Solution

Answer

The saturated solution of calcium hydroxide has a higher pH value (12.28) compared to the saturated solution of magnesium hydroxide (10.52).

Step 1: Write the balanced chemical equations

For the dissociation of magnesium hydroxide and calcium hydroxide, we have: $$\mathrm{Mg(OH)_2} \leftrightarrows \mathrm{Mg^{2+}} + 2 \mathrm{OH^-}$$ $$\mathrm{Ca(OH)_2} \leftrightarrows \mathrm{Ca^{2+}} + 2 \mathrm{OH^-}$$

Step 2: Find the concentration of hydroxide ions

For magnesium hydroxide, let the solubility of \(\mathrm{Mg(OH)_2}\) be \(s\). Then, the concentration of \(\mathrm{Mg^{2+}}\) is also \(s\), and the concentration of \(\mathrm{OH^-}\) is \(2s\). We can write the expression for \(K_{sp}\) as: $$K_{sp} = [\mathrm{Mg^{2+}}] [OH^-]^2 = s(2s)^2$$ Now, substitute the given value of \(K_{sp} = 5.6 \times 10^{-12}\) and solve for \(s\): $$5.6 \times 10^{-12} = s(2s)^2$$ $$s \approx 1.67 \times 10^{-4}$$ The concentration of \(\mathrm{OH^-}\) in the magnesium hydroxide solution is \(2s \approx 3.33 \times 10^{-4} \mathrm{M}\). For calcium hydroxide, let the solubility be \(s'\). Then, the concentration of \(\mathrm{Ca^{2+}}\) is \(s'\), and the concentration of \(\mathrm{OH^-}\) is \(2s'\). Write the solubility product expression as: $$K_{sp} = [\mathrm{Ca^{2+}}] [OH^-]^2 = s'(2s')^2$$ Substitute the given value of \(K_{sp} = 4.7 \times 10^{-6}\) and solve for \(s'\): $$4.7 \times 10^{-6} = s'(2s')^2$$ $$s' \approx 9.47 \times 10^{-3}$$ The concentration of \(\mathrm{OH^-}\) in the calcium hydroxide solution is \(2s' \approx 1.89 \times 10^{-2} \mathrm{M}\).

Step 3: Calculate the pH of both solutions

First, we will find the \(\mathrm{pOH}\) of the solutions using the equation \(\mathrm{pOH} = -\log_{10}[\mathrm{OH}^-]\), and then convert it to \(\mathrm{pH}\) using the equation \(\mathrm{pH} = 14 - \mathrm{pOH}\). For magnesium hydroxide: $$\mathrm{pOH} = -\log_{10}(3.33 \times 10^{-4}) \approx 3.48$$ $$\mathrm{pH} = 14 - 3.48 \approx 10.52$$ For calcium hydroxide: $$\mathrm{pOH} = -\log_{10}(1.89 \times 10^{-2}) \approx 1.72$$ $$\mathrm{pH} = 14 - 1.72 \approx 12.28$$

Step 4: Compare the pH values and identify the solution with the higher pH

Comparing the \(\mathrm{pH}\) values, we can see that the saturated solution of calcium hydroxide has a higher \(\mathrm{pH}\) (\(12.28\)) compared to the saturated solution of magnesium hydroxide (\(10.52\)).

Key Concepts

Dissociation Equation

In the process of dissolution, a compound breaks down into its constituent ions in a solution. For magnesium hydroxide, this process is represented by the chemical equation:\[ \text{Mg(OH)}_2 \leftrightarrows \text{Mg}^{2+} + 2 \text{OH}^- \]For calcium hydroxide, it follows the equation:\[ \text{Ca(OH)}_2 \leftrightarrows \text{Ca}^{2+} + 2 \text{OH}^- \]These equations show that for every mole of magnesium or calcium hydroxide that dissolves, two moles of hydroxide ions \( \text{OH}^- \) are produced.

• The dissociation equation helps us understand how many ions are formed in solution when the solid dissolves.
• Note that both compounds release two hydroxide ions per formula unit, which is crucial when calculating the hydroxide ion concentration.

By studying these equations, we can predict the behavior of the compounds in water, especially how basic or alkaline the resulting solution might become.

Hydroxide Ion Concentration

Calculating hydroxide ion concentration is key to understanding a solution’s basicity. For magnesium hydroxide, let's denote the solubility as \( s \). The concentration of \( \text{Mg}^{2+} \) ions becomes \( s \), and that of \( \text{OH}^- \) ions is \( 2s \). This stems from the dissociation equation where each mole of \( \text{Mg(OH)}_2 \) yields 2 hydroxide ions. The solubility product constant \( K_{sp} \) is defined as:\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 = s(2s)^2 \]By substituting the known \( K_{sp} \) value, we find \( s \), and hence \( 2s \) gives us the concentration of \( \text{OH}^- \) ions.For calcium hydroxide, a similar approach follows, where:\[ K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 = s'(2s')^2 \]Substituting the value of \( K_{sp} \) helps us determine the \( \text{OH}^- \) concentration as \( 2s' \).

• The higher the concentration of \( \text{OH}^- \), the more basic the solution.
• Magnesium hydroxide has a smaller \( K_{sp} \) compared to calcium hydroxide, resulting in less dissociation and thus a lower \( \text{OH}^- \) concentration.

pH Calculation

To assess a solution's acidity or basicity, pH is a useful scale. It relates to hydroxide ion concentration via the formula:\[ \text{pOH} = -\log_{10}([\text{OH}^-]) \]\[ \text{pH} = 14 - \text{pOH} \]For magnesium hydroxide:1. Calculate \( \text{pOH} \): \[ \text{pOH} = -\log_{10}(3.33 \times 10^{-4}) \approx 3.48 \]2. Find \( \text{pH} \): \[ \text{pH} = 14 - 3.48 = 10.52 \]For calcium hydroxide:1. Calculate \( \text{pOH} \): \[ \text{pOH} = -\log_{10}(1.89 \times 10^{-2}) \approx 1.72 \]2. Find \( \text{pH} \): \[ \text{pH} = 14 - 1.72 = 12.28 \]

• The solution with the higher \( \text{OH}^- \) concentration corresponds to a higher pH, indicating a stronger basicity.
• In this case, calcium hydroxide, with the greater \( \text{OH}^- \) concentration, exhibits a higher pH than magnesium hydroxide.