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Chapter 16: Problem 131

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 \(\left(K_{\mathrm{sp}}=4.7 \times 10^{-6}\right) ?\)

Short Answer

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

Step by step solution

01

Write the balanced chemical equations

For the dissociation of magnesium hydroxide and calcium hydroxide, we have: $$\mathrm{Mg(OH)_2} \leftrightarrows \mathrm{Mg^{2+}} + 2 \mathrm{OH^-}$$ $$\mathrm{Ca(OH)_2} \leftrightarrows \mathrm{Ca^{2+}} + 2 \mathrm{OH^-}$$
02

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 \(2s' \approx 1.89 \times 10^{-2} \mathrm{M}\).
03

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: $$\mathrm{pOH} = -\log_{10}(1.89 \times 10^{-2}) \approx 1.72$$ $$\mathrm{pH} = 14 - 1.72 \approx 12.28$$
04

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 (\(10.52\)).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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.

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Most popular questions from this chapter

A cook dissolves a teaspoon of baking soda (NaHCO \(_{3}\) ) in a cup of water and then discovers that the recipe calls for a tablespoon, not a teaspoon. If the cook adds two more teaspoons of baking soda to make up the difference, does the additional baking soda change the pH of the solution? Explain why or why not.

Chromium(III) hydroxide is amphiprotic. Write chemical equations showing how an aqueous suspension of this compound reacts to the addition of a strong acid and the addition of a strong base.

Calculate the \(\mathrm{pH}\) at the equivalence point for the titration of \(10.0 \mathrm{mL}\) of \(0.100 M\) formic acid \(\left(K_{\mathrm{a}}=1.77 \times 10^{-4}\right)\) with \(0.100 M\) NaOH and for the titration of 10.0 mL of boric acid \(\left(K_{a 1}=5.4 \times 10^{-10}\right)\) with the same solution. Should the same indicator be used for both titrations?

Antacid Tablets Antacids contain a variety of bases such as \(\mathrm{NaHCO}_{3}, \mathrm{MgCO}_{3}, \mathrm{CaCO}_{3},\) and \(\mathrm{Mg}(\mathrm{OH})_{2} .\) Only \(\mathrm{NaHCO}_{3}\) has appreciable solubility in water. a. Write a net ionic equation for the reaction of each base with aqueous HCl. b. Explain how substances sparingly soluble in water can act as effective antacids.

A pH 3.00 buffer is prepared by mixing solutions of nitrous acid and sodium nitrite. The total concentration of nitrous acid and sodium nitrate in the buffer is \(0.100 M\) Suppose 1.00 milliliter of \(1.00 M \mathrm{HCl}\) is added to a \(100 \mathrm{mL}\) sample of the buffer and 1.00 milliliter of \(1.00 M\) \(\mathrm{NaOH}\) is added to another \(100 \mathrm{mL}\) sample. In which sample would the addition of strong acid or base produce the greater change in \(\mathrm{pH}\) ? Explain your selection.
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