Chapter 15: Problem 58
A saturated solution of magnesium hydroxide is \(3.2 \times 10^{-4} \mathrm{M} \mathrm{Mg}(\mathrm{OH})_{2}\). What are the hydronium-ion and hydroxide-ion concentrations in the solution at \(25^{\circ} \mathrm{C} ?\)
Short Answer
Expert verified
\([\text{H}_3\text{O}^+] = 1.56 \times 10^{-11} \, \text{M}\), \([\text{OH}^-] = 6.4 \times 10^{-4} \, \text{M}\).
Step by step solution
01
Determine the Dissociation Equation
First, let's write the dissociation equation for magnesium hydroxide, \( \text{Mg(OH)}_2 \). It dissociates in water as follows:\[ \text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2\text{OH}^- (aq) \]From this equation, we see that one molecule of \( \text{Mg(OH)}_2 \) produces one \( \text{Mg}^{2+} \) ion and two \( \text{OH}^- \) ions.
02
Determine Hydroxide Ion Concentration
A saturated solution of \( \text{Mg(OH)}_2 \) has a magnesium ion concentration of \( [\text{Mg}^{2+}] = 3.2 \times 10^{-4} \, \text{M} \). According to the dissociation equation, for every mole of \( \text{Mg}^{2+} \), two moles of \( \text{OH}^- \) are produced.Thus, \([\text{OH}^-] = 2 \times 3.2 \times 10^{-4} \, \text{M} = 6.4 \times 10^{-4} \, \text{M}\).
03
Use the Water Ion Product
The water ion product at \( 25^\circ \text{C} \) is \( K_w = 1.0 \times 10^{-14} \). This relates the concentrations of hydronium ions \([\text{H}_3\text{O}^+]\) and hydroxide ions \([\text{OH}^-]\) as follows:\[ [\text{H}_3\text{O}^+] [\text{OH}^-] = K_w \]Using the previously calculated \([\text{OH}^-] = 6.4 \times 10^{-4} \, \text{M}\), rearrange to find \([\text{H}_3\text{O}^+]\):\[ [\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{6.4 \times 10^{-4}} \approx 1.56 \times 10^{-11} \, \text{M} \]
04
Verify the Logic and Units
Ensure the units in each calculation align. For \([\text{H}_3\text{O}^+]\), the value \(1.56 \times 10^{-11} \, \text{M}\) is reasonable considering the basicity due to generated \([\text{OH}^-] = 6.4 \times 10^{-4} \, \text{M}\). The calculations uphold the principle of chemical equilibrium and unit conservation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnesium Hydroxide Dissociation
Magnesium hydroxide, known chemically as \( \text{Mg(OH)}_2 \), dissociates in water through a reversible process. This dissociation is crucial to understand as it sets the stage for calculating ion concentrations in a solution. When \( \text{Mg(OH)}_2 \) enters water, it separates into magnesium ions \( (\text{Mg}^{2+}) \) and hydroxide ions \( (\text{OH}^-) \). It's represented by the following balanced chemical equation:
- \[ \text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2\text{OH}^- (aq) \]
Ion Product of Water
The ion product of water, often represented as \( K_w \), is a key concept in chemistry, especially for aqueous solutions. It is defined as the product of the molar concentrations of hydronium ions \( (\text{H}_3\text{O}^+) \) and hydroxide ions \( (\text{OH}^-) \) in pure water. At \( 25^\circ \text{C} \), this product always equals \( 1.0 \times 10^{-14} \):
- \[ [\text{H}_3\text{O}^+] [\text{OH}^-] = K_w = 1.0 \times 10^{-14} \]
Hydroxide Ion Concentration
In a saturated magnesium hydroxide solution, understanding the hydroxide ion concentration is vital. According to the dissociation of \( \text{Mg(OH)}_2 \), each mole of magnesium produces two moles of \( \text{OH}^- \). So, if the concentration of magnesium ions \( (\text{Mg}^{2+}) \) is \(3.2 \times 10^{-4} \, \text{M}\), the hydroxide ion concentration would be:
- \[ [\text{OH}^-] = 2 \times 3.2 \times 10^{-4} \, \text{M} = 6.4 \times 10^{-4} \, \text{M} \]
Hydronium Ion Concentration
Hydronium ion concentration \( ([\text{H}_3\text{O}^+]) \) is an essential parameter in characterizing a solution's acidity. In a typical calculation involving the ion product of water, if the hydroxide ion concentration \( ([\text{OH}^-]) \) is known, you can find the hydronium ion concentration using the formula:
- \[ [\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]} \]
- \[ [\text{H}_3\text{O}^+] = \frac{1.0 \times 10^{-14}}{6.4 \times 10^{-4}} \approx 1.56 \times 10^{-11} \, \text{M} \]