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Chapter 15: Problem 94

A solution is \(0.35 \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\). What are the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{OH}^{-}\) in this solution?

Short Answer

Expert verified
Concentration of \(\text{OH}^-\) is 0.70 M, and \(\text{H}_3\text{O}^+\) is \(1.43 \times 10^{-14} \, \text{M}\).

Step by step solution

01

Understand the Dissociation of Sr(OH)2

When strontium hydroxide, \( \text{Sr(OH)}_2 \), dissolves in water, it dissociates completely into one strontium ion \( \text{Sr}^{2+} \) and two hydroxide ions \( \text{OH}^- \) for each formula unit: \[ \text{Sr(OH)}_2 \rightarrow \text{Sr}^{2+} + 2\text{OH}^- \]. This means that for every \(1 \text{ M}\) of \(\text{Sr(OH)}_2\), we get \(2 \text{ M}\) of \(\text{OH}^-\).
02

Calculate the OH鈦� Concentration

Since the solution is \(0.35 \, \text{M} \text{Sr(OH)}_2 \), and each mole of \(\text{Sr(OH)}_2\) produces \(2 \text{ mol}\) of \(\text{OH}^-\), the concentration of \(\text{OH}^-\) ions will be: \[ \text{OH}^- \text{ concentration} = 2 \times 0.35 \, \text{M} = 0.70 \, \text{M} \].
03

Calculate the H鈧僌鈦� Concentration

We can use the water dissociation constant, \(K_w\), which is \(1.0 \times 10^{-14}\) at 25掳C, to find the \(\text{H}_3\text{O}^+\) concentration. Since \(\text{H}_3\text{O}^+ \times \text{OH}^- = K_w\), we rearrange to find \(\text{H}_3\text{O}^+\): \[ \text{H}_3\text{O}^+ = \frac{K_w}{\text{OH}^-} = \frac{1.0 \times 10^{-14}}{0.70} = 1.43 \times 10^{-14} \, \text{M} \].

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Key Concepts

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

Dissociation of Compounds
When compounds like strontium hydroxide (\( \text{Sr(OH)}_2 \)) are added to water, they tend to separate into their individual ions. This process is known as dissociation. For each molecule of \( \text{Sr(OH)}_2 \), it dissociates into one strontium ion (\( \text{Sr}^{2+} \)) and two hydroxide ions (\( \text{OH}^- \)).
  • The result is a complete break into separate ions once in a solution.
  • Each ion now behaves independently, contributing to the overall concentration of ions in the solution.
For example, if a solution has a concentration of \(0.35 \, \text{M} \) of \( \text{Sr(OH)}_2 \), it will produce double that concentration of \( \text{OH}^- \) ions, leading to \(0.70 \, \text{M} \) hydroxide ions. This principle of dissociation helps us calculate how much of each ion type is present in a solution.
Hydroxide Ion Concentration
Hydroxide ions (\( \text{OH}^- \)) play a crucial role in determining the basic nature of a solution. When a compound like \( \text{Sr(OH)}_2 \) dissolves, each molecule releases two hydroxide ions.
  • This directly multiplies the concentration of hydroxide ions in the solution by two.
  • Thus, the overall hydroxide concentration becomes twice that of the dissolved \( \text{Sr(OH)}_2 \) concentration.
In our example, with \(0.35 \, \text{M} \) \( \text{Sr(OH)}_2 \), the hydroxide ion concentration becomes \(0.70 \, \text{M} \). This increase is fundamental in calculating other properties of the solution, such as its pH level and the concentration of hydronium ions.
Water Dissociation Constant
The water dissociation constant, \(K_w\), is a key value that represents the product of the hydronium ion and hydroxide ion concentrations in water. At room temperature (25掳C), \(K_w\) has a value of \(1.0 \times 10^{-14}\).
  • It is used to understand the relation between \( \text{H}_3\text{O}^+ \) and \( \text{OH}^- \) concentrations in any aqueous solution.
  • Whenever we know one of these concentrations, we can calculate the other using \(K_w\).
For instance, if the hydroxide ion concentration is \(0.70 \, \text{M} \), the corresponding hydronium ion concentration can be calculated as:\[ \text{H}_3\text{O}^+ = \frac{K_w}{\text{OH}^-} = \frac{1.0 \times 10^{-14}}{0.70} = 1.43 \times 10^{-14} \, \text{M} \]This relationship explains how even in basic solutions, where \( \text{OH}^- \) is high, \( \text{H}_3\text{O}^+ \) still exists, and the product of the concentrations remains constant at \(K_w\).

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Most popular questions from this chapter

Give the conjugate base to each of the following species regarded as acids. a. \(\mathrm{HSeO}_{4}^{-}\) b. \(\mathrm{PH}_{4}^{+}\) c. HS \(^{-}\) d. \(\mathrm{HOCl}\)

Aqueous solutions of ammonia, \(\mathrm{NH}_{3}\), were once thought to be solutions of an ionic compound (ammonium hydroxide, \(\left.\mathrm{NH}_{4} \mathrm{OH}\right)\) in order to explain how the solutions could contain hydroxide ion. Using the Br酶nsted-Lowry concept, show how \(\mathrm{NH}_{3}\) yields hydroxide ion in aqueous solution without involving the species \(\mathrm{NH}_{4} \mathrm{OH}\)

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