/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Determine the hydrogen ion or hy... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine the hydrogen ion or hydroxide ion concentration in each of the following solutions, as appropriate. (a) a solution in which \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=4.5 \times 10^{-4} M\) (b) a solution in which \(\left[\mathrm{OH}^{-}\right]=8.33 \times 10^{-5} \mathrm{M}\)

Short Answer

Expert verified
(a) [OH^-] ≈ 2.22 × 10^{-11} M; (b) [H_3O^+] ≈ 1.20 × 10^{-10} M.

Step by step solution

01

Understanding the Problem

We need to find either the hydroxide ion concentration \([\text{OH}^-]\) from the given hydronium ion concentration \([\text{H}_3\text{O}^+]\), or the hydronium ion concentration from the given hydroxide ion concentration. The solutions are based on the ion-product constant for water at 25°C: \(K_w = 1.0 \times 10^{-14}\).
02

Calculating Hydroxide Ion Concentration (Part a)

For part (a), we are given \([\text{H}_3\text{O}^+] = 4.5 \times 10^{-4} \text{ M}\). We use the formula: \([\text{OH}^-] = \frac{K_w}{[\text{H}_3\text{O}^+]}\). Substituting the values, \([\text{OH}^-] = \frac{1.0 \times 10^{-14}}{4.5 \times 10^{-4}}\) which equals approximately \(2.22 \times 10^{-11} \text{ M}\).
03

Calculating Hydronium Ion Concentration (Part b)

For part (b), we are given \([\text{OH}^-] = 8.33 \times 10^{-5} \text{ M}\). We use the formula: \([\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]}\). Substituting the values, \([\text{H}_3\text{O}^+] = \frac{1.0 \times 10^{-14}}{8.33 \times 10^{-5}}\) which equals approximately \(1.20 \times 10^{-10} \text{ M}\).

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Key Concepts

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

Hydronium Ion Concentration
Hydronium ions, \(\text{H}_3\text{O}^+\), play a key role in determining the acidity of a solution. This concentration represents the presence of free hydrogen ions in water that are typically associated with \(\text{H}_2\text{O}\) molecules to form hydronium ions. Measuring \(\text{H}_3\text{O}^+\) concentrations helps us understand how acidic or basic a solution is.

When given the \(\text{OH}^-\) concentration, we can easily find the \(\text{H}_3\text{O}^+\) concentration using the ion-product constant for water. In part (b) of the exercise, you start with \(\left[\text{OH}^-\right] = 8.33 \times 10^{-5} \, \text{M}\). The hydronium ion concentration is calculated by using the formula:
  • \(\left[\text{H}_3\text{O}^+\right] = \frac{K_w}{\left[\text{OH}^-\right]}\)
Inserting the known values, \(\left[\text{H}_3\text{O}^+\right] = \frac{1.0 \times 10^{-14}}{8.33 \times 10^{-5}}\), which results in approximately \(1.20 \times 10^{-10} \, \text{M}\). This shows a very low concentration of hydronium ions, indicating that the solution is slightly basic.
Hydroxide Ion Concentration
Hydroxide ions, \(\text{OH}^-\), are crucial in determining the basicity of a solution. The balance between hydroxide and hydronium ions defines the pH of a solution.

In the exercise, given the hydronium ion concentration \(\left[\text{H}_3\text{O}^+\right] = 4.5 \times 10^{-4} \, \text{M}\) in part (a), you are asked to find \(\left[\text{OH}^-\right]\). The process uses the relationship based on the ion-product constant:
  • \(\left[\text{OH}^-\right] = \frac{K_w}{\left[\text{H}_3\text{O}^+\right]}\)
By substituting the given values into the equation, \(\left[\text{OH}^-\right] = \frac{1.0 \times 10^{-14}}{4.5 \times 10^{-4}}\), which results in approximately \(2.22 \times 10^{-11} \, \text{M}\). This low hydroxide concentration suggests that the solution is acidic.
Ion-Product Constant of Water
The ion-product constant of water, \(K_w\), is fundamental in understanding the self-ionization of water and its impact on hydronium and hydroxide ion concentrations. \(K_w\) is defined as the product of the molar concentrations of these ions in pure water at 25°C:
  • \(K_w = \left[\text{H}_3\text{O}^+\right] \times \left[\text{OH}^-\right] = 1.0 \times 10^{-14}\)
This constant value tells us that at 25°C, any aqueous solution will have the ion product of hydronium and hydroxide ions equal to \(1.0 \times 10^{-14} \, \text{M}^2\).

The constant is vital for solving both parts of the exercise since it allows for the calculation of one ion concentration when the other is known. This interrelationship ensures that when one concentration increases, the other decreases, maintaining the balance dictated by \(K_w\). Thus, water can self-ionize to a very limited extent, but this equilibrium impacts all aqueous acid-base reactions.

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Most popular questions from this chapter

Can a compound be an Arrhenius base and not a Brønsted-Lowry base? Explain your answer.

A solution is made by dissolving \(15.0 \mathrm{~g}\) sodium hydroxide in approximately \(450 \mathrm{~mL}\) water. The solution becomes quite warm, but after it is allowed to return to room temperature, water is added to bring the volume to \(500.0 \mathrm{~mL}\) of solution. (a) Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) in the final solution. (b) Why would we wait for it to return to room temperature? (c) If the mass of the water used to initially dissolve the sodium hydroxide were exactly \(450 \mathrm{~g}\) and the temperature of the water increased by \(8.865^{\circ} \mathrm{C}\), how much heat was given off by the dissolution of \(15.0 \mathrm{~g}\) of solute? Assume the specific heat of the solution is \(4.184 \mathrm{~J} / \mathrm{g} \cdot \mathrm{K}\). What is the molar heat change for the dissolution of sodium hydroxide (known as the enthalpy of solution, \(\left.\Delta H_{\mathrm{sol}}\right) ?\)

A hypothetical weak base has \(K_{b}=5.0 \times 10^{-4}\). Calculate the equilibrium concentrations of the base, its conjugate acid, and \(\mathrm{OH}^{-}\) in a \(0.15 \mathrm{M}\) solution of the base.

Choose from among the labels strongly acidic, weakly acidic, neutral, weakly basic, and strongly basic to estimate the \(\mathrm{pH}\) of the following solutions. (a) \(0.45 M \mathrm{NaCl}\) (b) \(0.18 M \mathrm{BaF}_{2}\) (c) \(0.25 \mathrm{M} \mathrm{KHSO}_{4}\) (d) \(0.33 \mathrm{M} \mathrm{NaNO}_{2}\)

Rank the following species in order of increasing acidity: \(\mathrm{NH}_{4}^{+}, \mathrm{H}_{2} \mathrm{O}, \mathrm{HF}, \mathrm{HSO}_{4}^{-}\)

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