Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 63

Calculate the concentration of all species present and the \(\mathrm{pH}\) of a \(0.020 M \mathrm{HF}\) solution.

Short Answer

Expert verified
In a 0.020 M HF solution, the equilibrium concentrations are approximately: [HF] ≈ 0.0162 M, [H+] ≈ 3.77 x 10⁻³ M, and [F-] ≈ 3.77 x 10⁻³ M. The pH of the solution is approximately 2.42.

Step by step solution

01

Write the dissociation equilibrium for HF and expression for Ka

The dissociation equilibrium for HF in water can be represented as follows: \( HF (aq) \rightleftharpoons H^+ (aq) + F^- (aq) \) Now, we can write the equilibrium constant expression (Ka) for the dissociation of HF: \(K_a = \frac{[H^+][F^-]}{[HF]}\)
02

Calculate the change in concentrations for reactants and products during the dissociation process

Let x be the change in concentrations during the equilibrium process, then: - The initial concentration of HF is 0.020 M. - At equilibrium, the concentration of HF has decreased by x M, so [HF] = 0.020 - x M. - At equilibrium, the concentration of H+ has increased by x M, so [H+] = x M. - At equilibrium, the concentration of F- has also increased by x M, so [F-] = x M.
03

Substitute equilibrium concentrations into the Ka expression and solve for x

The Ka value for HF is 7.1 x 10⁻⁴. Substituting the equilibrium concentrations into the Ka expression: \(7.1 \times 10^{-4} = \frac{x^2}{0.020 - x}\) To simplify the calculation, since x is very small compared to 0.020 M, we can assume (0.020 - x) ≈ 0.020: \(7.1 \times 10^{-4} ≈ \frac{x^2}{0.020}\) Now, we can solve for x: \(x \approx \sqrt{7.1 \times 10^{-4} \times 0.020} \approx \sqrt{1.42 \times 10^{-5}} \approx 3.77 \times 10^{-3}\)
04

Calculate the equilibrium concentrations of all species and the pH

Now, we can calculate the equilibrium concentrations of each species: - [HF] ≈ 0.020 - 3.77 x 10⁻³ ≈ 0.0162 M - [H+] ≈ 3.77 x 10⁻³ M - [F-] ≈ 3.77 x 10⁻³ M Finally, we can calculate the pH of the solution using the concentration of H+ ions: \(pH = -\log_{10}[H^+]\) \(pH \approx -\log_{10}(3.77 \times 10^{-3}) \approx 2.42\)
05

Present the solution

In the 0.020 M HF solution: - The equilibrium concentration of HF is ≈ 0.0162 M - The equilibrium concentration of H+ ions is ≈ 3.77 x 10⁻³ M - The equilibrium concentration of F- ions is ≈ 3.77 x 10⁻³ M - The pH of the solution is approximately 2.42.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What are the major species present in \(0.015 M\) solutions of each of the following bases? a. \(\mathrm{KOH}\) b. \(\mathrm{Ba}(\mathrm{OH})_{2}\) What is \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of each of these solutions?

What are the major species present in the following mixtures of bases? a. \(0.050 \mathrm{M} \mathrm{NaOH}\) and \(0.050 \mathrm{M} \mathrm{LiOH}\) b. \(0.0010 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) and \(0.020 \mathrm{M} \mathrm{RbOH}\) What is \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of each of these solutions?

Calculate \(\left[\mathrm{OH}^{-}\right]\) in a solution obtained by adding \(0.0100 \mathrm{~mol}\) solid \(\mathrm{NaOH}\) to \(1.00 \mathrm{~L}\) of \(15.0 \mathrm{M} \mathrm{NH}_{3}\).

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.12 \mathrm{M} \mathrm{KNO}_{2}\) c. \(0.40 \mathrm{M} \mathrm{NH}_{4} \mathrm{ClO}_{4}\) b. \(0.45 \mathrm{M} \mathrm{NaOCl}\)

Place the species in each of the following groups in order of increasing base strength. Give your reasoning in each case. a. \(\mathrm{IO}_{3}^{-}, \mathrm{Br} \mathrm{O}_{3}^{-}\) b. \(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}\) c. \(\mathrm{OCl}^{-}, \mathrm{OI}^{-}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free