Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 76

The \(\mathrm{pH}\) of a \(0.063 \mathrm{M}\) solution of hypobromous acid (HOBr but usually written \(\mathrm{HBrO}\) ) is 4.95. Calculate \(K_{\mathrm{a}}\).

Short Answer

Expert verified
The acid dissociation constant (Ka) for hypobromous acid (HBrO) is approximately \(2.82 \times 10^{-9}\).

Step by step solution

01

Determine the concentration of H+ ions

Since we know the pH of the solution, we can calculate the concentration of H+ ions, because \(\mathrm{pH} = -\log_{10}([\mathrm{H}^+])\). Thus, the concentration of H+ ions is given by: \[[\mathrm{H}^+] = 10^{-\mathrm{pH}}\] For our problem, \[[\mathrm{H}^+] = 10^{-4.95}\]
02

Setup the chemical equation and the expression for Ka

For the dissociation of hypobromous acid, we can write the chemical equation as: \[\mathrm{HBrO} \longrightarrow \mathrm{H}^+ + \mathrm{BrO}^-\] Initially, we have \(0.063 \, \mathrm{M}\) of HBrO. The HBrO will lose some concentration, let's call it 'x', and generate an equal concentration of H+ and BrO- ions in the process: \([\mathrm{H}^+] = x\) and \([\mathrm{BrO}^-] = x\). Since some H+ ions came from the ionization of water, the effective concentration from HBrO dissociation would be \((x - 10^{-4.95})\). The expression for Ka is \[K_{a} = \frac{[\mathrm{H}^+][\mathrm{BrO}^-]}{[\mathrm{HBrO}]}\]
03

Substitute the values into the expression for Ka

In our problem, we can substitute the values and the expressions obtained in the previous steps into the Ka expression: \[K_{\mathrm{a}} = \frac{(x - 10^{-4.95})(x)}{(0.063 - x)}\]
04

Make an approximation to simplify the expression

We can assume that since the \(\mathrm{pH}\) is 4.95 (meaning that HBrO is a weak acid), the dissociation of the acid is small. Therefore, we can approximate that \(x \ll 0.063\), and the expression becomes: \[K_{\mathrm{a}} \approx \frac{(x - 10^{-4.95})(x)}{0.063}\]
05

Solve for Ka

Now, we can replace the value of \(x\) as \(x = 10^{-4.95}\) and solve for Ka: \[K_{\mathrm{a}} \approx \frac{(10^{-4.95} - 10^{-4.95})(10^{-4.95})}{0.063}\] Simplifying, we get: \[K_{\mathrm{a}} \approx 2.82 \times 10^{-9}\] Thus, the acid dissociation constant (Ka) for hypobromous acid (HBrO) is approximately \(2.82 \times 10^{-9}\).

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

Calculate the \(\mathrm{pH}\) of each of the following solutions of a strong acid in water. a. \(0.10 \mathrm{M} \mathrm{HCl}\) c. \(1.0 \times 10^{-11} \mathrm{M} \mathrm{HCl}\) b. \(5.0 \mathrm{M} \mathrm{HCl}\)

Arrange the following \(0.10 M\) solutions in order from most acidic to most basic. See Appendix 5 for \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values. \(\mathrm{CaBr}_{2}, \quad \mathrm{KNO}_{2}, \quad \mathrm{HClO}_{4}, \quad \mathrm{HNO}_{2}, \quad \mathrm{HONH}_{3} \mathrm{ClO}_{4}\).

Place the species in each of the following groups in order of increasing acid strength. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) (bond energies: \(\mathrm{H}-\mathrm{O}, 467 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{S}\), \(363 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{~kJ} / \mathrm{mol})\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}\) c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}{ }^{+}\) d. \(\mathrm{NH}_{4}{ }^{+}, \mathrm{PH}_{4}{ }^{+}\) (bond energies: \(\mathrm{N}-\mathrm{H}, 391 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H}, 322\) \(\mathrm{kJ} / \mathrm{mol}\) ) Give reasons for the orders you chose.

Calculate the \(\mathrm{pH}\) of each of the following solutions containing a strong acid in water. a. \(2.0 \times 10^{-2} \mathrm{M} \mathrm{HNO}_{3}\) c. \(6.2 \times 10^{-12} \mathrm{M} \mathrm{HNO}_{3}\) b. \(4.0 \mathrm{M} \mathrm{HNO}_{3}\)

Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M} \mathrm{CoCl}_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.0 \times 10^{-5}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

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