Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 2

Practise using the Henderson-Hasselbalch equation. What are the relative proportions of the deprotonated \(\left(A^{-}\right)\)and protonated (HA) forms of each substance at the following \(\mathrm{pH}\) values: (a) ethanoic acid \(\left(\mathrm{p} K_{\mathrm{a}}=4.8\right.\) ) for use in an experiment at pH 3.8; (b) boric acid \(\left(\mathrm{p} K_{\mathrm{a}}=9.2\right)\) for use in an experiment at pH 9.5; (a) HEPES \(\left(p K_{a}=7.5\right)\) for use in an experiment at \(\mathrm{pH}\) 8.1?

Short Answer

Expert verified
The relative proportions of the deprotonated (A-) and protonated (HA) forms for each substance at the given pH values are as follows: for ethanoic acid, the ratio is 1 to 10;for boric acid, it's 2 to 1; and for HEPES, it's 4 to 1.

Step by step solution

01

Identify given and required parameters

In this problem, the given parameters for each item are: for ethanoic acid, pKa = 4.8, pH = 3.8; for boric acid, pKa = 9.2, pH = 9.5; and for HEPES, pKa = 7.5, pH = 8.1. The required parameters for each item are the relative proportions of A- and HA, which we can find by rearranging the Henderson-Hasselbalch equation to log([A-]/[HA]) = pH - pKa.
02

Apply the Henderson-Hasselbalch equation to find [A-]/[HA] for ethanoic acid

Plugging the given values for ethanoic acid into the rearranged Henderson-Hasselbalch equation, we get log([A-]/[HA]) = 3.8 - 4.8 = -1. Therefore, [A-]/[HA] = 10^(-1) = 0.1, which means that the ratio of the deprotonated form to the protonated form for ethanoic acid is 0.1 to 1 or 1 to 10.
03

Apply the Henderson-Hasselbalch equation to find [A-]/[HA] for boric acid

Using the given values for boric acid in the rearranged Henderson-Hasselbalch equation, we find log([A-]/[HA]) = 9.5 - 9.2 = 0.3. Therefore, [A-]/[HA] = 10^(0.3) ≈ 2, which means that the ratio of the deprotonated form to the protonated form for boric acid is 2 to 1.
04

Apply the Henderson-Hasselbalch equation to find [A-]/[HA] for HEPES

With the given values for HEPES, the Henderson-Hasselbalch equation gives log([A-]/[HA]) = 8.1 - 7.5 = 0.6. Therefore, [A-]/[HA] = 10^(0.6) ≈ 4, which indicates that the ratio of the deprotonated form to the protonated form for HEPES is 4 to 1.

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

Practise interconverting pH values and proton concentrations. Express all values to four significant figures. (a) What is a pH \(7.4\) expressed as \(\left[\mathrm{H}^{+}\right]\)in \(\mathrm{mol} \mathrm{L}^{-1} ?\) (b) What is a pH \(4.1\) expressed as \(\left[\mathrm{H}^{+}\right]\)in mol \(L^{-1} ?\) (c) What is the \(\mathrm{pH}\) of a solution of \(\left[\mathrm{H}^{+}\right]\)at \(2 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} ?\) (d) What is the pH of a solution of \(\left[\mathrm{H}^{+}\right]\)at \(2 \times 10^{-12.5} \mathrm{~mol} \mathrm{~L}^{-1} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

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