Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 97

Calculate \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right],\) and the \(\mathrm{pH}\) of \(0.20 M\) solutions of each of the following amines. a. triethylamine \(\left[\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{N}, K_{\mathrm{b}}=4.0 \times 10^{-4}\right]\) b. hydroxylamine (HONH \(\left._{2}, K_{\mathrm{b}}=1.1 \times 10^{-8}\right)\)

Short Answer

Expert verified
a. For the triethylamine solution: - [OH-] = 2.0 * 10^-2 M - [H+] = 5.0 * 10^-13 M - pH = 12.3 b. For the hydroxylamine solution: - [OH-] = 4.7 * 10^-4 M - [H+] = 2.1 * 10^-11 M - pH = 10.7

Step by step solution

01

Write the base-dissociation reaction and Kb expression for each amine

First, we need to write the base dissociation reaction and the Kb expression for each amine. a. For triethylamine, (C2H5)3N: Base dissociation reaction: (C2H5)3N + H2O -> (C2H5)3NH+ + OH- Kb expression: \(K_{b} = \frac{[(C_{2}H_{5})_{3}NH^{+}][OH^{-}]}{(C_{2}H_{5})_{3}N}\) b. For hydroxylamine, HONH2: Base dissociation reaction: HONH2 + H2O -> HONH3+ + OH- Kb expression: \(K_{b} = \frac{[HONH_{3}^{+}][OH^{-}]}{HONH_{2}}\)
02

Calculate the initial concentration of amine and dissociation

Initial concentration of each amine is given as 0.20 M. a. For triethylamine, let the dissociation per mole be x. Then, the concentration of (C2H5)3NH+ will be x, and the concentration of (C2H5)3N will be (0.20 - x) after dissociation. b. For hydroxylamine, let the dissociation per mole be x. Then, the concentration of HONH3+ will be x, and the concentration of HONH2 will be (0.20 - x) after dissociation.
03

Substitute the known values in Kb expressions and solve for x

Substitute the values in the Kb expression for each amine, and solve for x. a. For triethylamine: \(4.0 \times 10^{-4} = \frac{x^2}{0.20 - x}\) Assuming x << 0.20, we can approximate: \(4.0 \times 10^{-4} \approx \frac{x^2}{0.20}\) b. For hydroxylamine: \(1.1 \times 10^{-8} = \frac{x^2}{0.20 - x}\) Assuming x << 0.20, we can approximate: \(1.1 \times 10^{-8} \approx \frac{x^2}{0.20}\)
04

Find [OH-], [H+], and pH for each amine solution

Find [OH-], [H+], and pH for each amine solution using the calculated x values. a. For triethylamine, [OH-] = x: 1. x = 0.20 * sqrt(4.0 * 10^-4) = 2.0 * 10^-2 M 2. [H+] = Kw (10^-14) / [OH-] = 10^-14 / 2.0 * 10^-2 = 5.0 * 10^-13 M 3. pH = -log10(5.0 * 10^-13) = 12.3 b. For hydroxylamine, [OH-] = x: 1. x = 0.20 * sqrt(1.1 * 10^-8) = 4.7 * 10^-4 M 2. [H+] = Kw (10^-14) / [OH-] = 10^-14 / 4.7 * 10^-4 = 2.1 * 10^-11 M 3. pH = -log10(2.1 * 10^-11) = 10.7 So, the final answers are: a. [OH-] = 2.0 * 10^-2 M, [H+] = 5.0 * 10^-13 M, pH = 12.3 for the triethylamine solution. b. [OH-] = 4.7 * 10^-4 M, [H+] = 2.1 * 10^-11 M, pH = 10.7 for the hydroxylamine solution.

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

Write out the stepwise \(K_{\mathrm{a}}\) reactions for citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\) a triprotic acid.

Zinc hydroxide is an amphoteric substance. Write equations that describe \(\mathrm{Zn}(\mathrm{OH})_{2}\) acting as a Brönsted-Lowry base toward \(\mathrm{H}^{+}\) and as a Lewis acid toward \(\mathrm{OH}^{-}\).

Calculate \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right],\) and the \(\mathrm{pH}\) of \(0.40 M\) solutions of each of the following amines (the \(K_{\mathrm{b}}\) values are found in Table 13-3). a. aniline b. methylamine

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

A \(0.15-M\) solution of a weak acid is \(3.0 \%\) dissociated. Calculate \(K_{\mathrm{a}}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

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