Solved: Calculate [OH? ] for each solution.(a) pH = 4.25(b) pH = 12.53(c) pH = 1.50(d)

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

ISBN: 9780321910295 34

Solution for problem 76P Chapter 14

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition

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Problem 76P

Calculate $\mathrm{[OH^-]}$ for each solution.

(a) pH = 1.82

(b) pH = 13.28

(d) pH = 2.32

Equation Transcription:

$[O{H}^{-}]$

Text Transcription:

[OH^-]

Step-by-Step Solution:

Solution:

Step 1

The concentration of [OH-] ion and [H3O+] can be determined as follows ,

It is known that pH = -log [H3O+]

-pH = log [H3O+]

10 -pH = 10log [H3O+] or

[H3O+] = 10-pH

1.0 $\times{}$ 10-14

And [OH- ] = --------------

[H3O+]

Step 2 of 5

Chapter 14, Problem 76P is Solved

Step 3 of 5

Textbook: Introductory Chemistry

Edition: 5

Author: Nivaldo J Tro

ISBN: 9780321910295

Since the solution to 76P from 14 chapter was answered, more than 960 students have viewed the full step-by-step answer. The answer to “?Calculate $\mathrm{[OH^-]}$ for each solution.(a) pH = 1.82(b) pH = 13.28(c) pH = 8.29(d) pH = 2.32Equation Transcription:Text Transcription:[OH^-]” is broken down into a number of easy to follow steps, and 19 words. This full solution covers the following key subjects: calculate, solution. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. The full step-by-step solution to problem: 76P from chapter: 14 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5.