Calculate [OH? ] for each solution.(a) pH = 4.25(b) pH =

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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Problem 76P

Problem 75P

Calculate [OH− ] for each solution.

(a) pH = 4.25

(b) pH = 12.53

(d) pH = 8.25

Step-by-Step Solution:

Step 1 of 3

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

(a) pH = 4.25

We can calculate [OH− ], by using the following formula.

1.0 $\times{}$ 10-14

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

[H3O+]

Now the [H3O+] from pH = 4.25 can be calculated as follows,

[H3O+] = 10-pH or [H3O+] = antilog (- pH)

4.25 = - log [H3O+]

- 4.25 = log [H3O+]

[H3O+] = 10- 4.25 = 5.6 $\times{}$ 10-5 M

1.0 $\times{}$ 10-14 1.0 $\times{}$ 10-14

Thus, [OH- ] = -------------- = ----------- = 1.78 $\times{}$ 10-20 M

[H3O+] 5.6 $\times{}$ 10-5

Step-3

(b) pH = 12.53

We can calculate [OH− ], by using the following formula.

1.0 $\times{}$ 10-14

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

[H3O+]

Now the [H3O+] from pH = 12.53 can be calculated as follows,

[H3O+] = 10-pH or [H3O+] = antilog (- pH)

12.53 = - log [H3O+]

- 12.53 = log [H3O+]

[H3O+] = 10- 12.53 = 2.95 $\times{}$ 10-13 M

Step 2 of 3

Chapter 14, Problem 76P is Solved

Step 3 of 3

Textbook: Introductory Chemistry

Edition: 5

Author: Nivaldo J Tro

ISBN: 9780321910295

Since the solution to 76P from 14 chapter was answered, more than 270 students have viewed the full step-by-step answer. The answer to “Calculate [OH? ] for each solution.(a) pH = 4.25(b) pH = 12.53(c) pH = 1.50(d) pH = 8.25” is broken down into a number of easy to follow steps, and 18 words. This full solution covers the following key subjects: calculate, solution. This expansive textbook survival guide covers 19 chapters, and 2045 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.