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Determine the pOH of each solution.(a) [H3O+] = 8.3 × 10?10 M(b) [H3O+] = 1.6 × 10?7

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 82P Chapter 14

Introductory Chemistry | 5th Edition

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

Introductory Chemistry | 5th Edition

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Problem 82P

Determine the pOH of each solution.

(a) [H3O+] = 8.3 × 10−10 M

(b) [H3O+] = 1.6 × 10−7 M

(c) [H3O+] = 7.3 × 10−2 M

(d) [OH−] = 4.32 × 10−4 M

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to determine the pOH of the given solutions.

Step1:

In pure water, one H2O molecule donates proton and acts as an acid and another water molecule accepts a proton and acts as a base at the same time. The following equilibrium exists:

H2O(l) + H2O(l) ------> H3O+(aq) + OH-(aq)

The equilibrium constant for the reaction is given by:

                                Kw  = [H3O+][OH]        -----(1)

Where, Kw is called the ionic product of water and its value is 1 x 10-14 M2.

moreover, pH + pOH =14

Where, pOH =  - log[OH- ] and pH = - log[H+ ]

Step2:

  1. Given, [H3O+] = 8.3 x 10-10 M

From equation (1), [OH] = Kw / [H3O+]

                       [OH] = (1 x 10-14 M2) / (8.3 x 10-10 M)

                       [OH] = 1.2 x 10-5 M

Therefore pOH =  - log[OH- ]

                = - log (1.2 x 10-5)

                = - log 1.2 - log 10-5

= -0.079 + 5

   pOH = 4.921

    b)        Given, [H3O+] = 1.6 x 10-7 M

From equation (1), [OH] = Kw / [H3O+]

                       [OH] = (1 x 10-14 M2) / (1.6 x 10-7 M)

                       [OH] = 6.25 x 10-8 M

Therefore pOH =  - log[OH- ]

                = - log (6.25 x 10-8)

                = - log 6.25 - log 10-8

= -0.796 + 8

   pOH = 7.204

    c)        Given, [H3O+] = 7.3 x 10-2 M

From equation (1), [OH] = Kw / [H3O+]

                       [OH] = (1 x 10-14 M2) / (7.3 x 10-2 M)

                       [OH] = 1.37 x 10-13 M

Therefore pOH =  - log[OH- ]

                = - log (1.37 x 10-13)

                = - log 1.37 - log 10-13

= -0.137 + 13

   pOH = 12.863

    d)        Given, [OH] = 4.32 x 10-4 M

Therefore pOH =  - log[OH- ]

                = - log (4.32 x 10-4)

                = - log 4.32 - log 10-4

= -0.635 + 4

   pOH = 3.365

                                        ------------------------------

Step 2 of 3

Chapter 14, Problem 82P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. Since the solution to 82P from 14 chapter was answered, more than 436 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. The answer to “Determine the pOH of each solution.(a) [H3O+] = 8.3 × 10?10 M(b) [H3O+] = 1.6 × 10?7 M(c) [H3O+] = 7.3 × 10?2 M(d) [OH?] = 4.32 × 10?4 M” is broken down into a number of easy to follow steps, and 30 words. This full solution covers the following key subjects: determine, pOH, solution. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. The full step-by-step solution to problem: 82P from chapter: 14 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM.

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Determine the pOH of each solution.(a) [H3O+] = 8.3 × 10?10 M(b) [H3O+] = 1.6 × 10?7