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

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

Problem 75P Chapter 14

Introductory Chemistry | 5th Edition

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

Introductory Chemistry | 5th Edition

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Problem 75P

Calculate [OH− ] for each solution.

(a) pH = 4.25

(b) pH = 12.53

(c) pH = 1.50

(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  10-14 

And   [OH-  ]  =  --------------

                              [H3O+]                                                        

Step-2

(a) pH = 4.25

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

        1.0  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  10-5 M

                 1.0  10-14       1.0  10-14 

Thus,  [OH-  ]  =  -------------- =     -----------    = 1.78 10-20 M

         [H3O+]           5.6  10-5

Step-3

(b) pH = 12.53

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

        1.0  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  10-13 M  

Step 2 of 3

Chapter 14, Problem 75P is Solved
Step 3 of 3

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

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