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Calculate [H3O+] for each solution.(a) pH = 8.55(b) pH = 11.23(c) pH = 2.87(d) pH = 1.22

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

Solution for problem 71P 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 71P

Problem 71P

Calculate [H3O+] for each solution.

(a) pH = 8.55

(b) pH = 11.23

(c) pH = 2.87

(d) pH = 1.22

Step-by-Step Solution:

 Solution:

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 1

(a) pH = 8.25

pH = 8.25

The  [H3O+]  from pH =  8.25 can be calculated as follows,

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

 8.25 = - log [H3O+]

-  8.25 = log [H3O+]

[H3O+] = 10- 8.25  = 5.62 10-9 M  

                 1.0  10-14       1.0  10-14 

Thus,  [OH-]  =  -------------- =     -----------    = 1.710-6M

         [H3O+]           5.62 10-9

(b) pH = 11.23

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

        1.0  10-14 

  [OH-  ]  =  --------------

                       [H3O+]  

         

Now the  [H3O+]  from pH = 11.23can be calculated as follows,

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

11.23= - log [H3O+]

- 11.23= log [H3O+]

[H3O+] = 10- 11.23 = 15.9  10-9 M

                 1.0  10-14       1.0  10-14 

Thus,  [OH-  ]  =  -------------- =     -----------    = 1.8 *1010M

         [H3O+]           15.9  10-9 

(c) pH = 2.87

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

        1.0  10-14 

  [OH-  ]  =  --------------

                       [H3O+]  

         

Now the  [H3O+]  from pH = 2.87 can be calculated as follows,

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

2.87 = - log [H3O+]

- 2.87= log [H3O+]

[H3O+] = 10- 2.87 = 2.3 10-9 M

                 1.0  10-14       1.0  10-14 

Thus,  [OH-  ]  =  -------------- =     -----------    = 2.28*1011M

         [H3O+]           2.3 10-9

(d) pH = 1.22

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

        1.0  10-14 

  [OH-  ]  =  --------------

                       [H3O+]  

         

Now the  [H3O+]  from pH =1.22 can be calculated as follows,

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

1.22= - log [H3O+]

-1.22= log [H3O+]

[H3O+] = 10- 1.22 =2.28 10-9 M

                 1.0  10-14       1.0  10-14 

Thus,  [OH-  ]  =  -------------- =     -----------    = 2.28*1011M

         [H3O+]           2.8 10-9

                                               

Step 2 of 1

Chapter 14, Problem 71P is Solved
Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Calculate [H3O+] for each solution.(a) pH = 8.55(b) pH = 11.23(c) pH = 2.87(d) pH = 1.22