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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 14 - Problem 101p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 14 - Problem 101p

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# Complete the table. (The first row is completed for you.)[H3O+][OH?]pOHpHAcidic or ISBN: 9780321910295 34

## Solution for problem 101P Chapter 14

Introductory Chemistry | 5th Edition

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Problem 101P

Problem 101P

Complete the table. (The first row is completed for you.)

 [H3O+] [OH−] pOH pH Acidic or Basic 1.0 × 10−4 1.0 × 10−10 10.00 4.00 Acidic 5.5 × 10−3 ________ ________ ________ ________ ________ 3.2 × 10−6 ________ ________ ________ 4.8 × 10−9 ________ ________ ________ ________ ________ ________ ________ 7.55 ________

Step-by-Step Solution:
Step 1 of 3

Solution 101P: Here, we are going to complete the given table.

Step1:

pH is the negative logarithm to base 10 of hydrogen ion concentration whereas pOH is the negative logarithm to base 10 of hydroxyl ion concentration, i.e.,

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

And pH + pOH =14

We can distinguish acidic, neutral and basic aqueous solutions by the relative values of their pH

Acidic: pH < 7

Neutral: pH = 7

Basic: pH > 7

Step2:

 [H3O+] [OH-] pOH pH Acidic or Basic 1.0 x 10-4 1.0 x 10-10 10.00 4.00 Acidic 5.5 x 10-3 1.82 x 10-12 11.74 2.26 Acidic 3.13 x 10-9 3.2 x 10-6 5.495 8.505 Basic 4.8 x 10-9 2.08 x 10-6 5.681 8.319 Basic 2.82 x 10-8 3.55 x 10-7 6.45 7.55 Basic

1. Given, [H3O+] = 5.5 x 10-3

Therefore, pH = - log[H3O+ ]

= - log (5.5 x 10-3)

= - log 5.5 - log 10-3

= - 0.740 + 3

= 2.26

Therefore, pOH = 14 - pH

= 14 - 2.26

= 11.74

pOH =   - log[OH- ]

Therefore, [OH- ] = 10-pOH

= 10-11.74

= 1.82 x 10-12

Since, pH < 7, therefore the solution is acidic.

b)        Given, [OH-] = 3.2 x 10-6

Therefore, pOH = - log[OH-]

= - log (3.2 x 10-6)

= - log 3.2 - log 10-6

= - 0.505 + 6

= 5.495

Therefore, pH = 14 - pOH

= 14 - 5.495

= 8.505

pH =   - log[H3O+]

Therefore, [H3O+] = 10-pH

= 10-8.505

= 3.13 x 10-9

Since, pH > 7, therefore the solution is basic.

c)        Given, [H3O+] = 4.8 x 10-9

Therefore, pH = - log[H3O+ ]

= - log (4.8 x 10-9)

= - log 4.8 - log 10-9

= - 0.681 + 9

= 8.319

Therefore, pOH = 14 - pH

= 14 - 8.319

= 5.681

pOH =   - log[OH- ]

Therefore, [OH- ] = 10-pOH

= 10-5.681

= 2.08 x 10-6

Since, pH > 7, therefore the solution is basic.

d)        Given, pH = 7.55

pH =   - log[H3O+]

Therefore, [H3O+] = 10-pH

= 10-7.55

= 2.82 x 10-8

pOH = 14 - pH

= 14 - 7.55

= 6.45

pOH =   - log[OH- ]

Step 2 of 3

Step 3 of 3

##### ISBN: 9780321910295

Since the solution to 101P from 14 chapter was answered, more than 379 students have viewed the full step-by-step answer. The answer to “Complete the table. (The first row is completed for you.)[H3O+][OH?]pOHpHAcidic or Basic1.0 × 10?41.0 × 10?1010.004.00Acidic5.5 × 10?3________________________________________3.2 × 10?6________________________4.8 × 10?9________________________________________________________7.55________” is broken down into a number of easy to follow steps, and 22 words. This full solution covers the following key subjects: acidi, completed, complete, basic, pOH. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. The full step-by-step solution to problem: 101P from chapter: 14 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295.

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