1 Chapter 14 Acids and Bases 14.6 The pH Scale Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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Chapter 14 Acids and Bases

14.6

The pH Scale

Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

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pH Scale

The pH scale:

• Is used to indicate the acidity of a solution.

• Has values that usually range from 0 to 14.

• Is acidic when the values are less than 7.

• Is neutral with a pH of 7.

• Is basic when the values are greater than 7.

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pH of Everyday Substances

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Identify each solution as

1. acidic 2. basic 3. neutral

A. ___ HCl with a pH = 1.5

B. ___ Pancreatic fluid [H3O+] = 1 x 10−8 M

C. ___ Sprite soft drink pH = 3.0

D. ___ pH = 7.0

E. ___ [OH−] = 3 x 10−10 M

F. ___ [H3O+ ] = 5 x 10−12

Learning Check

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A. 1 HCl with a pH = 1.5

B. 2 Pancreatic fluid [H3O+] = 1 x 10−8 M

C. 1 Sprite soft drink pH = 3.0

D. 3 pH = 7.0

E. 1 [OH-] = 3 x 10−10 M

F. 2 [H3O+] = 5 x 10−12

Solution

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Testing the pH of Solutions

The pH of solutions can be determined using • pH meter, • pH paper, or • indicators that have specific colors at different

pH values.

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pH RangeTable 14.8

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pH is the negative log of the hydronium ion

concentration.

pH = - log [H3O+]

For a solution with [H3O+] = 1 x 10−4

pH = −log [1 x 10−4 ]

pH = - [-4.0]

pH = 4.0

Calculating pH

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Significant Figures in pH

When expressing log values, the number of

decimal places in the pH is equal to the number of

significant figures in the coefficient of [H3O+].

[H3O+] = 1 x 10-4 pH = 4.0

[H3O+] = 8.0 x 10-6 pH = 5.10

[H3O+] = 2.4 x 10-8 pH = 7.62

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Guide to pH Calculations

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Find the pH of a solution with a [H3O+] of 1.0 x 10−3:

STEP 1 Enter [H3O+]

Enter 1 x 10-3 by pressing 1 (EE) 3

The EE key gives an exponent of 10 and change sign (+/- key or – key)

STEP 2 Press log key and change sign

- log (1 x 10−3) = -[-3]

STEP 3 Adjust significant figures after decimal point

3 3.00 Two significant figures in 1.0 x 10−3

Calculating pH

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A. The [H3O+] of tomato juice is 2 x 10−4 M.

What is the pH of the solution? 1) 4.0 2) 3.7 3) 10.3

B. The [OH−] of a solution is 1.0 x 10−3 M. What is the pH of the solution?

1) 3.00 2) 11.00 3) -11.00

Learning Check

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A. 2) 3.7 pH = - log [ 2 x 10-4] = 3.7

2 (EE) 4 (+/-) log (+/-)

B. 2) 11.00 Use the Kw to obtain [H3O+] = 1.0 x 10−11

pH = - log [1.0 x 10−11] 1.0 (EE) 11 (+/-) log (+/-)

Solution

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pOH

The pOH of a solution • Is analogous to the pH value on the pH scale.• Is based on the [OH−].• Has high pOH values with low [OH−] and high

[H3O+].

• Is equal to the - log [OH-] pOH = - log [OH-].• Added to the pH value is equal to 14.0.

pH + pOH = 14.0

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Calculation pOH

Find the pOH of a solution with a [OH−] of 2.0 x 10−3.

STEP 1 Enter [OH−]

STEP 2 Press log key and change sign

- log (2.0 x 10−3) = -[-2.69897] = 2.69897

STEP 3 Adjust figures after decimal point

2.70 Two significant figures in 2.0 x 10−3

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Calculating [H3O+] from pH

The [H3O+] can be expressed by using the pH as the negative power of 10.

[H3O+] = 1 x 10−pH

If the pH is 3.0, the [H3O+] = 1 x 10−3

On a calculator

1. Enter the pH value 3.0

2. Change sign -3.0

3. Use the inverse log key (or 10x) to obtain

the [H30+]. = 1 x 10−3 M

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A. What is the [H3O+] of a solution with a pH of 10.0?

1) 1 x 10−4 M 2) 1 x 1010 M

3) 1 x 10−10 M

B. What is the [H3O+] of a solution with a pH of 5.75?

1) 1.8 x 10− 6 M 2) 1.778 x 10−6 M

3) 2 x 10−6 M

Learning Check

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A. What is the [H3O+] of a solution with a pH of 10.0?

3) 1 x 10-10 M 1 x 10-pH

B. What is the [H3O+] of a solution with a pH of 5.75?

1) 1.8 x 10− 6 M5.75 (+/-) inv log (or 10x) = 1.8 x 10-6

Note: Two decimal places (.75), indicate two significant figures in the coefficient of the answer.

Solution

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Learning Check

What is the pH and the pOH of coffee if the [H3O+] is 1 x 10−5 M?

1) pH = 5.0 pOH =7.0

2) pH = 7.0 pOH = 9.0

3) pH = 5.0 pOH = 9.0

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Solution

What is the pH and the pOH of coffee if the [H3O+] is 1 x 10−5M?

3) pH = 5.0 pOH = 9.0

pH = -log [1 x 10−5] = -(-5.0) = 5.0

pH + pOH = 14.0

pOH = 14.0 – pH = 14.0 – 5.0 = 9.0

or [OH-] = 1 x 10−9

pOH = - log [1 x 10−9] = -(-9.0) = 9.0