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16-3 The Self-Ionization of Water and the pH Scale

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Ion Product of Water

Kc= [H2O][H2O]

[H3O+][OH-]

H2O + H2O H3O+ + OH-

base acidconjugate

acid

conjugate

base

KW= Kc[H2O][H2O] = = 1.010-14[H3O+][OH-]

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pH and pOH

The potential of the hydrogen ion was defined in 1909 as the negative of the logarithm of [H+].

pH = -log[H3O+] pOH = -log[OH-]

-logKW = -log[H3O+]-log[OH-]= -log(1.010-14)

KW = [H3O+][OH-]= 1.010-14

pKW = pH + pOH= -(-14)

pKW = pH + pOH = 14

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pH and pOH Scales

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16-4 Strong Acids and Bases

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16-5 Weak Acids and Bases

Lactic Acid Glycine

General Chemistry: Chapter 16 Prentice-Hall © 2007

Acetic Acid

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Acetic Acid

Weak Acids

Ka= = 1.810-5

[CH3CO2H]

[CH3CO2-][H3O+]

pKa= -log(1.810-5) = 4.74

General Chemistry: Chapter 16 Prentice-Hall © 2007

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Weak Bases

Kb= = 4.310-4

[CH3NH2]

[CH3NH3+][HO-]

pKb= -log(4.210-4) = 3.37

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Table 16.3 Ionization Constants of Weak Acids and Bases

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Determining a Value of KA from the pH of a Solution of a Weak Acid. Butyric acid, HC4H7O2 (or CH3CH2CH2CO2H) is used to make compounds employed in artificial flavorings and syrups. A 0.250 M aqueous solution of HC4H7O2 is found to have a pH of 2.72. Determine KA for butyric acid.

HC4H7O2 + H2O C4H7O2 + H3O+

Ka = ?

EXAMPLE 16-5

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HC4H7O2 + H2O C4H7O2 + H3O+

Initial conc. 0.250 M 0 0

Changes -x M +x M +x M

Equilibrium (0.250-x) M x M x MConcentration

EXAMPLE 16-5

Solution:

For HC4H7O2 KA is likely to be much larger than KW. Therefore assume self-ionization of water is unimportant.

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Log[H3O+] = -pH = -2.72

HC4H7O2 + H2O C4H7O2 + H3O+

[H3O+] = 10-2.72 = 1.910-3 = x

[H3O+] [C4H7O2-]

[HC4H7O2] Ka=

1.910-3 · 1.910-3

(0.250 – 1.910-3)=

Ka= 1.510-5 Check assumption: Ka >> KW.

EXAMPLE 16-5

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Percent Ionization

HA + H2O H3O+ + A-

Degree of ionization =[H3O+] from HA

[HA] originally

Percent ionization =[H3O+] from HA

[HA] originally 100%

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Percent Ionization

Ka =[H3O+][A-]

[HA]

Ka =n

H3O+ A-n

HAn

1

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