Acids and Bases - Calculations

What if I have pH and want to find [H+]?• We use the antilog

[H+] = antilog(-pH)

In the calculator,

2nd LOG negative value )

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What is the [H+] in a healthy person’s blood that has a pH = 7.40? Assume that the temperature of the blood is 298 K.

What if I have pOH and want to find [OH-]?• We use the antilog

[OH-] = antilog(-pOH)

In the calculator,

2nd LOG negative value )

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What is the [OH-] in a healthy person’s blood that has a pOH = 6.6? Assume that the temperature of the blood is 298 K.

pH pOH

[H+] [OH-]

pH + pOH = 14

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

[H+] = antilog(-pH)

[OH-] = antilog(-pOH)

Ionization of water

Water is neutral with a pH = 7 so in solution this is what is happening:

Kw = [H+][OH-]

and Kw = 1.0 x 10-14

Example:

If [H+] = 1 x 10-5, what is [OH-] ?

Kw = [H+][OH-]

(1.0 x 10-14) = [H+] [OH-](1.0 x 10-14) = (1 x 10-5) [OH-](1.0 x 10-14) = [OH-] (1 x 10-5)

1x 10-9 = [OH-]

Example:

If [OH-] = 3.2 x 10-2 what is [H+] ?

Kw = [H+][OH-]

(1.0 x10-14) = [H+][OH-] (1.0 x10-14) = [H+](3.2 x10-2)

(1.0 x 10-14) = [H+] (3.2 x10-2)

3.13 x 10-13 = [H+]

pH pOH

[H+] [OH-]

pH + pOH = 14

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

[H+] = antilog(-pH)

[OH-] = antilog(-pOH)

Kw = [H+][OH-]

and Kw = 1.0 x 10-14