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Acids and Bases - Calculations

Acids and Bases - Calculations. What if I have pH and want to find [H + ]?. We use the antilog [H + ] = antilog(-pH) In the calculator, 2 nd LOG negative value ) -.

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Acids and Bases - Calculations

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  1. Acids and Bases - Calculations

  2. What if I have pH and want to find [H+]? • We use the antilog [H+] = antilog(-pH) In the calculator, 2nd LOG negative value ) -

  3. 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.

  4. What if I have pOH and want to find [OH-]? • We use the antilog [OH-] = antilog(-pOH) In the calculator, 2nd LOG negative value ) -

  5. 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.

  6. pH pH + pOH = 14 pOH [H+] = antilog(-pH) pH = -log[H+] pOH = -log[OH-] [OH-] = antilog(-pOH) [H+] [OH-]

  7. 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

  8. 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-]

  9. 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+]

  10. pH pH + pOH = 14 pOH [H+] = antilog(-pH) pH = -log[H+] pOH = -log[OH-] [OH-] = antilog(-pOH) [H+] [OH-] Kw = [H+][OH-] and Kw = 1.0 x 10-14

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