Bronsted-Lowry theory of Acids and Bases.

Bronsted-Lowry Definition of Acids and Bases

We will use the Bronsted-Lowry definitions for acids and bases:Acids are species that donate a proton (H⁺).andbases are species that accept a proton.Acid example:HNO_{3 (aq)} + H₂O  NO₃^-_(aq) + H₃O⁺_(aq)K_eq = a very large numberIn this example, HNO₃ is an acid and H₂O is acting as a base. NO₃^- is called the conjugate base of the acid HNO₃, and H₃O⁺ is the conjugate acid of the base H₂O.Base example:NH_{3 (aq)} + H₂O  NH₄⁺_(aq) + OH^-_(aq)K = 1.8x10^-5In this example, NH₃ is a base and H₂O is acting as an acid. NH₄⁺ is the conjugate acid of the base NH₃, and OH^- is the conjugate base of the acid H₂O.A compound that can act as either an acid or a base, such as the H₂O in the above examples, is called amphiprotic.

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H₂O + H₂O  H₃O⁺_(aq) + OH^-_(aq)
or
H₂O  H⁺_(aq) + OH^-_(aq)
K_w = [H₃O⁺][OH^-] = [H⁺][OH^-] = 1.00x10^-14 (at 25^oC)
(Using [H₃O⁺] is equivalent to using [H⁺].)
K_w is called the dissociation constant or ionization constant of water.
In pure water [H⁺] = [OH^-] = 1.00x10^-7 M.
pH is a shorthand notation for -log[H⁺]
and
pOH is a shorthand notation for -log[OH^-].
pH + pOH = 14.
Solutions are called
neutral when pH = 7, [H⁺] = [OH^-] = 1.00x10^-7
acidic when pH < 7, [H⁺] > 1.00x10^-7
basic when pH > 7, [H⁺] < 1.00x10^-7
Example: What is the pH of a solution of 0.025 M HNO₃? (See example 13.4 in text.)
HNO₃ is a strong acid and for all practical purposes dissociates completely.
HNO_3(aq) + H₂O  NO₃^-_(aq) + H₃O⁺_(aq)
[H⁺] = 0.025 M
pH = -log(0.025 M) = 1.6
What is the pOH of this solution? There are 2 ways to calculate pOH:
K_w = 1.00x10^-14 = [0.025 M][OH^-]
[OH^-] = 4.00x10^-13
pOH = -log(4.00x10^-13) = 12.40
or:
pOH = 14.00 - pH = 14.0 - 1.60 = 12.40
What are pH and pOH for a 0.0025 M solution of HNO₃?
pH = -log(0.0025 M) = 2.60
pOH = 14.00 - 2.60 = 11.40
Notice that pH and pOH change by 1 for a factor of 10 change in [H⁺] and [OH^-].

K_w, pH, and pOH