In any aqueous solution at 25 °C, the sum of pH and pOH is 14.0. Explain why this is so.

Solution 26Q:

Here, we are going to derive the the sum of pH and pOH is 14.0 at 25 °C.

Step 1:

Water is amphoteric which can act as either an acid or a base. Pure water can behave as an acid and base with itself. Thus the process is called autoionization.So, we can write,

H2O (l) --------> H+ (aq) + OH-(aq)

The Equilibrium constant for the reaction is

Kw = [H+(aq)][OH-(aq)]------(1)

where Kw is known as ion product constant for water and at 25oC.

For pure water, at 25oC, Kw = 1.0 x 10-14

Therefore, equation (1) ,

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

Step 2:

pH and pOH is defined as

pH= -log[H+]

pOH= -log[OH-]

From equation (2)

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

Taking log on both sides of equation ?(2),

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

10-14 = pH + pOH

pH + pOH =14