Question

Question regarding the following equation explanation:

Why is it that when finding the PH in qestions (B) you take the negative log of the M but to find the OH in question (A) you ALSO take the -log. Is there a rule as to when you use the negative log to get PH and when you use the negative log to get POH? Seems alot like an 'eenie meanie miney mo' thing going on here.

Please explain in detail!!

( A )

Determine the pH of a 4.3 x 10^-4 M NaOH solution.

pOH = -log[OH-] = -log(4.3 x 10^-4) = 3.37

pH = 14

Answer 1

The concept of acids and bases are to be considered here. Acids are said to be substances that give an H+ in aqueous medium and a base is that gives an OH- in aqueous medium. Water by itself undergoes self-ionization to form the conjugate acid and conjugate base of water by the equation,

2H2O <=====> H3O+ + OH-

The equilibrium constant for water can thus be written as,

Kw = [H3O+][OH-]/[H2O]

Since water is in large excess its concentration can be neglected in the equation,

Kw = [H3O+][OH-]

Or,

Kw = [H+][OH-]

The Kw for water is found to be 1 x 10^-14 , substituting this value we get,

[H⁺] [OH^-] = K_w = 1 X 10^-14

Take the -log of both sides of the equation

- log [H⁺] +(- log [OH^- ]) = - log [1 X 10^-14 ]

pH + pOH = 14

Thus, the pH which is defined as the activity of H+ in water and OH- activity of OH- in water is found by as given in the above given relation,

pH = -log[H+]

or from pOH it will be,

pH = 14 - pOH

and,

pOH = -log[OH-]

or from the pH,

pOH = 14 - pH

Thus we can clearly see that all the above relations are simply the rearranged forms of the original given equation employed for the calculation of pH and pOH of a substance in water.