How to find pOH

What is pOH?

pOH or potential of hydroxide is a criterion for measuring the hydroxide ion concentration in the solution. It is used to determine the acidity or alkalinity of a solution. pOH is equivalent to the negative log of hydroxide ion (OH^–) concentration.

pH = – log [OH^–]

At pOH 7, the solution is neutral. If the pOH value is less than 7, a solution will be alkaline, while a solution will be acidic if the pOH value is more than 7.

Table of Content

• How to find pOH
• pOH from Molarity
• pOH from pH
• pOH from pKb
• Frequently Asked Questions – FAQs

How to find pOH

pOH or potential of hydroxide is a criterion for measuring the hydroxide ion concentration in the solution. It is used to determine the acidity or alkalinity of a solution. pOH is equivalent to the negative log of hydroxide ion (OH^–) concentration.

pH = – log [OH^–]

At pOH 7, the solution is neutral. If the pOH value is less than 7, a solution will be alkaline, while a solution will be acidic if the pOH value is more than 7.

We can use numerous parameters to determine the pOH value.

• pOH from Molarity
• pOH from pH
• pOH from pKb

pOH from Molarity

We can use molarity to determine the pOH value. pOH is equivalent to the negative log of hydroxide ion (OH^–) concentration.

pOH = – log [OH^–]

By putting the value of the concentration of OH^– ion, we can determine the value of pOH.

Example:

• Calculate the pOH value of an aqueous solution whose molarity is 6.3 × 10^-5 M.

We know that,

pOH = – log [OH^–]

pOH = – log [ 6.3 × 10^-5 ]

pOH = 4.2

pOH from pH

We can use pH to determine the pOH value.

Consider dissociation of water,

H₂O ⇆ H⁺ + OH^–

Here dissociation constant, K_w would be,

K_w = [H⁺][OH^–]

Here, K_w is the dissociation constant, and [H⁺] and [OH^–] is hydrogen and hydroxide ion concentrations.

Take a negative logarithm of both sides.

– log K_w = – log [ H⁺ ] [OH^– ]

-log K_w = -(log [ H⁺ ] + [ OH^– ] )

-log K_w = – log [H⁺ ] – log [OH^– ] . . . A

We know that,

K_w = 1 X 10 ^-14 ( at 298 K )

-log [H⁺ ] = pH

-log [OH^– ] = pOH

Putting this value in equation A

-log 1 X 10 ^-14 = pH + pOH

14 = pH + pOH

pOH = 14 – pH

pOH from pKb

Using the Henderson equation of acidic buffer, we can determine pOH from pKb.

He states that,

pOH = pKb + log {[salt] / [Base]} . . . 1

At equivalence point, [salt] = [Acid],

So, [salt] / [Base] would be equal to 1.

Putting this value in equation 1.

pOH = pKb

Let [salt] / [Acid] be equal to 10.

Putting this value in equation 1.

pOH = pKb + log 10

pOH = pKb + 1

Let [salt] / [Acid] be equal to 1 / 10.

Putting this value in equation 1.

pOH = pKb + log 1 / 10

pOH = pKb + log 1 – log 10

pOH = pKb – 1