pOH of Strong Acids Calculations Chemistryy Tutorial

Key Concepts

The pOH of an aqueous solution of a strong monoprotic acid can be calculated if we know:
(i) the concentration of hydroxide ions in solution
or
(ii) the temperature¹ of the solution and either the pH of the solution or the concentration of hydrogen ions in solution², [H⁺_(aq)].

For an acidic solution:
pOH = -log₁₀[OH^-]
where [OH^-] = concentration of hydroxide ions in solution in mol L^-1
and pOH = pOH of the solution

For an aqueous acidic solution at 25°C
pOH = 14 - pH
where pH = the pH of the solution
and -log₁₀(10^-14) = 14 which is the negative logarithm (to base 10) of the dissociation constant for water at 25^o (also known as pK_w)³

For an aqueous acidic solution at 25°C
pOH = 14 - (-log₁₀[H⁺_(aq)])
where [H⁺_(aq)] = the concentration of hydrogen ions in solution in mol L^-1

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Calculating the pOH of Strong Monoprotic Acids

Before the acid is added to water, water molecules are in equilibrium with hydrogen ions and hydroxide ions:

water		hydrogen ions	+	hydroxide ions
H₂O		H⁺_(aq)	+	OH^-_(aq)

Only a very small number of water molecules dissociate into H⁺_(aq) and OH^-_(aq).
At 25°C, [H⁺_(aq)] = [OH^-_(aq)] ≈ 10^-7 mol L^-1 (a very low concentration)
and K_w = [H⁺_(aq)][OH^-_(aq)] = 10^-7 × 10^-7 = 10^-14

By calculating the logarithm (to base 10) of all the species above, we change the calculation to an addition instead of a multiplication:
K_w = [H⁺_(aq)][OH^-_(aq)]
log₁₀(K_w) = log₁₀[H⁺_(aq)] + log₁₀[OH^-_(aq)]

If we then multiply throughout by -1:
-log₁₀(K_w) = -log₁₀[H⁺_(aq)] + -log₁₀[OH^-_(aq)]
We can see that:
-log₁₀(K_w) = pH + pOH
since pH = -log₁₀[H⁺_(aq)]
and pOH = -log₁₀[OH^-_(aq)]

For water at 25^oC,
K_w = 10^-14
log₁₀K_w = log₁₀10^-14 = -14
-log₁₀K_w = -log₁₀10^-14 = 14
so, 14 = pH + pOH

When a strong monoprotic acid is added to water, the acid dissociates completely to form H⁺_(aq) and an aqueous anion:

acid	→	hydrogen ions	+	anions
HA	→	H⁺_(aq)	+	A^-_(aq)

Adding the acid to the water disturbs the water dissociation equilibrium:
H₂O H⁺_(aq) + OH^-_(aq)
By Le Chatelier's Principle, adding more H⁺_(aq) to the water will shift the equilibrium position to the left.
The water dissociation equilibrium system responds to the addition of more H⁺_(aq) by reacting some of the H⁺_(aq) with some of the OH^-_(aq) in order to re-establish equilibrium.
So, increasing the concentration of H⁺_(aq) in the water, reduces the concentration of OH^-_(aq), but, the water dissociation constant does not change⁴, K_w is still 10^-14.
So, K_w = [H⁺_(aq)][OH^-_(aq)] = 10^-14
and -log₁₀K_w = 14 = pH + pOH

[H⁺_(aq)] mol L^-1	[OH^-_(aq)] mol L^-1	K_w	At 25^oC	pH	pOH	-log(K_w) (pK_w)
10^-7	10^-7	10^-14	pure water (before acid added)	7	7	14
> 10^-7	< 10^-7	10^-14	aqueous solution (after acid added)	< 7	> 7	14

We can use the value of K_w or -log₁₀K_w (pK_w) and pH or [H⁺_(aq)] to calculate pOH at a given temperature:

If we know the pH of the acidic solution, we can calculate the pOH:
pH + pOH = -log₁₀K_w
So, pOH = -log₁₀K_w - pH
At 25°C : pOH = 14 - pH

If we know the concentration of hydrogen ions in solution, we can calculate the pOH:
K_w = [H⁺_(aq)][OH^-_(aq)]
By rearranging this equation (formula) we can determine the concentration of hydroxide ions in the aqueous solution:
[OH^-_(aq)] = K_w ÷ [H⁺_(aq)]
so, pOH = -log₁₀[OH^-_(aq)] = -log₁₀(K_w ÷ [H⁺_(aq)])
Both [H⁺_(aq)] and [OH^-_(aq)] must be in units of mol L^-1 (mol/L or M)

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Worked Examples

(based on the StoPGoPS approach to problem solving in chemistry.)

Question 1. Calculate the pOH of an aqueous solution of hydrochloric acid with a hydroxide ion concentration of 1.0 × 10^-12 mol L^-1.

What have you been asked to do?
Calculate the pOH
pOH = ?

What information (data) have you been given?
Extract the data from the question:
[OH^-_(aq)] = 1.0 × 10^-12 mol L^-1

What is the relationship between what you know and what you need to find out?
Write the equation (formula) for calculating pOH:
pOH = -log₁₀[OH^-_(aq)]

Subsitute in the value for the hydroxide ion concentration and solve:
pOH = -log₁₀[1.0 × 10^-12]
= 12

Is your answer plausible?
Use your calculated value for pOH to find [OH^-] and compare it to that given in the question:
[OH^-] = 10^-pOH = 10^-12
Since this value is the same as that given in the question we are confident our answer is correct.

State your solution to the problem:
pOH = 12

Question 2. Calculate the pOH of an aqueous solution of hydrochloric acid with a pH of 2.5 at 25°C.

What have you been asked to do?
Calculate the pOH
pOH = ?

What information (data) have you been given?
Extract the data from the question:
pH = 2.5
temperature = 25°C
so, K_w = 10^-14 (from Data Sheet)
and, -log₁₀K_w = -log₁₀10^-14 = 14

What is the relationship between what you know and what you need to find out?
Write the equation (formula) for calculating pOH:
pH + pOH = 14
Rearrange this equation (formula) to find pOH:
pOH = 14 - pH

Substitute in the pH value and solve:
pOH = 14 - 2.5
= 11.5

Is your answer plausible?
Use your calculated value for pOH to find pH and compare it to that given in the question:
pH + pOH = 14
pH + 11.5 = 14
pH = 14 - 11.5 = 2.5
Since this value is the same as that given in the question we are confident our answer is correct.

State your solution to the problem:
pOH = 11.5

Question 3. Calculate the pOH of 0.025 mol L^-1 HCl_(aq) at 25°C.

(i) Hydrogen ion concentration method:

What have you been asked to do?
Calculate the pOH
pOH = ?

What information (data) have you been given?
Extract the data from the question:
[HCl_(aq)] = 0.025 mol L^-1
temperature = 25^oC
so, K_w = 10^-14 (from Data Sheet)

What is the relationship between what you know and what you need to find out?
Calculate the concentration of hydrogen ions present in solution:
HCl_(aq) is a strong monoprotic acid, it dissociates fully in water:
HCl → H⁺_(aq) + Cl^-_(aq)
The stoichiometric ratio (mole ratio) of HCl to H⁺_(aq) is 1 : 1
so, [HCl_(aq)] = [H⁺_(aq)] = 0.025 mol L^-1
Write the equation (formula) for calculating the concentration of hydroxide ions in solution:
K_w = [H⁺_(aq)][OH^-_(aq)]
so, [OH^-_(aq)] = K_w/[H⁺_(aq)]

Substitute in the values for K_w and [H⁺_(aq)] and solve:
[OH^-_(aq)] = 10^-14/[0.025 mol L^-1] = 4.0 × 10^-13

Write the equation (formula) for calculating pOH:
pOH = -log₁₀[OH^-_(aq)]

Substitute in the hydroxide ion concentration and solve:
pOH = -log₁₀[4.0 × 10^-13] = 12.4

Is your answer plausible?
Use your calculated value for pOH to find [H⁺] and compare it to that given in the question:
pH = 14 - 12.4 = 1.6
[H⁺] = 10^-pH = 10^-1.6 = 0.025 mol L^-1 = [HCl_(aq)]
Since this value is the same as that given in the question we are confident our answer is correct.

State your solution to the problem:
pOH = 12.4

(ii) pH method:

What have you been asked to do?
Calculate the pOH
pOH = ?

What information (data) have you been given?
Extract the data from the question:
[HCl_(aq)] = 0.025 mol L^-1
temperature = 25°C
so, K_w = 10^-14 (from Data Sheet)

What is the relationship between what you know and what you need to find out?
Calculate the concentration of hydrogen ions present in solution:
HCl_(aq) is a strong monoprotic acid, it dissociates fully in water:
HCl → H⁺_(aq) + Cl^-_(aq)
The stoichiometric ratio (mole ratio) of HCl to H⁺_(aq) is 1 : 1
so, [HCl_(aq)] = [H⁺_(aq)] = 0.025 mol L^-1
Calculate the pH of the solution:
pH = -log₁₀[H⁺_(aq)] = -log₁₀[0.025] = 1.60

Write the equation (formula) for calculating pOH:
-log₁₀K_w = pH + pOH

Substitute in the values for K_w and pH and solve:
-log₁₀10^-14 = 1.60 + pOH
14 = 1.60 + pOH
pOH = 14 - 1.60
= 12.4
Is your answer plausible?
Use your calculated value for pOH to find [H⁺] and compare it to that given in the question:
pH = 14 - 12.4 = 1.6
[H⁺] = 10^-pH = 10^-1.6 = 0.025 mol L^-1 = [HCl_(aq)]
Since this value is the same as that given in the question we are confident our answer is correct.

State your solution to the problem:
pOH = 12.4

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1. The dissociation constant for water, K_w, varies with temperature. At 25^oC K_w ~10^-14.
The dissociation constant, or ionisation constant, for water should be provided to you either in the question or on a data sheet.
If no temperature is specified in a question, assume the temperature is 25^oC.

2. Hydrogen ions are being represented here as H⁺, or H⁺_(aq) for hydrogen ions in water (hydrated hydrogen ions).
You can use H₃O⁺ (oxidanium or hydronium or oxonium ion) instead to represented hydrogen ions in aqueous solution.
In this case the dissociation of water is 2H₂O H₃O⁺ + OH^-_(aq)
and K_w = [H₃O⁺][OH^-_(aq)] = 10^-14 (at 25^oC).

3. K_w = 10^-14 is only an approximation, but it is one that is widely used for convenience.

4. This assumes that the temperature of the water remains constant.
If the temperature changes, the value of K_w changes.