# Hydroxide Ion Concentration of Strong Acids Calculations Tutorial

## Key Concepts

• The concentration of hydroxide ions, OH-, in an aqueous solution of a strong monoprotic acid can be calculated if we know:
(i) the temperature of the solution
and either
(ii) the pH of the solution
OR the concentration of hydrogen ions in solution.
• For an aqueous acidic solution at 25°C1
[OH-] = 10-14 ÷ [H+]
where [OH-] = concentration of hydroxide ions in mol L-1
and [H+] = concentration of hydrogen ions2 in mol L-1
and 10-14 is the dissociation constant3 for water at 25°C
• For an aqueous acidic solution at 25°C
[OH-] = 10-14 ÷ [H+]
but [H+] = 10-pH
So, [OH-] = 10-14 ÷ [10-pH]
where [OH-] = concentration of hydroxide ions in mol L-1
and pH = the pH of the solution
and 10-14 is the dissociation constant for water at 25°C

No ads = no money for us = no free stuff for you!

## Calculating the Hydroxide Ion Concentration of Strong Monoprotic Acids

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

 water hydrogenions + 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 Kw = [H+(aq)][OH-(aq)] = 10-7 × 10-7 = 10-14

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

 acid → hydrogenions + 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, Kw is still 10-14.
So, Kw = [H+(aq)][OH-(aq)] = 10-14

[H+(aq)]mol L-1 [OH-(aq)]mol L-1 Kw At 25°C 10-7 10-7 10-14 > 10-7 < 10-7 10-14

We can use the value of Kw and [H+(aq)] to calculate [OH-(aq)] at a given temperature:
At 25°C: Kw = [H+(aq)][OH-(aq)] = 10-14
By rearranging this equation (formula) we can determine the concentration of hydroxide ions in the aqueous solution:
[OH-(aq)] = 10-14 ÷ [H+(aq)]
Both [H+(aq)] and [OH-(aq)] must be in units of mol L-1 (mol/L or M)

If we know the pH of the acidic solution, we know the concentration of hydrogen ions in the solution
because pH = -log10[H+(aq)]
By reaarranging this equation (formula) we can find the concentration of hydrogen ions in mol L-1:
[H+(aq)] = 10-pH
This value of [H+(aq)] can then be used to calculate [OH-(aq)] in mol L-1 using the appropriate value for Kw.

Do you know this?

Play the game now!

## Worked Examples

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

Question 1. What is the concentration of hydroxide ions in an aqueous solution of hyrochloric acid at 25°C with a hydrogen ion concentration of 0.028 mol L-1.

1. What have you been asked to do?
Calculate the concentration of hydroxide ions
[OH- = mol L-1?
2. What information (data) have you been given?
Extract the data from the question:
[H+(aq)] = 0.028 mol L-1
temperature = 25°C,
so Kw = 10-14 (see Data Sheet)
3. What is the relationship between what you know and what you need to find out?
Write the equation (formula) for Kw:
Kw = [H+(aq)][OH-(aq)]
10-14 = [H+(aq)][OH-(aq)]

Rearrange this equation (formula) to find [OH-(aq)] in mol L-1:
[OH-(aq)] = 10-14 ÷ [H+(aq)]

4. Substitute the value for [H+(aq)] into the equation and solve:
[OH-(aq)] = 10-14 ÷ [H+(aq)]
[OH-(aq)] = 10-14 ÷ 0.028
= 3.6 × 10-13 mol L-1
Use your calculated value for [OH-] to find [H+] and compare it to that given in the question:
Kw = [OH-][H+]
10-14 = 3.6 × 10-13[H+]
10-14 ÷ 3.6 × 10-13 = [H+]
0.028 mol L-1 = [H+]
Since this value is the same as that given in the question we are confident our answer is correct.
6. State your solution to the problem:
[OH-] = 3.6 × 10-13 mol L-1

Question 2. An aqueous solution of hydrochloric acid has a pH of 4.3 at 25°C.
What is the concentration of hydroxide ions in this solution in mol L-1?

1. What have you been asked to do?
Calculate the concentration of hydroxide ions
[OH- = mol L-1?
2. What information (data) have you been given?
Extract the data from the question:
pH = 4.3
temperature = 25°C,
so Kw = 10-14 (see Data Sheet)
3. What is the relationship between what you know and what you need to find out?
Calculate the concentration of hydrogen ions in the solution:
pH = -log10[H+(aq)]
So, [H+(aq)] = 10-pH
[H+(aq)] = 10-4.3 = 5.01 × 10-5 mol L-1

Write the equation (formula) for Kw:
Kw = [H+(aq)][OH-(aq)]
10-14 = [H+(aq)][OH-(aq)]

Rearrange this equation (formula) to find [OH-(aq)] in mol L-1:
[OH-(aq)] = 10-14 ÷ [H+(aq)]

4. Substitute the value of [H+(aq)] into the equation and solve:
[OH-(aq)] = 10-14 ÷ [H+(aq)]
[OH-(aq)] = 10-14 ÷ 5.01 × 10-5
= 2.0 × 10-10 mol L-1
Use your calculated value for [OH-] to find pH and compare it to that given in the question:
Kw = [OH-][H+]
for aqueous solutions at 25°C Kw = 10-14 (from Data Sheet)
so 10-14 = 2.0 × 10-10[H+]
10-14 ÷ 2.0 × 10-10 = [H+] = 5 × 10-5 mol L-1
pH = -log10[H+] = -log10[5 × 10-5] = 4.3
Since this value is the same as that given in the question we are confident our answer is correct.
6. State your solution to the problem:
[OH-] = 2.0 × 10-10 mol L-1

Question 3. Enough water is added to 10 mL of 0.10 mol L-1 HCl(aq) to make 150 mL of solution at 25°C.
Calculate the concentration of hydroxide ions in the diluted solution in mol L-1.

1. What have you been asked to do?
Calculate the concentration of hydroxide ions
[OH- = mol L-1?
2. What information (data) have you been given?
Extract the data from the question:

Before dilution:
c1 = initial HCl(aq) concentration = 0.10 mol L-1
volume of HCl(aq) solution initially = 10 mL
Convert volume in mL to volume in L by dividing by 1000 (equivalent to multiplying by 10-3)
V1 = volume of HCl(aq) solution= 10 × 10-3 L

After dilution:
volume of HCl(aq) solution finally = 150 mL
Convert volume in mL to volume in L by dividing by 1000 (equivalent to multiplying by 10-3)
V2 = 150 × 10-3 L

temperature = 25°C,
so Kw = 10-14 (see Data Sheet)

3. What is the relationship between what you know and what you need to find out?
Calculate the concentration of hydrochloric acid after dilution (final concentration):
initial HCl(aq)concentration × initial HCl(aq) volume = final HCl(aq) concentration × final HCl(aq) volume
c1V1 = c2V2
0.10 × 10 × 10-3 = c2 × 150 × 10-3
0.001 = c2 × 0.150
0.001 ÷ 0.150 = c2 = 6.67 × 10-3 mol L-1

Determine the concentration of hydrogen ions in solution:
HCl(aq) is a strong monoprotic acid.
HCl(aq) dissociates completely to form H+(aq) and Cl-(aq).
HCl → H+(aq) + Cl-(aq)
stoichiometric ratio (mole ratio) HCl(aq) : Cl-(aq) is 1 : 1
[HCl(aq)] = [H+] = 6.67 × 10-3 mol L-1

Write the equation (formula) for Kw:
Kw = [H+(aq)][OH-(aq)]
10-14 = [H+(aq)][OH-(aq)]

Rearrange this equation (formula) to find [OH-(aq)] in mol L-1:
[OH-(aq)] = 10-14 ÷ [H+(aq)]

4. Substitute the value of [H+(aq)] into the equation and solve:
[OH-(aq)] = 10-14 ÷ [H+(aq)]
[OH-(aq)] = 10-14 ÷ 6.67 × 10-3
= 1.5 × 10-12 mol L-1
Use your calculated value for [OH-] to find the intial concentration of HCl(aq) and compare it to that given in the question:
Kw = [OH-][H+]
for aqueous solutions at 25°C Kw = 10-14 (from Data Sheet)
so 10-14 = 1.5 × 10-12[H+]
10-14 ÷ 1.5 × 10-12 = [H+] = 6.7 × 10-3 mol L-1

Calculate moles H+ using final volume of solution
volume final = 150 × 10-3 L
c(H+(aq)) = moles(H+(aq)) ÷ volume(H+(aq))
moles(H+(aq)) = c(H+(aq)) × volume(H+(aq)) = 6.7 × 10-3 × 150 × 10-3 = 1 × 10-3 mol

From the balanced chemical equation:
HCl(aq) → H+(aq) + Cl-(aq)
moles(H+(aq)) = moles(HCl(aq)) = 1 × 10-3 mol

Calculate intial [HCl(aq)]
initial volume = 10 mL = 10 × 10-3 L
[HCl(aq)] = moles(HCl(aq)) ÷ volume(HCl(aq)) = (1 × 10-3) ÷ (10 × 10-3) = 0.1 mol L-1

Since this value is the same as that given in the question we are confident our answer is correct.

6. State your solution to the problem:
[OH-] = 1.5 × 10-12 mol L-1

Do you understand this?

Take the test now!

1. The dissociation constant for water, Kw, varies with temperature. At 25°C Kw ≈ 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°C.

2. Hydrogen ions are being represented here as H+, or H+(aq) for hydrogen ions in water (hydrated hydrogen ions).
You can use H3O+ (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 Kw = [H3O+][OH-(aq)] = 10-14 (at 25°C).

3. Kw = 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 Kw changes.