 # Hydrogen Ion Concentration of Strong Bases Chemistry Tutorial

## Key Concepts

• The concentration of hydrogen ions1, H+(aq), in an aqueous solution of a strong Arrhenius base can be calculated if we know :
(i) the temperature of the solution

and

(ii) either
(a) the pOH of the solution
or
(b) the concentration of hydroxide ions in solution.

• For an aqueous basic (alkaline) solution at 25°C2
[H+] = 10-14 ÷ [OH-]
where [H+] = concentration of hydrogen ions3 in mol L-1
and [OH-] = concentration of hydroxide ions in mol L-1
and 10-14 is the dissociation constant for water4 at 25°C
• For an aqueous basic (alkaline) solution at 25oC
[H+] = 10-14 ÷ [OH-]
but [OH-] = 10-pOH

So, [H+] = 10-14 ÷ [10-pOH]
where pOH = the pOH of the solution
and [H+] = concentration of hydrogren ions in mol L-1
and 10-14 is the dissociation constant for water at 25°C

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## Calculating the Hydrogen Ion Concentration of Strong Arrhenius Bases

Before the base 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 Arrhenius base is added to water, the base dissociates completely to form OH-(aq) and hydrated metal cations:

 Group 1 metal hydroxide: Group 2 metal hydroxide: base → hydroxideions + metal cations MOH → OH-(aq) + M+(aq) M(OH)2 → 2OH-(aq) + M2+(aq)

Adding the base to the water disturbs the water dissociation equilibrium:

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

By Le Chatelier's Principle, adding more OH-(aq) to the water will shift the equilibrium position to the left.
The water dissociation equilibrium system responds to the addition of more OH-(aq) by reacting some of the OH-(aq) with some of the H+(aq) in order to re-establish equilibrium.
So, increasing the concentration of OH-(aq) in the water, reduces the concentration of H+(aq), but, the value of the dissociation constant for the water does not change5, 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 [OH-(aq)] to calculate [H+(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 hydrogen ions in the aqueous solution:
[H+(aq)] = 10-14 ÷ [OH-(aq)]

Both [H+(aq)] and [OH-(aq)] must be in units of mol L-1 (mol/L or M)

If we know the pOH of the basic solution, we know the concentration of hydroxide ions in the solution

because pOH = -log10[OH-(aq)]

By rearranging this equation (formula) we can find the concentration of hydroxide ions in mol L-1:
[OH-(aq)] = 10-pOH

This value of [OH-(aq)] can then be used to calculate [H+(aq)] in mol L-1 using the appropriate value for Kw.

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## Worked Examples

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

Question 1. Calculate the concentration of hydrogen ions in an aqueous solution of sodium hydroxide with a pH of 12.5

1. What have you been asked to do?
Calculate concentration of hydrogen ions
[H+(aq)] = ? mol L-1
2. What information (data) have you been given in the question?
Extract the data from the question:
pH = 12.5
3. What is the relationship between what you know and what you need to find out?
Write the equation (formula) for calculating pH:
pH = -log10[H+]
Rearrange this equation to find [H+]:
[H+(aq)] = 10-pH
4. Substitute the value for pH into the equation and solve:
[H+(aq)] = 10-12.5 = 3.2 × 10-13 mol L-1
Check your answer by using your calculated value of hydrogen ion concentration to find the pH and make sure it is equal to 12.5:
pH = -log10[H+(aq)] = -log103.2 × 10-13 = 12.5
6. State your solution to the problem:
[H+(aq)] = 3.2 × 10-13 mol L-1

Question 2. The hydroxide ion concentration of an aqueous solution of barium hydroxide at 25°C is 4.2 × 10-2 mol L-1.
Calculate the concentration of hydrogen ions in this solution.

1. What have you been asked to do?
Calculate concentration of hydrogen ions
[H+(aq)] = ? mol L-1
2. What information (data) have you been given?
Extract the data from the question:
[OH-(aq)] = 4.2 × 10-2 mol L-1
temperature = 25oC
3. What is the relationship between what you know and what you need to find out?
Write the equation (formula) for the self-dissociation (ionisation) of water
Kw = [OH-][H+]
Kw = 1.0 × 10-14 (from Data Sheet)
Therefore:
1.0 × 10-14 = [OH-][H+]
Rearrange the equation to find [H+]:
[H+] = 1.0 × 10-14 ÷ [OH-]
4. Substitute in the values and solve:
[H+] = 1.0 × 10-14 ÷ 4.2 × 10-2
= 2.4 × 10-13 mol L-1
Check your answer by using your calculated value of [H+] to find the [OH-(aq)] and make sure it is 4.2 × 10-2 mol L-1.
Kw = [OH-][H+]
At 25°C, [OH-] = 10-14/[H+] = 10-14/2.4 × 10-13 = 0.042 = 4.2 × 10-2
6. State your solution to the problem:
[H+] = 2.4 × 10-13 mol L-1

3. The pOH of an aqueous solution of calcium hydroxide is 3.7 at 25°C.
Calculate the concentration of hydrogen ions in this solution.

1. What have you been asked to do?
Calculate concentration of hydrogen ions
[H+(aq)] = ? mol L-1
2. What information (data) have you been given?
Extract the data from the question:
pOH = 3.7
temperature = 25°C
3. What is the relationship between what you know and what you need to fin out?
Write the equation (formula) for finding hydroxide ion concentration:
pOH = -log10[OH-]
Rearrange to find [OH-]:
(i) [OH-] = 10-pOH

Write the equation (formula) for Kw:
Kw = [H+][OH-]
Rearrange this equation to find [H+]:
[H+] = Kw ÷ [OH-]
Kw = 1.0 × 10-14 (from Data Sheet)
Therefore:
(ii) [H+] = 1.0 × 10-14 ÷ [OH-]

4. Substitute the values into the equations and solve:
(i) [OH-] = 10-pOH
= 10-3.7
= 2.00 × 10-4 mol L-1

(ii) [H+] = 1.0 × 10-14 ÷ [OH-]
= 1.0 × 10-14 ÷ 2.00 × 10-4
= 5.0 × 10-11 mol L-1

Check your answer by using your calculated value of [H+] to find the pOH and make sure it is 3.7
pH = -log10[H+]
= -log105.0 × 10-11
= 10.3

At 25oC,
pOH = 14 - pH
= 14 - 10.3
= 3.7

6. State your solution to the problem:
[H+] = 5.0 × 10-11 mol L-1

Question 4. 0.25 g of calcium hydroxide has been dissolved fully in enough water to make exactly 2.0 L of solution at 25oC.
What is the concentration of hydrogen ions in the solution in mol L-1 ?

1. What have you been asked to do?
Calculate concentration of hydrogen ions
[H+(aq)] = ? mol L-1
2. What information (data) have you been given?
Extract the data from the question:
mass Ca(OH)2(s) = 0.25 g
volume of solution = 2.0 L
temperature = 25oC
so Kw = 1.0 x 10-14 (from Data Sheet)
3. What is the relationship between what you know and what you need to find out?
(i) Calculate the moles of Ca(OH)2(s) used:
moles = mass (g) ÷ molar mass (g mol -1)
mass Ca(OH)2(s) = 0.25 g
molar mass Ca(OH)2(s) = 40.08 + 2(16.00 + 1.008) = 74.096 g mol-1 (using the Periodic Table)
moles (Ca(OH)2(s)) = 0.25 ÷ 74.096 = 3.37 × 10-3 mol

(ii) Calculate the moles of OH- present in the solution:
Write the base dissociation equation:
Ca(OH)2(s) → Ca2+(aq) + 2OH-(aq)

Find the stoichiometric ratio (mole ratio)     Ca(OH)2(s) : OH-(aq)         is         1 : 2
Ca(OH)2(s) : OH-(aq) is 1 : 2

Use the stoichiometric ratio (mole ratio) to calculate moles of OH-(aq):
moles OH-(aq) = 2 × moles Ca(OH)2(s)
= 2 × 3.37 × 10-3
= 6.74 x 10-3 mol

(iii) Calculate the concentration of hydroxide ions in solution in mol L-1 (molarity or molar concentration):
concentration = moles ÷ volume (L)
[OH-(aq)] = mole (OH-(aq)) ÷ volume of solution
= 6.74 × 10-3 ÷ 2.0
= 3.37 × 10-3 mol L-1

(iv) Write the equation (formula) for the self-dissociation (ionisation) of water:
Kw = [OH-(aq)][H+(aq)]
and rearrange this equation to find [H+(aq)]:
[H+(aq)] = Kw ÷ [OH-(aq)]

4. Substitute in the values for Kw and [OH-(aq)]and solve:
[H+(aq)] = 1.0 × 10-14 ÷ 3.37 × 10-3
= 3.0 × 10-12 mol L-1
Check your answer by using your calculated value of [H+(aq)] to find the mass of Ca(OH)2 in 2 L of aqueous solution and make sure it is 0.25 g:
at 25°C: [OH-(aq)] = 10-14/[H+(aq)] = 10-14/3.0 × 10-12 = 3.33 × 10-3 mol L-1
For a strong base: Ca(OH)2(s) → Ca2+(aq) + 2OH-(aq)
2 mol L-1 OH- produced from 1 mol L-1 of Ca(OH)2(s) so 3.33 × 10-3 mol L-1 OH- is produced from 3.33 × 10-3/2 = 1.67 × 10-3 mol L-1 Ca(OH)2
In 2.0 L of solution, moles Ca(OH)2 = molarity (mol L-1) × volume (L) = 1.67 × 10-3 × 2.0 = 3.33 × 10-3 mol
mass Ca(OH)2 = moles × molar mass = 3.33 × 10-3 × 74.096 = 0.25 g
6. State your solution to the problem:
[H+(aq)] = 3.0 × 10-12 mol L-1

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1. 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 or oxidanium 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).

2. 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.

3. 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 oxidanium 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).

4. Kw = 10-14 is only an approximation, but it is one that is widely used for convenience.

5. This assumes that the temperature of the water remains constant.
If the temperature changes, the value of Kw changes.