Hydrogen Ion Concentration of Strong Acids Calculations Chemistry Tutorial

Key Concepts

Concentration of hydrogen ions in an aqueous solution can be calculated if the pH of the solution is known:
[H⁺] = 10^-pH
where pH is the pH of the aqueous solution
and [H⁺] = concentration of hydrogen ions¹ in mol L^-1
Square brackets, [ ], are Chemist's short-hand for concentration in mol L^-1 (molarity or molar concentration).

A strong acid is one that dissociates completely in water to form hydrogen ions:

acid → hydrogen
ions + anions

HA → H⁺ + A^-

For a strong monoprotic acid, an acid that contains only 1 hydrogen atom that can dissociate, the concentration of hydrogen ions in solution is only dependent on the concentration of the original acid.

monoprotic
acid → hydrogen
ions + anion

HA → H⁺ + A^-

[HA] = [H⁺] = [A^-]

This means that the concentration of hydrogen ions in solution is the same as the concentration of the monoprotic strong acid.

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Calculating the Hydrogen Ion Concentration of Strong Monoprotic Acids

pH is measure of the hydrogen ion concentration in a solution:

pH = -log₁₀[H⁺_(aq)]

We can rearrange this equation to find the concentration of hydrogen ions in a solution given its pH:

[H⁺] = 10^-pH

There are 2 steps for calculating the concentration of hydrogen ions (or oxonium or oxidanium or hydronium ions H₃O⁺) in a solution if you have been given the pH of the solution:

Step 1. Write the equation for finding [H⁺]:

[H⁺] = 10^-pH

Step 2. Substitute in the value for pH and solve to give the concentration of H⁺ in mol L^-1

Calculating the Concentration of Strong Monoprotic Acids

If we know the pH of a solution of strong acid, we can use this to calculate the concentration of the acid.

Step 1. Write the equation for finding [H⁺]:

[H⁺] = 10^-pH

Step 2. Substitute in the value for value for pH and solve to give the concentration of H⁺ in mol L^-1

Step 3. Write the equation for the complete dissociation of the strong monoprotic acid:

	monoprotic acid	→	hydrogen ions	+	anion
hydrochloric acid	HCl	→	H⁺	+	Cl^-
hydrobromic acid	HBr	→	H⁺	+	Br^-
hydroiodic acid	HI	→	H⁺	+	I^-
nitric acid	HNO₃	→	H⁺	+	NO₃^-
perchloric acid	HClO₄	→	H⁺	+	ClO₄^-

Strong Monoprotic Acids
hyrochloric acid	HCl
hydrobromic acid	HBr
hyroiodic acid	HI
nitric acid	HNO₃
perchloric acid	HClO₄

Step 4. Use the concentration of the hydrogen ions in solution to determine the concentration of the acid :

For a monoprotic acid, the stoichiometric ratio (mole ratio) of the acid, HA, to the hydrogen ions, H⁺, is 1 : 1

general monoprotic acid : HA → H⁺ + A^-

for 1 mole of acid : 1 mole HA → 1 mole H⁺ + 1 mole A^-

for 0.1 mole of acid : 0.1 mole HA → 0.1 mole H⁺ + 0.1 mole A^-

for 0.5 mole of acid : 0.5 mole HA → 0.5 mole H⁺ + 0.5 mole A^-

for 2.3 mole of acid : 2.3 mole HA → 2.3 mole H⁺ + 2.3 mole A^-

so for n mole of acid : n mole HA → n mole H⁺ + n mole A^-

Concentration in mol L^-1 (molarity or molar concentration) is calculated by dividing moles by volume in litres:

molarity = moles ÷ volume

The volume of the solution is the same for both the undissociated acid, HA, and for the hydrogen ions, H⁺, it produces.

general monoprotic acid : HA → H⁺ + A^-

for n mole of acid in 1 L of solution: [HA]=n/1 → [H⁺]=n/1 + [A^-]=n/1

for n mole of acid in 2 L of solution: [HA]=n/2 → [H⁺]=n/2 + [A^-]=n/2

for n mole of acid in 0.4 L of solution: [HA]=n/0.4 → [H⁺]=n/0.4 + [A^-]=n/0.4

for n mole of acid in 1.3 L of solution: [HA]=n/1.3 → [H⁺]=n/1.3 + [A^-]=n/1.3

so for n mole of acid in V L of solution: [HA]=n/V → [H⁺]=n/V + [A^-]=n/V

We can see that the concentration of the hydrogen ions produced by the strong monoprotic acid will be the same as the concentration of the acid.

[HA] = [H⁺] = 10^-pH

general monoprotic acid :	HA	→	H⁺	+	A^-
for 1 mole of acid :	1 mole HA	→	1 mole H⁺	+	1 mole A^-
for 0.1 mole of acid :	0.1 mole HA	→	0.1 mole H⁺	+	0.1 mole A^-
for 0.5 mole of acid :	0.5 mole HA	→	0.5 mole H⁺	+	0.5 mole A^-
for 2.3 mole of acid :	2.3 mole HA	→	2.3 mole H⁺	+	2.3 mole A^-
so for n mole of acid :	n mole HA	→	n mole H⁺	+	n mole A^-

general monoprotic acid :	HA	→	H⁺	+	A^-
for n mole of acid in 1 L of solution:	[HA]=n/1	→	[H⁺]=n/1	+	[A^-]=n/1
for n mole of acid in 2 L of solution:	[HA]=n/2	→	[H⁺]=n/2	+	[A^-]=n/2
for n mole of acid in 0.4 L of solution:	[HA]=n/0.4	→	[H⁺]=n/0.4	+	[A^-]=n/0.4
for n mole of acid in 1.3 L of solution:	[HA]=n/1.3	→	[H⁺]=n/1.3	+	[A^-]=n/1.3
so for n mole of acid in V L of solution:	[HA]=n/V	→	[H⁺]=n/V	+	[A^-]=n/V

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Worked Examples

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

Question 1. Find the concentration of hydrogen ions in an aqueous solution of hydrochloric acid with a pH of 2.0

What have you been asked to do?
Calculate the concentration of hydrogen ions
[H⁺] = ? mol L^-1

What information (data) have you been given?
Extract the data from the question:
pH = 2.0

What is the relationship between what you know and what you need to find out?
Write the equation (formula) for finding [H⁺]:
[H⁺] = 10^-pH

Substitute in the value for pH and solve:
[H⁺] = 10^-pH
[H⁺] = 10^-2.0
= 0.010 mol L^-1

Is your answer plausible?
Use your calculated value for [H⁺] to find pH and compare it to that given in the question:
pH = -log₁₀[H⁺] = -log₁₀[H⁺] = 2
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:
[H⁺] = 0.010 mol L^-1

Question 2. An aqueous solution of hydrochloric acid has a pH of 3.6
Calculate the concentration of the acid in mol L^-1.

What have you been asked to do?
Calculate the concentration of the hydrochloric acid
[HCl_(aq)] = ? mol L^-1

What information (data) have you been given?
Extract the data from the question:
pH = 3.6

What is the relationship between what you know and what you need to find out?
Write the equation (formula) for finding the concentration of hydrogen ions in solution:
[H⁺] = 10^-pH
Substitute in the pH value and solve:
[H⁺] = 10^-3.6 = 2.5 × 10^-4 mol L^-1

Write the balanced chemical equation for the dissociation of the acid:
HCl → H⁺_(aq) + Cl^-_(aq)

Find the mole ratio (stoichiometric ratio) :
H⁺ : HCl
1 : 1

Determine the concentration of the acid using the mole ratio (stoichiometric ratio):
1 mole per litre H⁺ is produced by 1 mole per litre HCl
So, 2.5 × 10^-4 mol L^-1 H⁺ is produced by 2.5 × 10^-4 mol L^-1 HCl
Concentration of the acid is 2.5 × 10^-4 mol L^-1

Is your answer plausible?
Use your calculated value for [HCl_(aq)] to find pH and compare it to that given in the question:
HCl_(aq) → H⁺_(aq) + Cl^-_(aq)
[HCl_(aq)] = [H⁺] = 2.5 × 10^-4 mol L^-1
pH = -log₁₀[H⁺] = -log₁₀[2.5 × 10^-4] = 3.6
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:
[HCl_(aq)] = 2.5 × 10^-4 mol L^-1

3. 0.25 L of an aqueous solution of hydrochloric acid has a pH of 3.5
Calculate the moles of hydrogen ions present in the solution.

What have you been asked to do?
Calculate the moles of hydrogen ions
n(H⁺) = ? mol

What information (data) have you been given?
Extract the data from the question:
pH = 3.5
volume = V = 0.25 L

What is the relationship between what you know and what you need to find out?
Write the equation (formula) for finding the concentration of hydrogen ions in solution:
[H⁺] = 10^-pH
Substitute in the pH value and solve:
[H⁺] = 10^-3.5 = 3.2 × 10^-4 mol L^-1

Write the equation (formula) for finding moles given concentration and volume
moles = concentration (mol L^-1) × volume (L)
concentration of hydrogen ions = [H⁺] = 3.2 × 10^-4 mol L^-1
volume of solution = 0.25 L

Substitute the values into the equation and solve:
moles = concentration (mol L^-1) × volume (L)
moles(H⁺) = 3.2 × 10^-4 mol L^-1 × 0.25
= 8.0 × 10^-5 mol
Is your answer plausible?
Use your calculated value for moles of H⁺ to find pH and compare it to that given in the question:
[H⁺] = n(H⁺) ÷ V(solution) = 8.0 × 10^-5 mol ÷ 0.25 L = 3.2 × 10^-4 mol L^-1
pH = -log₁₀[H⁺] = -log₁₀[3.2 × 10^-4] = 3.5
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:
n(H⁺) = 8.0 × 10^-5 mol

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1. A hydrogen ion is a hydrogen atom that has lost an electron.
Given that the most common naturally occurring isotope of hydrogen contains just 1 proton and NO neutrons in its nucleus, and 1 "orbiting" electron, then, when this isotope loses an electron it is just a proton!
Because naturally occuring hydrogen is also made up of very tiny amount of other isotopes of hydrogen, the term hydron is preferred by IUPAC to represent H⁺, but most Chemists will still use the term proton when referring to H⁺.
Most hydrogen ions, H⁺, are just a naked proton!
A naked proton is very reactive, so, in practice an H⁺ ion "jumps" onto a water molecule to form the hydronium (or oxidanium or oxonium) ion, H₃O⁺.
For this reason H₃O⁺ is also known as a hydrated hydrogen ion or hydrated proton.
When Chemists refer to hydrogen ions, hydrons, or H⁺ in aqueous solutions, they really mean H₃O⁺.
Should you write H⁺ or H₃O⁺ when talking about aqueous solutions?
Generally speaking, it doesn't matter, but it would be better to refer to H⁺_(aq), rather than H⁺, so that there is no confusion about the nature of the proton.
We use H⁺ and H⁺_(aq) here because it highlights the fact that pH relates to H⁺ concentration.
If you decide to use H₃O⁺ instead of H⁺_(aq), then the equation for finding H₃O⁺ concentration becomes:
[H₃O⁺] = 10^-pH