pH of Strong Acids Calculations Chemistry Tutorial

Key Concepts

A strong acid is one that dissociates completely in water to form hydrogen ions¹:

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^-]

Square brackets, [ ], are Chemist's short-hand for concentration in mol L^-1 (molarity or molar concentration).

pH is a measure of the hydrogen ion, or proton, (H⁺) concentration in a solution.

pH of a solution is calculated using the formula (equation):
pH = -log₁₀[H⁺]

where [H⁺] is the concentration of hydrogen ions in mol L^-1 (mol/L or M)²

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Calculating the pH of Strong Monoprotic Acids

Reagent bottles containing strong acids are usually labelled with the name and/or formula and the concentration of the acid they contain,
for example³:

Hydrochloric acid
[HCl_(aq)] = 1.0 mol L^-1

Sometimes we need to be able to convert this "concentration of acid" into a pH.
The steps you can follow to do this are given below:

Step 1. 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
hydrochloric acid	HCl
hydrobromic acid	HBr
hyroiodic acid	HI
nitric acid	HNO₃
perchloric acid	HClO₄

Step 2. Use the concentration of the acid to determine the concentration of hydrogen ions in solution:

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⁺]

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

Step 3. Use the concentration of hydrogen ions in mol L^-1, [H⁺], to calculate the pH of the solution:

pH = -log₁₀[H⁺]

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Worked Examples

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

Question 1. Find the pH of 0.2 mol L^-1 HCl_(aq).

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

What information (data) have you been given?
Extract the data from the question:
[HCl_(aq)] = 0.2 mol L^-1

What is the relationship between what you know and what you need to find out?
Write the balanced chemical equation for the dissociation of HCl in water:
HCl → H⁺_(aq) + Cl^-_(aq)
Use the balanced chemical equation to calculate the concentration of hydrogen ions in solution in mol L^-1:
[strong monoprotic acid] = [H⁺]
[HCl_(aq)] = [H⁺] = 0.2 mol L^-1

Write the equation (formula) for calculating pH:
pH = -log₁₀[H⁺]

Substitute the value for [H⁺] into the equation and solve:
pH = -log₁₀[H⁺]
pH = -log₁₀[H⁺]
= -log₁₀[0.2]
= 0.7

Is your answer plausible?
Round-off the concentration of the acid given in the question, that is
[HCl_(aq)] ≈ 0.1 mol L^-1 = 10^-1 mol L^-1
HCl fully dissociates (ionises) in water so
[HCl_(aq)] = [H⁺(aq)] ≈ 10^-1 mol L^-1
pH = -log₁₀[H⁺] = -log₁₀[10^-1] = 1
Since this value is about the same as the one we calculated carefully, we are confident our answer is correct.

State your solution to the problem:
pH = 0.7

Question 2. 0.0010 mol of hydrogen chloride is dissolved in water to make 0.10 L of solution.
Calculate the pH of the aqueous hydrochloric acid solution.

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

What information (data) have you been given?
Extract the data from the question:
moles of HCl_(aq) = n(HCl) = 0.0010 mol
volume of solution = V = 0.10 L

What is the relationship between what you know and what you need to find out?
Write the balanced chemical equation for the dissociation of HCl in water:
HCl → H⁺_(aq) + Cl^-_(aq)
Use the balanced chemical equation to calculate the moles of hydrogen ions in solution:
stoichiometric ratio of HCl : H⁺ is 1 : 1
0.0010 moles HCl produces 0.0010 moles H⁺

Calculate the concentration of hydrogen ions in solution in mol L^-1:
[H⁺_(aq)] = moles H⁺ ÷ volume of solution in L
[H⁺_(aq)] = 0.0010 ÷ 0.10 = 0.010 mol L^-1

Write the equation (formula) for calculating pH:
pH = -log₁₀[H⁺]

Substitute the value for [H⁺] into the equation and solve:
pH = -log₁₀[H⁺]
pH = -log₁₀[H⁺]
= -log₁₀[0.010]
= 2.0

Is your answer plausible?
Use your calculated pH value to find the moles of acid in 0.1 L of solution and compare it to that given in the question:
pH = 2.0 = -log₁₀[H⁺]
so, [H⁺] = 10^-pH = 10^-2.0 mol L^-1
HCl_(aq) is a strong acid, it fully dissociates (ionises) so
[HCl_(aq)] = [H⁺] = 10^-2.0 mol L^-1
[HCl_(aq)] = moles(HCl) ÷ volume(solution)
moles(HCl) = [HCl_(aq)] × volume(solution) = 10^-2 × 0.10 = 10^-3 mol = 0.0010 mol
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:
pH = 2.0

Question 3. 0.12 g of hydrogen chloride is dissolved in enough water to make 0.25 L of solution.
Calculate the pH of the hydrochloric acid solution.

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

What information (data) have you been given?
Extract the data from the question:
mass of HCl_(aq) = m(HCl) = 0.12 g
volume of solution = V = 0.25 L

What is the relationship between what you know and what you need to find out?
Calculate the moles of HCl:
moles HCl = mass HCl in grams ÷ molar mass of HCl in g mol ^-1
molar mass HCl = 1.008 + 35.45 = 36.458 g mol ^-1 (use Periodic Table to find molar mass of each element)
moles HCl = 0.12 g ÷ 36.458 g mol ^-1
= 3.29 × 10^-3 mol
Write the balanced chemical equation for the dissociation (ionisation) of HCl in water:
HCl → H⁺_(aq) + Cl^-_(aq)

Use the balanced chemical equation to calculate the moles of hydrogen ions in solution:
stoichiometric ratio of HCl : H⁺ is 1 : 1
3.29 × 10^-3 moles HCl produces 3.29 × 10^-3 moles H⁺

Calculate the concentration of hydrogen ions in solution in mol L^-1:
[H⁺_(aq)] = moles H⁺ ÷ volume of solution in L
[H⁺_(aq)] = 3.29 × 10^-3 ÷ 0.25
= 0.013 mol L^-1

Write the equation (formula) for calculating pH:
pH = -log₁₀[H⁺]

Substitute the value for [H⁺] into the equation and solve:
pH = -log₁₀[H⁺]
= -log₁₀[0.013]
= 1.9

Is your answer plausible?
Round off the values given in the question to find an approximate, "ball-park" value for pH and compare it to your calculated value:
mass(HCl) = 0.12 g ≈ 0.1 g
molar mass(HCl) ≈ 1 + 35 ≈ 50
moles(HCl) ≈ 0.1 ÷ 50 = 0.002
volume(solution) = 0.25 L ≈ 0.2 L
[HCl] ≈ 0.002 ÷ 0.2 = 0.01 = 10^-2 mol L^-1
HCl is a strong acid, it fully dissociates (ionises)
[HCl_(aq)] = [H⁺_(aq)] = 10^-2 mol L^-1
pH = -log₁₀[H⁺_(aq)] = -log₁₀[10^-2] =2
Since this value is similar to (in the ball-park of) the result of our careful calculations we are confident our answer is correct.

State your solution to the problem:
pH = 1.9

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1. A hydrogen ion, H⁺, is a naked proton!
A naked proton is very reactive, so, in practice an H⁺ ion "jumps" onto a water molecule to form the hydronium (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 or H⁺ in aqueous solutions, they really mean H₃O⁺.
Should you write H⁺ or H₃O⁺?
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.
We use H⁺ and H⁺_(aq) here because it highlights the fact that pH relates to H⁺ concentration.

2. If you decide to use H₃O⁺ instead of H⁺_(aq), then the equation for finding pH becomes:
pH = -log₁₀[H₃O⁺]

3. The label will probably contain other information as well, such as a "corrosive" symbol and information about the "purity" or "grade" of the reagent.