go to the AUS-e-TUTE homepage

pH of Strong Acids Calculations Tutorial

Key Concepts

Please do not block ads on this website.
No ads = no money for us = no free stuff for you!

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 acidHCl
hydrobromic acidHBr
hyroiodic acidHI
nitric acidHNO3
perchloric acidHClO4

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 : HAH+ + A-
for 1 mole of acid : 1 mole HA1 mole H++ 1 mole A-
for 0.1 mole of acid : 0.1 mole HA0.1 mole H++ 0.1 mole A-
for 0.5 mole of acid : 0.5 mole HA0.5 mole H++ 0.5 mole A-
for 2.3 mole of acid : 2.3 mole HA2.3 mole H++ 2.3 mole A-
so for n mole of acid : n mole HAn 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 : HAH+ + 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+]

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

pH = -log10[H+]

Do you know this?

Join AUS-e-TUTE!

Play the game now!

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).

  1. What have you been asked to do?
    Calculate the pH
    pH = ?
  2. What information (data) have you been given?
    Extract the data from the question:
    [HCl(aq)] = 0.2 mol L-1
  3. 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+]

  4. Substitute the value for [H+] into the equation and solve:
    pH = -log10[H+]
    pH = -log10[H+]
    = -log10[0.2]
    = 0.7
  5. 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.
  6. 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.

  1. What have you been asked to do?
    Calculate the pH
    pH = ?
  2. 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
  3. 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+]

  4. Substitute the value for [H+] into the equation and solve:
    pH = -log10[H+]
    pH = -log10[H+]
    = -log10[0.010]
    = 2.0
  5. 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.
  6. 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.

  1. What have you been asked to do?
    Calculate the pH
    pH = ?
  2. 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
  3. 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+]

  4. Substitute the value for [H+] into the equation and solve:
    pH = -log10[H+]
    = -log10[0.013]
    = 1.9
  5. 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.
  6. State your solution to the problem:
    pH = 1.9

Do you understand this?

Join AUS-e-TUTE!

Take the test now!


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 (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.