Strong vs Weak Acids: Concentration, pH, and Practice

A weak-acid solution can have the same pH as a strong-acid solution. It can also have a lower pH than a different strong-acid solution. The three solutions below show both comparisons without changing what “strong” and “weak” mean.

Keep acid strength, total acid concentration, and hydronium concentration separate. A correct logarithm can still produce the wrong pH if you put the wrong concentration into it.

Strong doesn't mean concentrated

A strong acid ionizes essentially completely in water; a weak acid ionizes partially. “Dilute” and “concentrated” describe the amount of acid per volume of solution. AQA's strong and weak acids specification separates these terms. Diluting a strong acid doesn't make it a weak acid.

Use three separate entries when setting up a problem:

Quantity	Meaning
Acid strength	The acid's tendency to ionize in water
Analytical acid concentration, C	Total acid added per liter, including its ionized and un-ionized forms
Equilibrium [H₃O⁺]	Hydronium concentration after the solution reaches equilibrium

M means mol/L. Square brackets give the concentration of a particular species. For the simple weak acid HA used below, C = [HA] + [A⁻]. The analytical concentration counts both forms; [HA] counts only the acid remaining un-ionized.

Try the three-solution comparison

This is an original hypothetical worksheet. All three solutions are ideal aqueous solutions at 25 °C, each containing one monoprotic acid dissolved in water, with no added conjugate-base salt. A monoprotic acid can donate one proton per acid molecule. We neglect water's contribution to [H₃O⁺] and activity effects throughout the worksheet.

Solution	Acid classification	Analytical concentration C	Given equilibrium ionization
A	Strong monoprotic acid	0.0010 M	Essentially 100%
B	Hypothetical weak monoprotic acid HA	0.010 M	10%
C	Strong monoprotic acid	0.00010 M	Essentially 100%

Before reading the working, calculate [H₃O⁺] and pH for each row. Then identify which solutions share a pH and whether B or C has the lower pH.

B's 10% is supplied equilibrium data for this one solution. It isn't a general rule for weak acids or a percentage to reuse after dilution.

Work from the fraction that ionizes

For B, the equilibrium is:

HA + H₂O ⇌ H₃O⁺ + A⁻

Each HA that ionizes produces one H₃O⁺ and one A⁻. Write the fraction ionized as α. Under our assumptions:

[H₃O⁺] ≈ αC

Percent ionization ≈ ([H₃O⁺]/C) × 100

The denominator is the total C, including the HA that remains. Convert percentages to fractions before multiplying: 10% is 0.10, and 100% is 1.00. OpenStax's acid-strength section defines percent ionization from equilibrium composition and distinguishes it from Kₐ.

For pH, use pH ≈ −log₁₀[H₃O⁺], inserting the numerical concentration in mol/L. Its inverse is [H₃O⁺] ≈ 10^(−pH) M. These are the concentration calculations in OpenStax's pH section. The IUPAC definition of pH uses hydrogen-ion activity; our ideal model uses concentration as its approximation.

Solution	Hydronium calculation	Equilibrium [H₃O⁺]	pH
A	1.00 × 0.0010 M	0.0010 M	3.00
B	0.10 × 0.010 M	0.0010 M	3.00
C	1.00 × 0.00010 M	0.00010 M	4.00

For A and B, −log₁₀(10⁻³) = 3. For C, −log₁₀(10⁻⁴) = 4. If the powers slow you down, try the separate scientific notation practice.

A and B share a pH. B has ten times A's analytical concentration, but only one tenth of B's acid ionizes. The products αC match.

B has ten times C's hydronium concentration, so its pH is one unit lower. The weak-acid solution is more acidic in this comparison. A strength label alone can't rank solutions with different analytical concentrations.

Keep track of the HA that remains

In B, 0.0010 M of the original 0.010 M acid has ionized. The equilibrium composition is:

• [HA] = 0.010 − 0.0010 = 0.0090 M.
• [A⁻] ≈ [H₃O⁺] ≈ 0.0010 M.
• C = [HA] + [A⁻] = 0.010 M.

Ionization divides the analytical total between HA and A⁻. The conjugate acid–base pairs guide explains how those partners differ by one H⁺.

Putting 0.010 into −log₁₀C gives 2.00. The arithmetic works, but it assumes complete ionization and contradicts B's given 10%. Putting 0.0090 into the logarithm also fails: that's the remaining HA concentration. The pH calculation needs 0.0010 M, the hydronium concentration.

What happens when B is diluted?

B's equilibrium gives a concentration-based acid ionization constant:

Kₐ ≈ [H₃O⁺][A⁻]/[HA] = (0.0010 × 0.0010)/0.0090 ≈ 1.11 × 10⁻⁴

For the same acid at a fixed temperature, Kₐ stays constant in this ideal model. Percent ionization depends on concentration. OpenStax explains this distinction.

Suppose B is diluted tenfold on paper, giving a new C of 0.0010 M. Keeping the old α = 0.10 would predict [H₃O⁺] = 0.00010 M and pH 4.00. That prediction would give:

[H₃O⁺][A⁻]/[HA] = (0.00010 × 0.00010)/(0.0010 − 0.00010) ≈ 1.11 × 10⁻⁵

This is ten times smaller than B's Kₐ. The assumed 10% no longer fits the equilibrium.

Let x be the new equilibrium [H₃O⁺]. Then [A⁻] ≈ x and [HA] = 0.0010 − x. Use numerical concentrations in mol/L to solve:

x²/(0.0010 − x) = Kₐ

Rearranging gives x² + Kₐx − KₐC = 0, with C = 0.0010. The physically meaningful root is:

x = (√(Kₐ² + 4KₐC) − Kₐ)/2 ≈ 0.0002824 M

The other root is negative, so it cannot represent a concentration. Using the unrounded Kₐ from B gives:

Quantity after tenfold dilution of B	Result
[H₃O⁺]	About 2.8 × 10⁻⁴ M
Fraction ionized, x/C	About 0.28, or 28%
pH, calculated before rounding x	3.55

Keep C − x in the denominator: about 28% of the diluted acid ionizes, so treating all of C as remaining HA would distort the calculation.

The ionized fraction increased, while hydronium concentration decreased from 0.0010 M. A larger fraction of a smaller total can still give a smaller concentration.

For comparison, tenfold dilution of A gives [H₃O⁺] ≈ 0.00010 M and pH 4.00 under the same assumptions. A's pH rises by one unit; B's rises by about 0.55. Neither dilution changes the acid's strength classification.

Strong vs weak acids practice

Cover the answers and write down the quantity you need before calculating. Every question uses the same ideal monoprotic, acid-only aqueous model at 25 °C. The other weak-acid examples have their own supplied equilibrium data; they aren't new concentrations of B.

1. Two solutions each have C = 0.0050 M. One contains a strong acid; the other contains a weak acid that is 20% ionized at equilibrium. Find both hydronium concentrations and pH values.
2. A hypothetical weak-acid solution has C = 0.020 M and pH 3.00. Find its equilibrium percent ionization. Does sharing A's pH make it a strong acid?
3. Two samples of the same strong acid both have C = 0.0040 M. Their volumes are 25 mL and 100 mL. Find the total moles of acid in each. Which sample has the lower pH?
4. A hypothetical weak-acid solution has C = 0.015 M and equilibrium [H₃O⁺] = 0.00075 M. Find its percent ionization and remaining [HA].
5. A student calculates B's pH as −log₁₀(0.010) = 2.00. The logarithm is correct. Which chemical assumption is wrong?
6. After dilution of a simple weak-acid solution, its percent ionization increases. A student concludes that [H₃O⁺] must have increased too. Explain why that conclusion doesn't follow.
7. A question supplies only “an acid solution has pH 2.50.” Can you classify the acid as strong or weak? State what you can calculate and what information is missing.

Answers with the reasoning

1. Strong: [H₃O⁺] ≈ 0.0050 M, pH 2.30. Weak: [H₃O⁺] ≈ 0.0010 M, pH 3.00. For the strong acid, −log₁₀(0.0050) ≈ 2.30. For the weak acid, 0.20 × 0.0050 = 0.0010 M, then −log₁₀(0.0010) = 3.00. Equal analytical concentrations make the effect of their different ionization fractions clear.
2. 5.0% ionized; it remains weak. pH 3.00 gives [H₃O⁺] ≈ 0.0010 M. Then (0.0010/0.020) × 100 = 5.0%. Sharing A's pH means sharing its hydronium concentration, while this solution has a different C and ionization fraction.
3. 0.00010 mol and 0.00040 mol; neither has a lower pH. Use n = CV: 0.0040 × 0.025 = 0.00010 mol and 0.0040 × 0.100 = 0.00040 mol. Both samples have [H₃O⁺] ≈ 0.0040 M and pH ≈ −log₁₀(0.0040) = 2.40. More moles in a larger sample don't imply a higher concentration.
4. 5.0% ionized; [HA] ≈ 0.014 M. The percentage is (0.00075/0.015) × 100 = 5.0%. Subtraction gives [HA] = 0.015 − 0.00075 = 0.01425 M before rounding to the supplied precision. Dividing by that remaining [HA] would use the wrong denominator for percent ionization.
5. They assumed complete ionization. B's C counts HA and A⁻ together. Its supplied 10% gives [H₃O⁺] ≈ 0.10 × 0.010 = 0.0010 M and pH 3.00.
6. [H₃O⁺] depends on αC, and C has decreased. The fraction alone doesn't settle the comparison. In B's worked dilution, α rises from 0.10 to about 0.28, yet [H₃O⁺] falls from 0.0010 M to about 2.8 × 10⁻⁴ M.
7. You can't classify the acid from that pH alone. You can calculate [H₃O⁺] ≈ 10^(−2.50) M ≈ 3.2 × 10⁻³ M. To find the fraction ionized in this model, you also need C. An acid identity or Kₐ can establish its strength; the supplied pH doesn't provide either.

Save the distinction you missed

After checking an answer, make a small card for the step you skipped. These front/back prompts focus on three different errors:

Front	Back
Weak monoprotic HA: C = 0.010 M, given equilibrium ionization 10%. What goes into the pH calculation?	[H₃O⁺] ≈ 0.10 × 0.010 = 0.0010 M; pH 3.00. C includes un-ionized HA.
Two acid solutions share pH 3.00. Must they share acid strength or C?	No. They share [H₃O⁺] ≈ 0.0010 M in this model. Different C and ionization fractions can produce it.
Same weak acid after dilution: can its old percent ionization be reused?	No. Use new equilibrium data or calculate with Kₐ and the new C.

Choose the prompt that matches your error, then redo the calculation without the answer visible. The chemistry flashcard guide explains how to keep focused recall cards alongside full worked problems.

For a fresh transfer question, suppose a worksheet gives C = 0.030 M for an unnamed weak monoprotic acid, with no pH, Kₐ, or percent ionization. Name the missing information before doing a logarithm. You need enough equilibrium data to find [H₃O⁺]; the word “weak” doesn't supply a numerical fraction.