Worked example 1
Calculate the pH of 0.0045 M Ba(OH)₂ at 25 C.
Try it first: Count how many hydroxide ions each formula unit releases.
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Acids, Bases & Aqueous EquilibriaAcids and BasesContent level: Core 18 min
Strong Acid and Strong Base Calculations
What you'll be able to do: Calculate the pH of strong acid and strong base solutions, including dilutions and mixtures that neutralize.
Introduction
Strong acids and bases ionize completely, which makes their pH calculations pure bookkeeping. The skill worth building here is tracking moles through dilution and neutralization.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Recognize the common strong acids and strong bases
- Calculate pH directly from a strong acid or strong base concentration
- Handle dilutions with M₁V₁ = M₂V₂
- Find the pH of a mixture by finding the excess in moles
Lesson
Complete ionization
For a strong acid the reaction HA + H₂O → H₃O⁺ + A⁻ is written with a single arrow. There is no equilibrium constant because effectively no HA remains, so 0.025 M HNO₃ simply gives [H₃O⁺] = 0.025 M and pH = 1.60.
Six strong acids are worth memorizing: HCl, HBr, HI, HNO₃, HClO₄ and H₂SO₄.
Strong bases and stoichiometry
NaOH and KOH give one OH⁻ per formula unit, but Ba(OH)₂ and Ca(OH)₂ give two. A 0.010 M Ba(OH)₂ solution therefore has [OH⁻] = 0.020 M, pOH = 1.70 and pH = 12.30.
[OH⁻] = n x M(base), where n is the number of OH⁻ groups
Dilution
Diluting changes the concentration but not the number of moles of solute, so M₁V₁ = M₂V₂. Each tenfold dilution of a strong acid raises the pH by exactly one unit, until the solution becomes so dilute that water itself matters.
M₁V₁ = M₂V₂
Mixing acid with base
Strong acid plus strong base gives a straightforward neutralization: H₃O⁺ + OH⁻ → 2 H₂O. Convert both to moles, subtract, and divide whatever is left over by the combined volume. If they cancel exactly, the solution is neutral.
Never average two pH values. pH is logarithmic; moles are what add and subtract.
Very dilute acids
For 1 × 10⁻⁸ M HCl, a naive calculation gives pH 8, which would make an acid basic. The error is ignoring the 10⁻⁷ M of H₃O⁺ that water supplies. Below about 10⁻⁶ M you must account for autoionization, and the true pH approaches 7 from the acidic side.
Water matters for extremely dilute acid
For ordinary strong-acid concentrations, [H₃O⁺] is approximately the acid's analytical concentration. Near 10⁻⁷ M at 25 °C, water's own autoionization is comparable, so the approximation can even predict an impossible basic pH for an added acid. With monoprotic strong acid concentration C, charge balance and Kw give [H₃O⁺] = (C + sqrt(C²+4Kw))/2.
At the water background scale, check whether the simple approximation remains valid.
Key ideas
Rule
Strong acid pH
[H₃O⁺] equals the nominal acid concentration; pH = -log of it.
Rule
Polyhydroxy bases
Multiply the base molarity by the number of OH⁻ groups.
Rule
Dilution
M₁V₁ = M₂V₂; moles of solute are conserved.
Rule
Mixtures
Work in moles, find the excess, divide by total volume.
Rule
Dilute limit
Below about 10⁻⁶ M, water autoionization cannot be neglected.
Equation
Strong acid
[H₃O⁺] = C(acid)
- C(acid) = molarity of a monoprotic strong acid
Equation
Dilution
M₁V₁ = M₂V₂
- V = volume in any consistent unit
Equation
Excess after mixing
n(excess) = n(H₃O⁺) - n(OH⁻)
- n = moles
Worked examples
Worked example 2
25.0 mL of 0.100 M HCl is mixed with 15.0 mL of 0.100 M NaOH. Find the pH.
Try it first: Convert both to moles before doing anything else.
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Worked example 3
10.0 mL of 0.50 M HNO₃ is diluted to 250.0 mL. What is the pH?
Try it first: Find the new concentration first, then take the log.
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Worked example 4
At 25 °C, estimate the pH of 1.0×10⁻⁸ M HCl using Kw=1.0×10⁻¹⁴.
Try it first: Do not take -log(10⁻⁸=8 as the answer.
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Common mistakes
Forgetting that Ba(OH)₂ supplies two hydroxide ions.
Why it's wrong: It halves the calculated [OH⁻] and shifts the pH by 0.30.
Check instead: Read the formula before computing.
Averaging the pH of two mixed solutions.
Why it's wrong: pH is logarithmic and does not add.
Check instead: Subtract moles, then divide by total volume.
Using only the acid volume after mixing.
Why it's wrong: The final concentration depends on the combined volume.
Check instead: Add the two volumes.
Reporting pH 8 for 1 × 10⁻⁸ M HCl.
Why it's wrong: An acid can never be basic; water autoionization dominates at that dilution.
Check instead: Recognize the dilute limit and expect a pH just under 7.
Writing an equilibrium expression for HCl.
Why it's wrong: Strong acids ionize completely, so there is no equilibrium to solve.
Check instead: Use Ka only for weak acids.
After the lesson · Question bank
Independent practice
Work through new problems on this topic on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
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What you should now know
Strong acids (HCl, HBr, HI, HNO₃, HClO₄, H₂SO₄ in its first proton) ionize essentially completely, so [H₃O⁺] equals the acid concentration; there is no equilibrium to solve. Strong bases are the group 1 hydroxides plus Ca(OH)₂, Sr(OH)₂ and Ba(OH)₂, and the group 2 hydroxides release two OH⁻ per formula unit. For dilutions use M₁V₁ = M₂V₂, and for mixtures work in moles: compute moles of H₃O⁺ and OH⁻, cancel the smaller against the larger, and divide the excess by the total volume. When an acid is more dilute than about 10⁻⁶ M, the autoionization of water contributes measurably and a simple -log of the acid concentration is no longer valid.
- Strong acids and bases ionize completely, so no Ka is needed
- Group 2 hydroxides supply two OH⁻ per formula unit
- Dilution follows M₁V₁ = M₂V₂ and shifts pH by log of the dilution factor
- Mixtures are solved in moles, then divided by total volume
- Below about 10⁻⁶ M the autoionization of water matters
Sources and further reading
This lesson is original Chem Help content. No external sources were adapted.