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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What you'll be able to do: Calculate the pH of strong acid and strong base solutions, including dilutions and mixtures that neutralize.
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.
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.
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
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
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.
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.
[H₃O⁺] equals the nominal acid concentration; pH = -log of it.
Multiply the base molarity by the number of OH⁻ groups.
M₁V₁ = M₂V₂; moles of solute are conserved.
Work in moles, find the excess, divide by total volume.
Below about 10⁻⁶ M, water autoionization cannot be neglected.
[H₃O⁺] = C(acid)
M₁V₁ = M₂V₂
n(excess) = n(H₃O⁺) - n(OH⁻)
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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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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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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Why it's wrong: It halves the calculated [OH⁻] and shifts the pH by 0.30.
Check instead: Read the formula before computing.
Why it's wrong: pH is logarithmic and does not add.
Check instead: Subtract moles, then divide by total volume.
Why it's wrong: The final concentration depends on the combined volume.
Check instead: Add the two volumes.
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.
Why it's wrong: Strong acids ionize completely, so there is no equilibrium to solve.
Check instead: Use Ka only for weak acids.
No practice questions are available for this topic yet. You can still practice the whole unit.
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.
This lesson is original Chem Help content. No external sources were adapted.