What you'll be able to do: Interpret and calculate points on a titration curve and choose an appropriate indicator.
A titration curve is the whole unit in one picture: strong acid regions, buffer regions, hydrolysis at the equivalence point. Learning to read it ties everything together.
These are recommended, not required. You can start this lesson at any time.
Titrant is added from a burette and pH is recorded. The curve is flat where the solution resists change and steep where a small addition swamps whatever was buffering it. The steep region is what makes the endpoint detectable.
The initial pH is simply -log of the acid concentration. Before equivalence you divide the excess moles of acid by the total volume; at equivalence only water and spectator ions remain, so pH = 7.00; after equivalence you divide the excess moles of base by the total volume. The jump spans roughly pH 3 to 11.
pH = 7.00 at the equivalence point of a strong-strong titration
Region 1, before any base, is a weak acid ICE problem. Region 2 is a buffer, handled with Henderson-Hasselbalch on moles. Region 3, the half-equivalence point, has equal moles of acid and conjugate base, so pH = pKa and the curve is at its flattest. Region 4, equivalence, is a solution of the conjugate base alone, so it is basic. Region 5 is governed by excess strong base.
At equivalence all the weak acid has become A⁻, which then removes a proton from water. Compute its concentration by dividing moles by the total volume, get Kb from Kw/Ka, and run the weak base calculation. The result is typically pH 8 to 9.
An indicator is itself a weak acid whose two forms differ in colour, and it changes over roughly pKa plus or minus 1. Phenolphthalein (range 8.3 to 10.0) suits a weak acid titration, methyl red (4.4 to 6.2) suits a weak base titration, and either works for strong-strong because the jump is so large.
HIn + H₂O ⇌ H₃O⁺ + In⁻
H₂CO₃ or H₃PO₄ titrated with base give one steep rise per proton, with a buffer plateau between each. Each half-equivalence point reveals a successive pKa, and the last proton is often too weak to show a usable jump at all.
Moles of titrant equal moles of analyte, by stoichiometry.
Where the indicator changes colour; ideally very close to equivalence.
pH = pKa for a weak acid titrated with strong base.
7.00 for strong-strong, above 7 for weak acid, below 7 for weak base.
Pick a pKa within about one unit of the equivalence pH.
M(a)V(a) = M(b)V(b) for a 1:1 titration
pH = pKa + log(n(A⁻) / n(HA))
pH = pKa
25.0 mL of 0.100 M CH₃COOH (Ka = 1.8 × 10⁻⁵ is titrated with 0.100 M NaOH. Find the pH after 10.0 mL of base has been added.
Try it first: Work out whether you are in the buffer region before choosing a method.
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For the same titration, find the pH at the equivalence point.
Try it first: Find the total volume before computing any concentration.
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A weak acid titration shows pH 5.15 after exactly half the equivalence volume of base has been added. What are pKa and Ka?
Try it first: Recall what is special about the half-equivalence point.
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Why it's wrong: Only strong-strong titrations produce a neutral salt.
Check instead: Ask whether a hydrolysing ion is left behind.
Why it's wrong: The endpoint is an indicator observation and only approximates equivalence.
Check instead: Choose an indicator whose range brackets the equivalence pH.
Why it's wrong: Volume is changing throughout the titration.
Check instead: Use moles; the volume cancels in the ratio.
Why it's wrong: The conjugate base has been diluted by the added titrant.
Check instead: Add the analyte and titrant volumes.
Why it's wrong: It changes near pH 5, far below the equivalence pH near 9.
Check instead: Phenolphthalein matches that curve.
No practice questions are available for this topic yet. You can still practice the whole unit.
A titration curve plots pH against added titrant. For a strong acid titrated with a strong base, the pH rises slowly, jumps steeply through pH 7 and levels off; the equivalence point is at pH 7.00 because the salt formed is neutral. For a weak acid with a strong base the curve starts higher, passes through a buffer region where the Henderson-Hasselbalch equation applies, reaches the half-equivalence point where pH = pKa exactly, and finishes at an equivalence point above 7 because the conjugate base hydrolyses. A weak base titrated with strong acid mirrors this with an equivalence point below 7. The equivalence point is the stoichiometric point where moles of titrant equal moles of analyte; the endpoint is where the indicator changes colour, and a good indicator has a pKa within about one unit of the equivalence pH so those two nearly coincide. Polyprotic acids give one steep rise per ionizable proton.
This lesson is original Chem Help content. No external sources were adapted.