4.6 Intro to Titration

Introduction to Titration

Pick from 4 reactions (precipitation, redox 1:5, acid-base, diprotic 2:1) and add titrant by drops or auto-run. Live mole bars track analyte consumed, and V_eq updates from moles — not color.

Mole Ratio (k)4 ReactionsV at EquivalenceEndpoint vs Eq
Topic 4.6

Introduction to Titration

Identify the equivalence point in a titration based on the amounts of the titrant and analyte, assuming the titration reaction goes to completion.

A titration determines the amount of an analyte in solution by reacting it with a titrant of known concentration that reacts specifically and quantitatively with it.

Two terms that the CED distinguishes carefully, and that the exam tests:

  • Equivalence point — the point at which the analyte is totally consumed by the reacting species in the titrant. This is a stoichiometric definition. It is a fact about moles, not something you can see.
  • Endpoint — the observable event, such as an indicator color change, that signals the equivalence point has been reached.

A well-chosen indicator makes the endpoint fall as close to the equivalence point as possible. The difference between them is the titration error.

The core calculation at the equivalence point:

moles of titrant added × (stoichiometric ratio) = moles of analyte originally present

Then divide by the analyte's volume to get its concentration. The stoichiometric ratio is critical: titrating H₂SO₄ with NaOH requires 2 mol of base per mole of acid.

Procedure and error sources. The buret is read to two decimal places (±0.01 mL) at the bottom of the meniscus, both before and after. Adding water to the analyte flask does not change the moles of analyte, so it does not affect the result. Failing to rinse the buret with titrant, however, dilutes the titrant, requiring a larger volume and overstating the analyte.

Key points

  • Equivalence point = stoichiometric; endpoint = observed. Never use them interchangeably.
  • At equivalence, moles of titrant × the stoichiometric ratio = moles of analyte.
  • Extra water in the flask does not change the moles of analyte, so it does not change the result.
  • A buret not rinsed with titrant gives a titrant that is too dilute and an analyte result that is too high.

Equations

  • not on the sheeta/b is the stoichiometric ratio of analyte to titrant from the balanced equation.

Common mistakes

  • Ignoring the stoichiometric ratio. A 1:1 assumption fails for diprotic acids and Group 2 hydroxides.
  • Confusing equivalence point with the half-equivalence point (see 8.5).
  • Thinking dilution of the analyte matters. Moles of analyte are unchanged by added water.
  • Reading the buret from the top. Burets are numbered downward; final minus initial gives volume delivered.

Worked example

A 25.00 mL sample of H₂SO₄ requires 32.45 mL of 0.1050 M NaOH to reach the equivalence point. Calculate the concentration of the acid.

Balanced reaction: H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

Moles of titrant:
n(NaOH) = (0.1050 mol/L)(0.03245 L) = 3.407 × 10⁻³ mol

Apply the ratio (1 mol H₂SO₄ per 2 mol NaOH):
n(H₂SO₄) = 3.407 × 10⁻³ ÷ 2 = 1.704 × 10⁻³ mol

Concentration:
M = 1.704 × 10⁻³ ÷ 0.02500 L = 0.06814 M H₂SO₄

Note the ratio is doing real work here: forgetting it would double the answer.

Full notes for topic 4.6 →