Concentration from Titration Data

Introduction

Titration uses a measured amount of known solution to infer an unknown concentration. The route is titrant cV, then balanced mole ratio, then division by the original analyte aliquot volume. The final flask volume after addition is not the denominator for the original analyte concentration.

Core explanation

Suppose 25.00 mL HCl requires 20.00 mL of 0.1000 M NaOH at the stated equivalence point. Delivered NaOH moles are 0.02000 L × 0.1000 mol L⁻¹ = 0.002000 mol. For HCl + NaOH → NaCl + H₂O, the 1:1 ratio gives 0.002000 mol HCl originally in the aliquot. Its original molarity is 0.002000/0.02500 = 0.08000 M. The final mixed volume might be about 45 mL, but using it would calculate a different final-mixture concentration, not the original HCl concentration.

Balanced coefficients are essential. For complete neutralization H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, the acid mole amount is half the NaOH amount. If 0.00400 mol NaOH was delivered, it corresponds to 0.00200 mol H₂SO₄ under that specified endpoint. Treating the ratio as 1:1 would double the inferred acid concentration.

The burette volume delivered is final reading minus initial reading. If the reading moves from 1.30 mL to 21.30 mL, delivered volume is 20.00 mL, not 21.30 mL. Convert delivered mL to L for cV. A titration endpoint is an observed signal; equivalence is the chemical amount condition. An indicator should be selected so its observed endpoint closely represents the intended equivalence for the required accuracy.

Dilution before titration adds another layer. If a stock analyte was diluted and a portion of the diluted solution titrated, this calculation gives concentration of the diluted liquid first. The original stock concentration follows from the dilution factor, assuming no loss or reaction during dilution. Never insert the original stock volume directly into the aliquot denominator if the measured aliquot came from the diluted flask.

Interfering substances can consume titrant. The inferred amount belongs to the named analyte only when the selected reaction is sufficiently specific or the interference is accounted for. Reporting a numerical answer should state the assumed reaction and endpoint.

Step-by-step reasoning

1. Write the balanced titration reaction and specified endpoint. 2. Subtract burette readings for titrant delivered volume. 3. Calculate titrant moles with cV. 4. Apply analyte-to-titrant coefficient ratio. 5. Divide analyte moles by original analyte aliquot volume.

Visual explanation

Draw a burette with initial and final readings over a flask labeled “25.00 mL unknown.” An arrow shows delivered titrant moles, a balanced-equation arrow gives analyte moles, and the last divides by 0.02500 L.

Real-world analogy

If each unknown package requires two measured tokens to match it, counting used tokens reveals packages only after dividing by two. A balanced titration equation supplies that matching rule.

Real-world example

Acetic acid in a diluted vinegar sample can be determined using NaOH and a specified endpoint. The measured aliquot concentration must be multiplied by the correct dilution factor before comparing with the original product.

Why?

Why use the original aliquot volume? The requested concentration describes analyte moles per litre of the solution before titrant was added. Added titrant changes mixture volume after the sampled analyte amount was fixed.

Common misconception

“Any acid and base titrate 1:1.” Polyprotic acids and multihydroxide bases can have different formula ratios. Use the specified balanced equation.

Worked example

An unknown H₂SO₄ sample of 20.0 mL needs 24.0 mL of 0.150 M NaOH for complete neutralization. NaOH amount is 0.0240 × 0.150 = 0.00360 mol. The acid amount is 0.00360/2 = 0.00180 mol. Acid concentration in the original 0.0200 L aliquot is 0.00180/0.0200 = 0.0900 M.

Quick check

1. What is the analyte-volume denominator when reporting its original molarity? Answer: The volume of the original analyte aliquot, before titrant addition, is the denominator.

Exam focus

Record burette difference, mL-to-L conversion and coefficient ratio separately. Check whether the analyte aliquot came from a diluted stock.

Advanced insight

Different titration endpoints can correspond to different stoichiometric stages for polyprotic systems. An observed endpoint should be interpreted through the chemistry and detection method, not merely through a color change.

Summary

Unknown concentration follows titrant cV, balanced mole ratio and original aliquot volume. Endpoint, dilution history and interferences determine whether that ratio accurately represents the named analyte.

Practice questions

1. A 10.0 mL HCl aliquot needs 15.0 mL of 0.100 M NaOH. Find HCl molarity. Answer: NaOH moles are 0.00150; 1:1 HCl amount is 0.00150 mol; c = 0.00150/0.0100 = 0.150 M. 2. Why subtract initial burette reading from final reading? Answer: The scale shows positions, and only their difference is titrant volume delivered. 3. If acid requires two base moles per acid mole, what acid moles correspond to 0.0060 mol base? Answer: The balanced ratio gives 0.0060/2 = 0.0030 mol acid at that equivalence.