Titration Calculator — Find Unknown Concentration

Titration Technique and Calculations Guide

What do I need to know about The Four-Step Titration Calculation?

Always follow these steps: Step 1: moles of titrant = concentration × volume (in litres). Step 2: moles of analyte = moles of titrant × (ratio from equation). Step 3: concentration of analyte = moles ÷ volume of analyte (litres). Step 4: check units and significant figures. Example: 23.45 mL of 0.100 mol/L NaOH neutralises 25.00 mL HCl. Moles NaOH = 0.100 × 0.02345 = 0.002345 mol. Ratio 1:1, so moles HCl = 0.002345 mol. Concentration HCl = 0.002345 ÷ 0.025 = 0.0938 mol/L.

What should I know about Titration Technique for Accurate Results?

Rinse burette with titrant solution (not water) before filling — water dilutes the solution and changes concentration. Rinse pipette with analyte before use. Fill burette to a known level (not necessarily 0.00 mL) — read to 2 decimal places. Add indicator appropriate to the titration (phenolphthalein for weak acid/strong base; methyl orange for strong acid/weak base). Add titrant dropwise near the endpoint. Record the rough titre, then repeat two or more concordant titres (within 0.10 mL of each other) rather than relying on a single reading, since a single titre could be affected by an undetected error. Use at least two, ideally three, concordant titres and calculate the mean from those alone — discarding any outlier that doesn't agree with the others rather than including it in the average.

What's the key thing to understand about Concordant Titres and Significant Figures?

A concordant titre means two or more titration results agree to within 0.10 mL. If rough titre is 23.80 mL and repeats are 23.45 mL and 23.48 mL — use the mean of the two concordant results: (23.45 + 23.48)/2 = 23.465 → record as 23.47 mL. Never include the rough titre in the mean. Significant figures: titre readings to 2 decimal places (0.01 mL precision). Concentration answers to 3 significant figures unless data warrants otherwise. Volume in mL must be converted to litres (÷ 1000) before calculating molarity, since the standard concentration formula (c = n/V) requires volume in litres, not millilitres — forgetting this conversion is one of the most common arithmetic errors in titration calculations, typically producing an answer exactly 1,000 times too large.

What should I know about Common Titration Pairs?

Acid-base: HCl + NaOH (1:1 ratio, any indicator), H₂SO₄ + NaOH (1:2 ratio), HCl + Na₂CO₃ (2:1 ratio). Redox: KMnO₄ (purple) + Fe²⁺ in H₂SO₄ (1:5 ratio, KMnO₄ is self-indicating — turns from purple to colourless then pink at endpoint). Iodine + sodium thiosulfate (1:2 ratio, starch indicator turns blue-black then colourless at endpoint). Complexometric: EDTA with metal ions — used in water hardness determination. Back titration: add excess standard reagent, then titrate the excess — used when the analyte reacts too slowly, or interferes with the indicator, for a direct titration to work well.

What's the key thing to understand about Back Titration?

Back titration is used when a direct titration is impractical — typically because the analyte is insoluble, reacts too slowly, or the endpoint is hard to detect directly. Method: react the analyte with a known excess of a reagent (so all of the analyte is consumed), then titrate the unreacted excess reagent against a second standard solution. Moles of analyte = moles of reagent initially added minus moles of reagent found unreacted. Example: finding the purity of impure calcium carbonate (which reacts slowly and incompletely in a direct titration). Add excess HCl of known concentration to dissolve all the CaCO3, let the reaction go to completion, then titrate the leftover HCl with standard NaOH. Moles HCl reacted with CaCO3 = moles HCl added minus moles HCl left over — from this, the mass and purity of the original CaCO3 sample can be calculated using the CaCO3 + 2HCl stoichiometry (1:2 ratio).

What should I know about Sources of Error in Titration?

Parallax error reading the burette meniscus introduces a small random error in each volume reading — reading at eye level, with the meniscus bottom aligned to the graduation, minimises this. Air bubbles trapped in the burette tip, if dislodged during the titration, cause an apparent volume of titrant to be recorded that was never actually delivered — always check and expel bubbles before starting. Overshooting the endpoint (adding titrant too quickly, or stopping on the first permanent colour change rather than the precise point of change) systematically inflates the recorded titre; good technique adds titrant dropwise near the expected endpoint. Indicator choice itself introduces a tiny systematic error (the indicator's colour change does not occur at exactly the stoichiometric point for every acid-base pair), which is why indicator choice should match the pH range at the equivalence point. Temperature variations between standardising a solution and using it can slightly change its concentration, though this is usually negligible for routine school-level titrations.

What do I need to know about Titration Curve Shapes?

A strong acid titrated with a strong base produces a titration curve with a sharp, near-vertical jump in pH right at the equivalence point, which sits at pH 7. A weak acid titrated with a strong base instead shows a more gradual rise, a distinct buffering 'plateau' region before the equivalence point (where the weak acid and its conjugate base coexist), and an equivalence point that sits above pH 7, since the resulting salt solution is itself slightly basic. Recognising the expected curve shape helps identify whether an unknown acid or base is strong or weak well before any calculation is done, just from the shape of the plotted data.

Worked Example: Finding an Unknown Concentration from Titration Data

A 25.00 mL sample of HCl of unknown concentration is titrated with 0.100 M NaOH. The endpoint is reached after 22.40 mL of NaOH has been added. The neutralization reaction is HCl + NaOH → NaCl + H2O, a 1:1 mole ratio. Moles of NaOH used: 0.100 mol/L × 0.02240 L = 0.002240 mol. Because the mole ratio is 1:1, moles of HCl present must also equal 0.002240 mol. Concentration of the HCl: moles ÷ volume = 0.002240 mol ÷ 0.02500 L = 0.0896 M. This four-step pattern — find moles of titrant, apply the mole ratio from the balanced equation, divide by the analyte's volume — is identical regardless of which acid-base pair is used, only the mole ratio changes for reactions that aren't 1:1.

Reference Table: Common Indicators and Their Color Change Ranges

Choosing the right indicator means picking one whose color-change pH range brackets the titration's equivalence point:

Phenolphthalein is the standard choice for strong acid-strong base titrations because the equivalence point sits at pH 7, comfortably below its 8.2-10.0 switching range but still within the steep part of the titration curve; methyl orange, by contrast, is better suited to titrations with a lower equivalence point, such as titrating a weak base with a strong acid.

Worked Example: A Back Titration Calculation

Back titration is used when a direct titration isn't practical — for example, finding the purity of an impure, insoluble solid. A 2.00 g sample of impure calcium carbonate is reacted with 50.00 mL of 0.500 M HCl (added in excess). After the reaction, the leftover (excess) HCl is titrated with 0.400 M NaOH, requiring 30.00 mL to reach the endpoint. Moles of HCl added initially: 0.500 × 0.05000 = 0.02500 mol. Moles of NaOH used to neutralize the excess HCl: 0.400 × 0.03000 = 0.01200 mol, which equals the moles of excess (unreacted) HCl. Moles of HCl that actually reacted with the CaCO₃: 0.02500 − 0.01200 = 0.01300 mol. Since CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ consumes 2 mol HCl per mol CaCO₃, moles of CaCO₃ = 0.01300 / 2 = 0.00650 mol, or 0.00650 × 100.09 ≈ 0.651 g. The purity of the original sample is 0.651 / 2.00 × 100 ≈ 32.5%.

Checking a Dilution Before Titration

Titration problems sometimes start with a diluted stock solution rather than a solution used directly, and getting that dilution step right is essential before the titration math can be trusted. The dilution formula is M₁V₁ = M₂V₂. If 10.0 mL of a 2.00 M stock solution is diluted to a final volume of 100.0 mL, the new concentration is M₂ = M₁V₁/V₂ = (2.00 × 10.0) / 100.0 = 20.0 / 100.0 = 0.200 M. That diluted 0.200 M concentration, not the original 2.00 M stock value, is what should be carried into any subsequent titration calculation.