Titration Calculations: From a Buret Volume to the Amount of Analyte

The three numbers every titration calculation needs

In a titration, a solution of known concentration (the titrant) is added from a buret to the substance being measured (the analyte) until the reaction between them is complete. Every calculation uses three pieces of information: the titrant's concentration, the titrant volume delivered, and the mole ratio from the balanced equation.

The route is always the same. Convert the titrant volume to liters and multiply by its concentration to get moles of titrant, n = c × V. Use the mole ratio to convert to moles of analyte. Then convert to whatever the question asks for: a concentration, a mass, or a percentage of a sample.

The equivalence point is the point where the titrant added exactly matches the analyte in the stoichiometric ratio. The endpoint is what you actually observe, such as an indicator changing color. A good method makes the two nearly coincide, and the small difference between them is called the titration error.

Standardizing sodium hydroxide with potassium hydrogen phthalate

You cannot make an accurate sodium hydroxide solution by weighing the solid. NaOH pellets absorb water and carbon dioxide from the air, so their mass is not all NaOH. Instead, a roughly prepared solution is standardized: titrated against a primary standard, a substance pure and stable enough to be weighed directly.

Potassium hydrogen phthalate, KHC₈H₄O₄ (often called KHP), is a common choice. It is a stable, non-hygroscopic solid with one acidic hydrogen, and NIST certifies a grade of it as an acidimetric primary standard (SRM 84L). Its large molar mass also means that a typical sample weighs about half a gram, so the relative weighing error is small.

Worked example 1. A 0.5106 g sample of KHP needs 24.68 mL of NaOH solution to reach a phenolphthalein endpoint. The reaction is 1:1: KHC₈H₄O₄(aq) + NaOH(aq) → KNaC₈H₄O₄(aq) + H₂O(l). Using IUPAC standard atomic weights (K 39.098, C 12.011, H 1.008, O 15.999), M(KHP) = 204.22 g/mol. n(KHP) = 0.5106 g ÷ 204.22 g/mol = 2.500 × 10⁻³ mol, which equals n(NaOH). c(NaOH) = 2.500 × 10⁻³ mol ÷ 0.02468 L = 0.1013 mol/L, reported to four significant figures because the mass and the volume each have four.

Diprotic acids change the mole ratio

Sulfuric acid gives up two protons per molecule to hydroxide: H₂SO₄(aq) + 2 NaOH(aq) → Na₂SO₄(aq) + 2 H₂O(l). Each mole of acid therefore consumes two moles of base, and the mole ratio must come from this equation, not from a habit of treating titrations as 1:1.

Worked example 2. A 25.00 mL sample of sulfuric acid is titrated with the 0.1013 mol/L NaOH from example 1, and the endpoint comes at 18.45 mL. n(NaOH) = 0.1013 mol/L × 0.01845 L = 1.869 × 10⁻³ mol. n(H₂SO₄) = 1.869 × 10⁻³ mol ÷ 2 = 9.345 × 10⁻⁴ mol. c(H₂SO₄) = 9.345 × 10⁻⁴ mol ÷ 0.02500 L = 0.03738 mol/L. Using a 1:1 ratio by mistake would give 0.07476 mol/L, exactly twice the true value.

Weaker polyprotic acids need more care. With phosphoric acid, the protons come off in clearly separate steps, so the ratio depends on which equivalence point your indicator detects. State that endpoint before you choose the mole ratio.

Redox titrations: permanganate and iron(II)

Titrations are not limited to acids and bases. In acidic solution, permanganate oxidizes iron(II): MnO₄⁻(aq) + 5 Fe²⁺(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 5 Fe³⁺(aq) + 4 H₂O(l). Manganese goes from +7 to +2, gaining five electrons, while each iron loses one, so one permanganate ion reacts with five iron(II) ions.

Permanganate acts as its own indicator. The MnO₄⁻ ion is intensely purple and Mn²⁺ is almost colorless, so the first drop in excess turns the flask a persistent pale pink. Potassium permanganate is not a primary standard, and its solutions slowly decompose, so they are standardized before use, commonly against sodium oxalate.

Worked example 3. A 1.2084 g iron ore sample is dissolved and all of its iron is reduced to Fe²⁺. Titration needs 26.34 mL of 0.02150 mol/L KMnO₄. n(MnO₄⁻) = 0.02150 mol/L × 0.02634 L = 5.663 × 10⁻⁴ mol. n(Fe²⁺) = 5 × 5.663 × 10⁻⁴ mol = 2.832 × 10⁻³ mol. Using Fe 55.845 g/mol, the iron mass is 0.1581 g, and the ore is 0.1581 g ÷ 1.2084 g × 100 = 13.09 percent iron by mass.

Back titrations for solids that react slowly

Some samples do not dissolve or react quickly enough for a direct titration. Calcium carbonate in an antacid tablet is an example: it reacts with acid, but it is a solid and the reaction releases carbon dioxide, which makes a sharp endpoint hard to see. In a back titration you add a measured excess of one reagent, let the reaction finish, and then titrate whatever is left over.

Worked example 4. A 0.5000 g portion of a crushed antacid tablet is treated with 50.00 mL of 0.2000 mol/L HCl, more than enough to react with all of its CaCO₃: CaCO₃(s) + 2 HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). After the solution is boiled to drive off CO₂, the leftover acid needs 21.38 mL of 0.1000 mol/L NaOH. HCl added: 0.2000 × 0.05000 = 1.000 × 10⁻² mol. HCl left over, equal to the NaOH used: 0.1000 × 0.02138 = 2.138 × 10⁻³ mol. HCl consumed by the tablet: 1.000 × 10⁻² − 2.138 × 10⁻³ = 7.862 × 10⁻³ mol.

The carbonate reacts 1:2 with acid, so n(CaCO₃) = 7.862 × 10⁻³ ÷ 2 = 3.931 × 10⁻³ mol. With M(CaCO₃) = 100.09 g/mol, the mass is 0.3935 g (0.39345 g before rounding), which is 78.69 percent of the 0.5000 g portion. The subtraction is the step most often skipped. Without it, you would be calculating the acid left in the flask, not the acid the tablet used.

Indicator choice and buret technique

An indicator changes color over a range of roughly its pKa ± 1, so choose one whose range falls on the steep part of the titration curve. For a strong acid with a strong base at about 0.1 mol/L, the equivalence point is at pH 7.00 and the pH jumps across several units within a drop or two, so several common indicators give a sharp endpoint, phenolphthalein among them. For a weak acid with a strong base, the equivalence point is basic: OpenStax calculates pH 8.72 for 0.100 mol/L acetic acid titrated with 0.100 mol/L NaOH. Phenolphthalein, which turns pink above about pH 8.3, suits that case. Methyl orange (Ka = 4.0 × 10⁻⁴ in OpenStax, so pKa ≈ 3.4) would change color far too early.

A 50 mL buret is read to 0.01 mL, and each delivered volume is the difference of two readings. An uncertainty of about ±0.02 mL is 0.08 percent of a 25 mL delivery but 0.4 percent of a 5 mL one. That is why methods are designed so that the titrant volume falls in the middle of the buret's range.

• Rinse the buret with a few milliliters of titrant before filling it, so leftover water does not dilute it.
• Remove the air bubble from the tip before the first reading.
• Read the bottom of the meniscus at eye level, and record both readings, not just the difference.
• Run a rough titration first, then repeat until two careful titrations agree closely.

Titration arithmetic errors and their checks

Most wrong answers come from the mole ratio or the units, not the chemistry. Before calculating, write the balanced equation and circle the coefficients of the titrant and the analyte.

• Using C₁V₁ = C₂V₂ for a titration. That shortcut is for dilution and assumes a 1:1 ratio, which fails for H₂SO₄, Ba(OH)₂ and MnO₄⁻/Fe²⁺.
• Leaving volumes in milliliters while concentrations are in mol/L. Convert to liters, or use millimoles consistently.
• Forgetting the subtraction in a back titration.
• Using the formula mass of the wrong species, such as NaOH instead of KHP in a standardization.
• Reporting more significant figures than the buret and balance support. Four is typical for careful volumetric work.
• Assuming the solution is neutral at every equivalence point. That holds only for strong acid with strong base.