EDTA Titration: Complexometric Calculations, Indicators and Back Titration

By Aniket Bhardwaj · 12 November 2026 · Analytical Chemistry

EDTA titration, also called complexometric titration, is the standard method for finding metal ions such as Ca²⁺, Mg²⁺, Zn²⁺ and Ni²⁺ in a solution. It appears in analytical chemistry questions for IIT-JAM, GATE, CSIR-NET and CUET-PG, usually as a hardness-of-water numerical or as a question on indicators and pH. The arithmetic is easy, because EDTA always reacts in a 1:1 ratio. The marks are lost in the details: which indicator, which pH, and which mole ratio.

Why EDTA gives a clean 1:1 reaction

EDTA (ethylenediaminetetraacetic acid, H₄Y) is a hexadentate ligand. Its fully deprotonated form Y⁴⁻ wraps around a metal ion using two nitrogen atoms and four carboxylate oxygens. Because it occupies six positions, one EDTA binds one metal ion, whatever the charge of the ion.

Mⁿ⁺ + Y⁴⁻ → MY⁽ⁿ⁻⁴⁾
At the end point: moles of EDTA = moles of metal ion
M_EDTA × V_EDTA = M_metal × V_sample

In practice the disodium salt Na₂H₂Y·2H₂O is used, because it dissolves well in water. Both the 1:1 ratio and the volume ratio are all you need for a direct titration.

Role of pH and the buffer

Only the Y⁴⁻ form binds strongly, and its fraction rises with pH. Below about pH 8 most EDTA is in protonated forms, so the complex with Ca²⁺ and Mg²⁺ is incomplete. For this reason hardness titrations are carried out in an ammonia–ammonium chloride buffer at about pH 10. The buffer also stops the pH from falling during the titration, because each reaction releases H⁺.

Indicators: how the colour change works

A metal-ion indicator is itself a weak complexing agent that has a different colour when free and when bound to the metal. It must bind the metal less strongly than EDTA does, so that EDTA takes the metal away at the end point.

Indicator	Typical use	pH	Colour change
Eriochrome Black T (EBT)	Total hardness (Ca²⁺ + Mg²⁺)	about 10	Wine red → blue
Murexide	Ca²⁺ alone	about 12	Pink/red → purple

At pH 12, magnesium precipitates as Mg(OH)₂ and does not react with EDTA, so only calcium is titrated. Subtracting this volume from the total-hardness volume gives magnesium by difference.

Types of EDTA titration

Direct titration: the metal solution is titrated straight with EDTA (hardness of water).
Back titration: a known excess of EDTA is added, and the leftover EDTA is titrated with a standard Zn²⁺ or Mg²⁺ solution. Used when the metal reacts slowly or no good indicator exists.
Masking: a reagent that ties up an interfering ion so that it does not react with EDTA (for example, triethanolamine is commonly used to mask Fe³⁺ and Al³⁺).

Worked examples

Example 1: standardising EDTA. 0.4000 g of pure CaCO₃ (molar mass 100.09 g mol⁻¹) is dissolved in acid and made up to 250.0 mL. A 25.00 mL portion needs 20.15 mL of EDTA solution. Find the molarity of EDTA.

Moles of CaCO₃ in 250 mL = 0.4000 ÷ 100.09 = 3.9964 × 10⁻³ mol.
Molarity of Ca²⁺ = 3.9964 × 10⁻³ ÷ 0.2500 = 0.015986 M.
Moles Ca²⁺ in 25.00 mL = 0.015986 × 0.02500 = 3.9964 × 10⁻⁴ mol.
M_EDTA = 3.9964 × 10⁻⁴ ÷ 0.02015 L = 0.01983 M.

Example 2: total hardness. A 50.00 mL water sample at pH 10 with EBT needs 18.40 mL of 0.0100 M EDTA. Express the hardness as mg L⁻¹ of CaCO₃.

Moles EDTA = 0.0100 × 0.01840 = 1.840 × 10⁻⁴ mol = moles of (Ca²⁺ + Mg²⁺).
Concentration = 1.840 × 10⁻⁴ ÷ 0.05000 L = 3.68 × 10⁻³ mol L⁻¹.
Hardness = 3.68 × 10⁻³ × 100.09 g mol⁻¹ × 1000 mg g⁻¹ = 368.3 mg L⁻¹ as CaCO₃.

(This is ppm as CaCO₃, assuming the sample density is about 1 g mL⁻¹.)

Example 3: calcium and magnesium separately. Another 50.00 mL portion of the same water, at pH 12 with murexide, needs 11.20 mL of the 0.0100 M EDTA. Find Ca²⁺ and Mg²⁺ in mg L⁻¹.

Ca: 0.0100 × 11.20 ÷ 50.00 = 2.240 × 10⁻³ mol L⁻¹; × 40.078 × 1000 = 89.8 mg L⁻¹ Ca²⁺ (224.2 mg L⁻¹ as CaCO₃).
Mg by difference: 18.40 − 11.20 = 7.20 mL; 0.0100 × 7.20 ÷ 50.00 = 1.440 × 10⁻³ mol L⁻¹; × 24.305 × 1000 = 35.0 mg L⁻¹ Mg²⁺ (144.1 mg L⁻¹ as CaCO₃).

Check: 224.2 + 144.1 = 368.3 mg L⁻¹, equal to the total hardness in Example 2.

Example 4: back titration. To 25.00 mL of a Ni²⁺ solution, 20.00 mL of 0.0500 M EDTA (excess) is added. The unreacted EDTA needs 12.50 mL of 0.0200 M Zn²⁺. Find the concentration of Ni²⁺.

EDTA added = 20.00 × 0.0500 = 1.000 mmol.
EDTA left over = Zn²⁺ used = 12.50 × 0.0200 = 0.250 mmol.
EDTA used by Ni²⁺ = 1.000 − 0.250 = 0.750 mmol = Ni²⁺ present.
Concentration = 0.750 mmol ÷ 25.00 mL = 0.0300 M (mass of Ni = 0.750 × 58.693 = 44.0 mg in the portion, i.e. 1.76 g L⁻¹).

Common mistakes

Using a 2:1 or 1:2 ratio from the charge of the ion. EDTA is always 1:1.
Mixing mL with L, or mmol with mol, inside one line.
Reporting hardness as the mass of Ca²⁺ when the question asks for “as CaCO₃”, or the reverse.
Using EBT without a buffer, or at the wrong pH, so the colour change is faint.
Forgetting that in a back titration the Zn²⁺ volume measures the excess EDTA, not the metal.
Saying murexide measures total hardness. At pH 12 it measures calcium only.

Exam relevance

Exam	What is usually tested
IIT-JAM	1:1 stoichiometry, hardness in ppm, simple indicator questions
GATE Chemistry	Analytical methods: pH dependence, masking, back titration
CSIR-NET	Conditional formation constant ideas, choice of indicator, selectivity by pH
CUET-PG	Direct hardness numericals

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