Water hardness by EDTA titration

UKChO Round 2 · IChO · unreviewed

Direct EDTA titration of the Ca²⁺ and Mg²⁺ in a water sample; express as CaCO₃.

Worked example

A 100.0 cm³ sample of tap water, buffered at pH 10 with Eriochrome Black T indicator, needed 27.18 cm³ of 0.0100 mol dm⁻³ EDTA to titrate its Ca²⁺ and Mg²⁺. Express the total hardness as grams of CaCO₃ per dm³ (M = 100.09 g mol⁻¹).

1. Concentration from moles and volume: solve for n(EDTA) = n(Ca²⁺ + Mg²⁺)
$n = cV$
= 2.718 × 10⁻⁴ mol
2. Concentration from moles and volume: solve for c(M²⁺)
$c = \dfrac{n}{V}$
= 0.002718 mol dm⁻³
3. Mass concentration: solve for Hardness as CaCO₃
$\rho = cM$
= 0.2720 g dm⁻³

Answer: Hardness (as CaCO₃) = 0.272 g dm⁻³

Method

EDTA forms 1 : 1 complexes with every M²⁺, whatever its charge-balancing ions. n(M²⁺) = n(EDTA) = cV; divide by the sample volume, then express the concentration as if it were all CaCO₃ by multiplying by 100.09 g mol⁻¹. (×1000 gives mg dm⁻³, often quoted as ppm.)

Formulas used

Concentration from moles and volume
$c = \dfrac{n}{V}$
Mass concentration
$\rho = cM$

Practise this with fresh numbers

Water hardness by EDTA titration

Worked example

Method

Formulas used

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