50.0 mL of a hard water sample is titrated with 0.0105 M EDTA (buffer pH 10, EBT) and requires 11.3 mL. Taking CaCO3=100 and EDTA to metal ion as 1:1, find the total hardness in ppm of CaCO3 (1 ppm = 1 mg L^-1), to the nearest integer.

JEE Main levelNumerical answerDifficulty 2 of 3eq208

$50.0 mL$ of a hard water sample is titrated with $0.0105 M$ EDTA (buffer pH 10, EBT) and requires $11.3 mL$ . Taking $CaCO X_{3} = 100$ and EDTA to metal ion as $1 : 1$ , find the total hardness in $ppm$ of $CaCO X_{3}$ ( $1 ppm = 1 mg L^{- 1}$ ), to the nearest integer.

Numerical answer type, answer in ppm, tolerance 1.5.

Mmol of EDTA

= 11.3 \times 0.0105 = 0.11865

= mmol of

Ca X^{2 +} + Mg X^{2 +}

= mmol

CaCO X_{3}

equivalent. Mass

= 0.11865 \times 100 = 11.865

mg in

0.0500 L

. Hardness

= 11.865/0.0500 = 237.3 mg L^{- 1}