50.0 mL of a water sample whose only anions are chloride and sulphate is passed through a cation-exchange resin in H+ form, which exchanges every Ca^2+ or Mg^2+ for two H+. The effluent and washings need 12.4 mL of 0.0100 M NaOH. The hardness as ppm of CaCO3 (molar mass 100) is

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JEE Advanced levelSingle correct optionDifficulty 3 of 3eq108

$50.0 mL$ of a water sample whose only anions are chloride and sulphate is passed through a cation-exchange resin in $H X^{+}$ form, which exchanges every $Ca X^{2 +}$ or $Mg X^{2 +}$ for two $H X^{+}$ . The effluent and washings need $12.4 mL$ of $0.0100 M$ $NaOH$ . The hardness as ppm of $CaCO X_{3}$ (molar mass 100) is

A124 ppm
B248 ppm
C62 ppm
D12.4 ppm

Answer: A

mol H+ = 0.0124 x 0.0100 = 1.24e-4. Divalent ions = 1.24e-4/2 = 6.2e-5 mol in 50.0 mL, i.e. 1.24e-03 mol/L. Hardness = 1.24e-03 x 100 x 1000 = 124 mg/L = 124 ppm.

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50.0 mL of a water sample whose only anions are chloride and sulphate is passed through a cation-exchange resin in HX+ form, which exchanges every CaX2+ or MgX2+ for two HX+. The effluent and washings need 12.4 mL of 0.0100 M NaOH. The hardness as ppm of CaCOX3​ (molar mass 100) is

Answer: A

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