Question

Hardness in groundwater is due to the presence of metal ions, primarily Mg2 and Ca2 . Hardness is generally reported as ppm CaCO3 or mmol/L Ca2 . To measure water hardness, a sample of groundwater is titrated with EDTA, a chelating agent, in the presence of the indicator eriochrome black T, symbolized here as In. Eriochrome black T, a weaker chelating agent than EDTA, is red in the presence of Ca2 and turns blue when Ca2 is removed. redblue Ca(In)2+ + EDTA --> Ca(EDTA)^2+ + In A 50.00-mL sample of groundwater is titrated with 0.0600 M EDTA. Assume that Ca2 accounts for all of the hardness in the groundwater. If 12.00 mL of EDTA is required to titrate the 50.00-mL sample, what is the hardness of the groundwater in molarity and in parts per million of CaCO3 by mass?

Answer 1

Answer:

• 0.0144 M

• 1440 ppm

Explanation:

Ca(In)²⁺ + EDTA → Ca(EDTA)²⁺ + In

We use the volume of EDTA consumed in the titration to calculate the moles of Ca⁺² ions:

• 0.012 L * 0.0600 M * $\frac{1molCa^{+2}}{1molEDTA}$ = 7.20x10⁻⁴ mol Ca⁺²

Now we calculate the molarity:

• 7.20x10⁻⁴ mol Ca⁺² / 0.050 L = 0.0144 M

To calculate in ppm, we use the moles of Ca⁺² and convert to mg of CaCO₃:

• 7.20x10⁻⁴ mol Ca⁺² = 7.20x10⁻⁴ mol CaCO₃

• 7.20x10⁻⁴ mol CaCO₃ * 100g/mol * $\frac{1000mg}{1g}$ = 72 mg CaCO₃

Finally, the concentration in ppm is:

• 72 mg CaCO₃ / 0.050L = 1440 ppm