Question

Hardness in groundwater is due to the presence of metal ions, primarily Mg2+ and Ca2+ . Hardness is generally reported as ppm CaCO3 . 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 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. red blue Ca(In)2+ + EDTA ⟶ Ca(EDTA)2+ + In A 50.00 mL sample of groundwater is titrated with 0.0450 M EDTA . If 13.70 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? Assume that Ca2+ accounts for all of the hardness in the groundwater.

Answer 1

Answer:

[Ca⁺²] = 0.0123 M

1234 ppm of CaCO₃

Explanation:

Ca(In)²⁺ + EDTA ⟶ Ca(EDTA)²⁺ + In

To solve this problem we need to keep in mind the definition of Molarity (moles/L) and parts per million (mg CaCO₃ / L water). First we calculate the moles of EDTA:

• 0.0450 M * 13.70 mL = 0.6165 mmol EDTA

Looking at the reaction we see that one mol of EDTA reacts with one mol of Ca(In)²⁺, so we have as well 0.6165 mmol Ca(In)²⁺ (which for this case is the same as Ca⁺²). We calculate the molarity of Ca⁺²:

• [Ca⁺²] = 0.6165 mmol / 50 mL = 0.0123 M

For calculating the concentration in ppm, we use the moles of Ca⁺² and the molecular weight of CaCO₃ (100.09 g/mol):

• 0.6165 mmol Ca⁺² * $\frac{1mmolCaCO_{3}}{1mmolCa^{+2}}$ * 100.09 mg/mmol = 61.70 mg CaCO₃

That is the mass of CaCO₃ present in 50 mL (or 0.05 L) of water, so the concentration in ppm is:

• 61.70 mg / 0.05 L = 1234 ppm