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.

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 * [tex]\frac{1mmolCaCO_{3}}{1mmolCa^{+2}}[/tex] * 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

nizhalspark nizhalspark · Answer 2 · 2021-08-09T12:03:25+02:00

Answer:

The hardness of the groundwater in molarity and in parts per million of [tex]CaCO_3[/tex] by mass is 1440 ppm.

Explanation:

The given chemical equation is

[tex]Ca(In)^2^+ + EDTA\ gives\ Ca(EDTA)^2^+ + In[/tex]

Step 1:

We use the volume of EDTA consumed in the titration to calculate themoles of ions:

[tex]$0.012 \mathrm{~L}\times 0.0600 \mathrm{M} \times \frac{1 \mathrm{molCa}^{+2}}{1 \mathrm{molE} D T A}=7.20 \times 10^{-4} \mathrm{~mol} \mathrm{Ca}^{+2}$[/tex]

Step 2:

To calculate in ppm, we use the moles of and convert to mg of :

[tex]7.20 \times 10^{-4} \mathrm{~mol} \mathrm{Ca}^{+2}[/tex]

[tex]=7.20 \times 10^{-4} \mathrm{~mol} \mathrm\ {CaCO}_{3} 7.20 \times 10^{-4} \mathrm{~mol} \mathrm\ {CaCO}_{3} \times$100 \mathrm{~g} / \mathrm{mol} \times \frac{1000 \mathrm{mg}}{1 g}[/tex]

[tex]=72 \mathrm{mg} \mathrm\ {CaCO}_{3}$[/tex]

Step 3:

Finally, the concentration in ppm is:

[tex]72 mg \ CaCO_3 / 0.050L = 1440 ppm[/tex]

To learn more, refer:

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