A 50.00 mL sample of groundwater is titrated with 0.0800 M EDTA . If 10.90 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 CaCO 3 by mass

0.01744 M is the hardness of the groundwater.

1,744 ppm is the hardness of the groundwater in parts per million of calcium carbonate by mass.

Explanation:

In = Indicator

Concentration of calcium ion in groud water =

Volume of ground water =

Concentration of EDTA solution =

Volume of EDTA solution =

So.

0.01744 M is the hardness of the groundwater.

0.01744 Moles of calcium carbonate are present in 1 Liter of solution.

Mass of 0.01744 moles of calcium carbonate =

1 g = 1000 g

1.744 g = 1.744 × 1000 mg = 1,744 mg

The hardness of the groundwater in parts per million :

1,744 ppm is the hardness of the groundwater in parts per million of calcium carbonate by mass.

i think that the answer is c