Question

A 50.00 mL sample of groundwater is titrated with 0.0300 M EDTA . Of 12.40 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

Answer 1

Answer:

7.44x10⁻³ mol/L and 744 ppm

Explanation:

Let's assume that the hardness of the water is totally from Ca⁺² ions only(the hardness is the measure of Ca⁺² and Mg⁺² ions). The titration with EDTA will form a complex. The EDTA is always in 1:1 proportion, so the number of moles of it will be the number of moles of Ca⁺², which will be the number of moles of CaCO₃.

n = 0.0124 L * 0.0300 mol/L

n = 3.72x10⁻⁴ mol

The molarity is the number of moles divided by the volume (0.05 L)

M = 3.72x10⁻⁴/0.05

M = 7.44x10⁻³ mol/L

1 part per million = 1 mg/L. The molar mass of the CaCO₃ is 100 g/mol, so the mass of it is:

m = 3.72x10⁻⁴ mol * 100 g/mol

m = 0.0372 g = 37.2 mg

Then, the ppm:

37.2/0.05 = 744 ppm