Question

A 50.00 mL sample of groundwater is titrated with 0.0750 M EDTA . If 10.30 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:

$Molarity_{CaCO_3}=0.01545\ M$

Concentration = 1545 ppm

Explanation:

Hardness of the water is generally due to the presence of $CaCO_3$ in water.

At equivalence point

Moles of $CaCO_3$ = Moles of EDTA

Considering

$Molarity_{CaCO_3}\times Volume_{CaCO_3}=Molarity_{EDTA}\times Volume_{EDTA}$

Given  that:

$Molarity_{EDTA}=0.0750\ M$

$Volume_{EDTA}=10.30\ mL$

$Volume_{Water\ or\ CaCO_3}=50.00\ mL$

So,  

$Molarity_{CaCO_3}\times 50=0.0750\times 10.30$

$Molarity_{CaCO_3}=0.01545\ M$

It means that 0.01545 moles are present in 1 L of the solution.

Molar mass of $CaCO_3$ = 100 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus, moles are:

$0.01545= \frac{Mass}{100\ g/mol}$

$Mass= 1.545\ g$

Also, 1 g = 0.001 mg

So, mass = 1545 mg

Concentration = 1545 mg/L

1 mg/L = 1 ppm

So, Concentration = 1545 ppm