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

To determine the hardness of the groundwater in molarity and parts per million of CaCO3, we can use the titration data. The molar concentration of Ca²+ ions in the groundwater is 0.0750 M. The hardness of the groundwater is 7509 ppm of CaCO3.

Step-by-step explanation:

To determine the hardness of the groundwater in molarity and parts per million of CaCO3, we need to use the titration data. Given that it took 10.30 mL of 0.0750 M EDTA to titrate a 50.00 mL sample of groundwater, we can calculate the molar concentration of Ca2+ ions in the groundwater.

The balanced chemical equation for the reaction is:

Ca²+ + EDTA → [Ca(EDTA)]²—

Since the molar ratio between Ca²+ and EDTA is 1:1, the molar concentration of Ca²+ ions in the groundwater is 0.0750 M.

To convert this to parts per million of CaCO3, we can use the molar mass of CaCO3 (100.09 g/mol). The conversion factor is 1000 mg/g:

0.0750 mol/L × 100.09 g/mol × 1000 mg/g = 7509 mg/L = 7509 ppm

Answer 2

`Molarity_(CaCO_3)=0.01545\ M`

Concentration = 1545 ppm

Step-by-step 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)* Volume_(CaCO_3)=Molarity_(EDTA)* 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)* 50=0.0750* 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 = (Mass\ taken)/(Molar\ mass)`

Thus, moles are:

`0.01545= (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