Question

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

Answer 1

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.

Step-by-step explanation:

`Ca(In)^(2+)(red)+EDTA\rightarrow Ca(EDTA)^(2+)+In(blue)`

In = Indicator

Concentration of calcium ion in groud water =
`C_1=?`

Volume of ground water =
`V_1=50.00 mL`

Concentration of EDTA solution =
`C_2=0.0800 M`

Volume of EDTA solution =
`V_2=10.90 mL`

`C_1V_1=C_2V_2`

`C_1=(C_2V_2)/(V_1)=(0.0800 M* 10.90 mL)/(50.00 mL)=0.01744 M`

`CaCO_3(sq)\rightarrow Ca^(2+)(aq)+CO_3^(2-)(aq)`

`[CaCO_3]=[Ca^(2+)]`

So.
`[CaCO_3]=0.01744 M`

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 =

`0.01744 mol* 100=1.744 g`

1 g = 1000 g

1.744 g = 1.744 × 1000 mg = 1,744 mg

`ppm=\frac{\text{Mass of solute) mg)}}{\text{Volume of solution}{L}}`

The hardness of the groundwater in parts per million :

`(1,744 mg)/(1 L)= 1,744`

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