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Zina [86]
3 years ago
14

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
Chemistry
1 answer:
vitfil [10]3 years ago
4 0

Answer:

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:

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=\frac{C_2V_2}{V_1}=\frac{0.0800 M\times 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\times 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 :

\frac{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.

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