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

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.