Question

A water sample was tested for water hardness. A 50.00 mL sample, prepared and buffered

to pH 10, required 23.55 mL of 0.0250 M EDTA. A second 50.00 mL aliquot, prepared with

NaOH, was titrated with 14.53 mL of the same EDTA solution.

a. Calculate the total hardness of the water sample, express in ppm CaCO3

b. Calculate the concentration of CaCO3 in the sample in ppm

c. Calculate the concentration of MgCO3 in the sample in ppm

Answer 1

Calculate ppm $CaCO_3$ in standard solution:

$\dfrac{0.4985g}{500ml}\times\dfrac{1000mg}{1g}\times\dfrac{1000ml}{1L}=\dfrac{997.0mg}{L}CaCO_3$

Now Calculate concentration EDTA in mg $CaCO_3$ per mL EDTA:

$\dfrac{997mgCaCO_3}{L}\times0.02500L=24.925mgCaCO_3$

$\dfrac{24.925mgCaCO_3}{23.62mlEDTA}=\dfrac{1.055mgCaCO_3}{1mlEDTA}$

Now calculate the mg $CaCO_3$ in the water sample:

$\dfrac{1.055mgCaCO_3}{1.0mlEDTA}\times30.13mlEDTA=31.79mgCaCO_3$

Now calculate ppm $CaCO_3$ in the water sample:

$\dfrac{31.79mgCaCO_3}{0.100Lwater}=317.9$ ppm $CaCO_3$

We have to use experiment data as it is

There is no proof of those experimental values.