Question

I have sample = 50 ml and EDTA TITRATION is 2,3 ml and 2,5 ml (2 titrations), we should take its average the concentration of EDTA is 0,0200 N.

I wanna have the result in mg/L of CaCO_3, please could u tell how we could convert?

Answer 1

V1, Volume of CaCO3 sample solution = 50 mL

V2, Average Volume of EDTA solution used = 2.4 mL

N2, Normality of EDTA solution used = 0.0200 N

N1, Normality of CaCO3 sample can be found using N1V1 = N2V2 equation and rearranging it and substituting the above values into the equation as shown below:

N1 = N2V2/V1

   = 2.4 x 0.02 / 50 = 0.00096 N

Normality​ of CaCO3 solution used = 0.00096 N

Volume of CaCO3 solution in L = 0.050 L

Since

Normality = Number of equivalent wts/ Volume in L

Number of equivalent wts​ = Normality x Volume in L

So Number of CaCO3 equivalent wts = 0.00096 x 0.05 = 0.000048​ =

Since

Equivalent wt = molar mass/no. of Hydrogens in an acid

CaCO3 = Mol. Wt = 100 g/mol

Equivalent wt of CaCO3 = molar mass/no. of Hydrogens in carbonic acid

Equivalent wt of CaCO3 =​ 100 /2 = 50

Since

Number of equivalent wts = Mass used/ equivalent wt and rearranging it one can get mass used as below:

Mass used = Number of equivalent wts / equivalent wt

Mass of CaCO3 used = 0.000048 / 50 = 0.00000096 g = 9.6 x 10^-7 g = 9.6 x 10^​-​10​ mg/L

Therefore

Mass of CaCO3 used in making 50 mL of 0.00096 N solution = 9.6 x 10^​-​10​ mg/L