Question

Determine the %(p/p) of CaCO3 of a 25 mg chalk sample dissolved in 25 mL water, 12.5 mL HCl 0.1 M, titrated with 8.6 mL NaOH 0.1 M.

Answer 1

The equation of the above reaction is as follows:

NaOH + HCl NaCl + H2O

No. of moles of NaOH= 0.1 x 8.6/1000 = 0.00086 moles

Therefore, 0.00086 moles of NaOH reacts with 0.00086 moles of HCl.

Actual No. of moles of HCl = CV = 0.1M x 37.5/1000 L= 0.00375 (Total volume 12.5+25mL)

Therefore, no. of moles of HCl that reacted with CaCO3 in the equation given below = 0.00375-0.00086 =0.00289 moles

CaCO3 (s) + 2HCl(aq) CaCl2 (aq) + H2O + CO2 (g)

1mol of CaCO3 reacts with 2 mols of HCl , therefore, 0.00289/2 moles of CaCO3 reacts with 0.00289 moles of HCl= 0.00144 moles, Hence 0.00144 moles of CaCO3 was presnt in chalk .

molar mass of CaCO3 = 100 g/mol

therefore, mass of CaCO3 = nM = 0.00144 mole x 100g/mol = 0.144 g

Are you asking in percentage of parts per million units? if yes, Then we divide 0.144 with 10-6 .

Answer : 0.144 x 10-6 ppm

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