Question

The alkalinity of natural waters is usually controlled by OH- , CO3 2- , and HCO3 - , which may be present alone or in combination. Titrating a 100.0 mL sample to a pH of 8.3 requires 18.70 mL of a 0.0281 M solution of HCl. A second 100.0 mL aliquot requires 48.20 mL of the same titrant to reach a pH of 4.5. Calculate the concentrations of CO3 2- and HCO3 - in ppm.

Answer 1

As mentioned there is no hydroxide ion concentration in the solution

the 100mL sample required 18.70mL of HCl solution to reach the pH =8.3

Hence, The first equivalence point is due to half neutralization of carbonate

CO3^-2 + H+ ---> HCO3^-

concentration of bicarbonate = concentration of acid used X volume of acid used / total volume

concentration of bicarbonate = 0.0281 x 18.70 / 100 = 0.0053 Moles / Litre

Second equivalence point is due to complete neutralization of HCO3- and bicarbonate, the pH = 4.5

CO3^-2 + H+ ---> HCO3^-

HCO3- + H+ --> H2O + CO2

so total concentration of carbonate and bicarbonate ions will be

volume of natural water X concentration = volume of HCl used X concentration of HCl

Concentration of natural water = 48.20 x 0.0281 / 100 = 0.0135

[CO₃^-2] = 0.0053 moles / Litres = 0.0053 X molecular weight (g / L) = 0.318 g/L = 318 mg / L = 318ppm

2X [CO3^-2] + [HCO3-] = 0.0135

[HCO3-] = 0.0029 moles / Litres = 0.0029 X 61 g /L = 0.177 g /L = 177 mg / L = 177ppm