Question

A water sample has a pH of 8.21 and a total Ca2+ concentration of 143 ppm. a)What are the molar concentrations of CO32− and HCO3−? b)What is the total alkalinity of the water?

Answer 1

a)

HCO3- + H2O <-> H+ + CO3-2

assume

pH = pKa + log(CO3-2/HCO3-)

pKa2 = 10.33

if pH = 8.21, then

8.21 = 10.33 + log(CO3-2/HCO3-)

(CO3-2/HCO3-) = 10^(8.21-10.33) = 0.00758

then

CaCO3 = Ca+2 + CO3-2

CO3-2+ H2O -- >HCO3+ + H+

assume initially,

[Ca+2] = 143 ppm = 143 mg / L = (143*10^-3/40) / L = 0.003575 mol of Ca+2 / L

[CO3-2] = 0.003575 M initially

now..

[CO3-2] + [HCO3-] = 0.003575 M

[CO3-2]/[HCO3-] = 0.00758

0.00758 *[HCO3-] + [HCO3-] = 0.003575 M

[HCO3-] = 0.003575 /0.00758 = 0.4716

now..

alkalinity of water --> 143 ppm or 0.003575 M