Question

You have a water sample with an alkalinity of 2.3 x 10-3 eq/L and a pH of 8.10. Assume that the inorganic carbon species are the main contributors to alkalinity (true in most natural water systems). Calculate the [H2CO3*], [HCO3 - ] [CO3 2- ] and [OH- ] concentrations

Answer 1

Given that, pH = 8.10

i.e. pH = (pKa - LogC)/2, where C = [H₂CO₃]

i.e. C = 10^{(pKa - 2pH)}

Here, pKa = 6.35

Therefore, C = [H₂CO₃] = 10^-9.85 eq/L (i.e. equivalents/L)

i.e. pOH = 14 - 8.1 = 5.9 (since pH + pOH = 14)

i.e. -Log[OH^-] = 5.9

i.e. [OH^-] = 10^-5.9 eq/L

Also given that the alkalinity of water sample = 2.3*10^-3 eq/L

i.e. 2[CO₃^2-] + [HCO₃^-] = 2.3*10^-3

If you find some how the value of any of the two, carbonate or bicarbonate concentration, you can easily find the other from the above relation.

All the best !!