VICAIRE - Module 2

Qualitative Hydrology - Chapter 2

exercise 1 - solution

1) -log(4·10^-3) = -(-3+0.60) = 2.40

2) pH = -log [H⁺] -log [H⁺] = 4.60 log[H⁺] = -4.60 [H⁺] = 10-4,60 = 2.5·10^-5 mol/L.

3) pH+pOH = 14 pOH = 14-pH = 14-10.8 = 3.20 -log[OH^-] = 3.20 log[OH^-] = -3.20 [OH^-] = 10^-3.20 = 6.31·10^-4 mol/L.

exercise 2 - solution

1) The hydrogen ion (H⁺) concentration in solution if defined as:

-log[H⁺] = 4,18 log[H⁺] = - 4,18 [H⁺] = 6,61·10^-5 mol/L.

The value of the ion-product constant (K1) is determined in Table 1.

The concentration of [HCO₃^-] ions is determined as: [HCO₃^-] = 4,45·10^-7·10^-7/ (6,61·10^-5) = 6,67·10^-5 mol/L.

In the same way the value of the ion - product (K2) is determined in Table 1.

The concentration of [CO₃^2-] ions is determined as: [CO₃^2-] = 4,69·10^-11·6,73·10^-5/(6,61·10^-5) = 4,8·10^-11 mol/L