Question

A drinking water sample has pH of 7.9, 173 of mg/HCO₃^- L, 3mg/L of CO_2(aq) and negligible CO₃^2- . Calculate the alkalinity of the water in unit of mg/L as CaCO₃

Answer 1

Alkalinity is a water characteristic that quantifies the capacity of water to neutralize acids, namely, accepts hydrogen ions H+. Alkalinity of natural water is mainly due to the presence of two forms of the carbonate ions denoted as HCO3(-) and CO3(2-) that act as a buffer system. Alkalinity prevents sudden changes in the acidity level of water and, hence, is important for fish and other aquatic life. It is measured in mg/L of CaCO3. Naturally occurred alkalinity is in the range from 400 to 500 mg/L.

calculate alkalinity if 1 L of water contains 0.173g of HCO3(-) and negligible of CO3(2-) carbonate ions.

Calculate the molar mass of HCO3(-),CO3(2-) and CaCO3 as the sum of mass of all atoms in the molecule. Atomic weights of corresponding elements are given in the periodic table of the chemical elements (see Resources). Molar mass (HCO3(-)) = M(H) + M(C) + 3 x M(O) = 1+12 + 3 x 16 = 61 g/mole. Molar mass (CO3(2-)) = M(C) + 3 x M(O) = 12+ 3 x 16 = 60 g/mole. Molar mass (CaCO3) = M(Ca) + M(C) + 3 x M(O) = 40 + 12 + 3 x 16 = 100 g/mole.

Divide the molar mass by the ion charge or oxidation number (for CaCO3) to determine equivalent (Eq.) weights. Eq. weight (HCO3(-)) = 61 / 1 (charge) = 61 g/Eq. Eq. weight (CO3(2-)) = 60 / 2 (charge) = 30 g/Eq. Eq. weight (CaCO3) = 100 / 2 (oxidation state) = 50 g/Eq.

Divide masses of HCO3(-) and CO3(2-) by their equivalent (Eq.) weights to calculate a number of equivalents. In our example, Number of Eq. (HCO3(-)) = 0.173g / 61 g/Eq = 0.0028 Eq. Number of Eq. (CO3(2-)) = 0.0g / 30 g/Eq = 0.000 Eq. Equivalents are needed to reflect the following fact. Each ion HCO3(-) reacts with one hydrogen proton H+, but each CO3(-2) ion can accept two protons or two equivalents.

Add up equivalents of HCO3(-) and CO3(2-) to calculate the alkalinity expressed in equivalents of CaCO3. In our example, Number of Eq. (CaCO3) = 0.0028 Eq + 0.000 Eq = 0.0028 Eq/L. Multiply it by 1,000 to get it in milliequivalents: 0.0028 Eq/L x 1,000 = 2.8 mEq/L.

Multiply alkalinity in “Eq/L” by the equivalent weight of CaCO3 to calculate it in g/L. In our example, Alkalinity as CaCO3 = 0.0028 Eq/L x 50 g/Eq = 0.141 g/L = 141 mg/L.