Question

(a) Sketch the titration curve for an alkalinity test in which the water sample had an initial pH of 6.35. In that test, 15 mL of 0.1 N acid was added to a 150-mL sample.

(b) Calculate the alkalinity in meq/L and mg/L as CaCO₃. Answer: 10 meq/L, 500 mg/L as CaCO₃

_{I know the answer is correct but I am having a problem on a few of the steps, please show all work for a "thumbs up"! Thanks!}

Answer 1

• Naturally occurred alkalinity is in the range from 400 to 500 mg/L. As an example, calculate alkalinity if 1 L of water contains 035 g of HCO3(-) and 0.12 g 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) = 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(-)) = 0.35g / 61 g/Eq = 0.0057 Eq.
  Number of Eq. (CO3(2-)) = 0.12g / 30 g/Eq = 0.004 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) = 0.0057 Eq + 0.004 Eq = 0.0097 Eq/L. Multiply it by 1,000 to get it in milliequivalents: 0.0097 Eq/L x 1,000 = 9.7 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 = 0.0097 Eq/L x 50 g/Eq = 0.485 g/L = 485 mg/L.