Question: 2. Prepare an excel sheet for example 2.3 to calculate the ionic strength for C. Also, determine the minimum solubility of calcium if K.p -

 2. Prepare an excel sheet for example 2.3 to calculate the
ionic strength for C. Also, determine the minimum solubility of calcium if
K.p - 7x109 EXAMPLE 2-3 Determine the Activity Coefficients and Solubility of

2. Prepare an excel sheet for example 2.3 to calculate the ionic strength for C. Also, determine the minimum solubility of calcium if K.p - 7x109 EXAMPLE 2-3 Determine the Activity Coefficients and Solubility of Calcium Carbonate Determine the activity coefficients for the mono and divalent ions in the wastewater given in Example 2-2. Using the value of the activity coefficient for a divalent ion, esti- mate the equilibrium concentration of calcium in solution needed to satisfy the solubil- ity product for calcium carbonate (CaCO,) at 25C. The value of the solubility product constant K,, at 25C is 5 x 10 Solution 1. Determine the ionic strength of the wastewater using Eq. (2-11). a. Prepare a computation table to determine the summation term in Eq. (2-10) using the data from Example 2-2. Conc. Cx 105, lon $, mg/ mole/L 72 CZ2 x 103 Ca2 82.2 2.051 8.404 Mg?- 179 0.736 2.944 No 46.4 2017 ! 2017 K 15.5 0.396 1 0.397 HCO; 220 3.607 1 3.607 SO? 98.3 1.024 4.096 CI 78.0 2.200 1 2.200 NO; 25.6 0.413 0.413 23.876 4 4 4 Sum b. Determine the ionic strength for the concentration C. 1= $c2 = (23876 x 10) = 11.938 x 10*% - 2. Determine the activity coefficients for Ca+ and Co-. Because both species have a valence (charge) of 2, the activity of each will be the same. a. For monovalent ions 0.5(2)VI 0.5(1) V11.938 x 10-3 log y = 0.0492 + Vi 1 + V11.938 X 100 y = 0.893 b. For divalent ions 0.5(Z) VI 0.5(2) V11.938 x 10-3 logy itvi -0.1968 1 + V11.938 x 10- y = 0.636 3. Determine the minimum solubility of calcium using Eq. (2-15). a. Because the molar concentrations of calcium and carbonate ions are the same. Eq (2-15) can be written as follows: 7C] = kg b. Solve for the concentration C. = Ky 5 X 100 C= y? 11.1 X 10 role/L (0.636) c. Convert the molar concentration of calcium carbonate to mg/L. Ca = 11.1 X 10-3 mole/L x 40,000 mg/mole = 4.45 mg/L

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