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Superminds cefr year 1 unit 1 At school
Superminds cefr year 1 unit 1 At school
Exercise 1: Alkalinity
Q1– Approximate Alkalinity
Calculate the approximate alkalinity (in mg/L as CaCO 3) of a water containing 120.0 mg/L of bicarbonate ion and 15.00 mg/L of carbonate Bicarbonate ion (1-), thus n = 1. While the molecular ion. weight is (1) + (12.01) + (3*16) = 61.01
Thus, the EW CO32- is 60 / 1 = 30 mg/meq Solution:
Carbonate ion (2-), thus n = 2. While the molecular weight is (12) + (3*16) = 60. Thus, the EWCO32- is 60 / 2 = 30 mg/meq CaCO3, n =2 as it needs 2 hydrogen ions to replace Ca2+ to form carbonic acid, H2CO3. Thus, the EW CaCO3 100.09/2 = 50.05 g/eq or mg/meq
Q2– Exact alkalinity Calculate the exact alkalinity of the water in Q1 if the pH is 9.43.
Solution
Step 1: Convert HCO3- to mg/L as CaCO 3 (similar to Q1) mg/L as CaCO 3 120 (50.04/61.016) = 98.41 mg/L as CaCO3 Step 2: Convert CO 32- to mg/L as CaCO 3 (similar to Q1) mg/L as CaCO3 15.0 (50.04/30.004) = 25.02 mg/L as CaCO3 Step 3: Convert pH to H (Given in question) [H+] = 10-9.43 = 3.715 x 10-10 mole/L Step 4: Convert mole/L to mg/L mg/L = (3.715 x 10-10 mole/L)(1.0079 x 103 mg/mole) = 3.745 x 10-7 Step 5: Calculate in mg/L as CaCO 3 (for H) (3.745 x 10-7) (50.04/1.0079) = 1.86 x 10-5
Atomic W H =1.00794
Step 6: Convert pH to [OH -] pOH = 14.00 - 9.43 = 4.57 [OH-] = 10-4.57 = 2.692 x 10-5 mole/L Step 7: Convert from mole/L to mg/L mg/L = (2.692 x 10-5)(17.007 x 103 mg/mole) = 0.4578 Step 8: Calculate in mg/L as CaCO 3 (for OH) 0.4578 (50.04/17.007) = 1.347
Atomic W H =1.00794 Atomic W O = 16
Step 9: Lastly, sum them up for exact alkalinity (all in mg/L as CaCO3) ALK = HCO3- + CO32- + OH- - H+ ALK = 98.41 + 25.02 + 1.347 - (1.86 x 10-5) ALK = 124.78 mg/L as CaCO 3
Exercise 2 Calculation of ThOD
Q1 Calculate the ThOD of glucose having the formula C 6H12O6 Solution:
Molecular weight of C6H12O6 is (6x12) + (12 X1) + (6x 16) = 180 g The equation: C6H12O6 + 6 O2 6 CO2 + 6 H2O 180
192
Thus, ThOD = 192 g O2 / 180 g glucose = 1.07 g O2/g glucose
Q2 From the answer computed in Q1, what is the ThOD for a glucose concentration of 200 mg/L?
Solution:
ThOD = 1.07 g O2 / g glucose For 200 mg/L of glucose, ThOD = 1.07 x 200 mg/L = 214 mg/L
Nitrogenous oxygen demand (NOD) is a quantitative measure of the amount of dissolved oxygen required for the biological oxidation of nitrogenous material, for example, nitrogen in ammonia, and organic nitrogen in waste water.
In short, we only interested in the OD of the N of the organic compound, not the NH3 which is the intermediate product.
The nitrogenous compounds NH 3 and Organic N exert an oxygen demand when they are oxidized by nitrifying bacteria. Typically, one mg/L of NH 3 or organic N requires 4.57 mg/L of oxygen to be completely oxidized.
Fraction of nitrogen mass in C 5H7NO2 = 14/113 Thus, nitrogenous oxygen demand = 4.57 x 14/113 = 0.56 mg / mg of cell tissue
