Quiz No. 1
Pollution Monitoring Techniques I
1. What, in your own words, do you understand when it is stated that MPN of a certain water sample is 25/100 mL.
Solution:
MPN of a certain water sample is 25/100 mL does not mean that the sample necessarily has 25 microorganisms. Rather is means that the probability that the number of microorganisms in the sample is 25 is the highest among probabilities corresponding to different possible numbers of microorganisms in the sample.
MPN test was performed on a wastewater sample, and the results given in the table below obtained.

1:10^{5} Dilution 
1:10^{6} Dilution 
1:10^{7} Dilution 
1:10^{8} Dilution 
1. 
+ 
+ 
+ 
+ 
2. 
+ 
+ 
 
+ 
3. 
+ 
+ 
 
 
4. 
+ 
 
 
 
5. 
+ 
 
 
 
Describe in
detail how you would determine the MPN of the sample based on the above
results.
Solution:
Let the MPN of the wastewater sample be _{}.
With 1:10^{5} dilution,
Probability of a negative result is, _{}
Probability of a positive result is, _{}
With 1:10^{6} dilution,
Probability of a negative result is, _{}
Probability of a positive result is, _{}
With 1:10^{7} dilution,
Probability of a negative result is, _{}
Probability of a positive result is, _{}
With 1:10^{8} dilution,
Probability of a negative result is, _{}
Probability of a positive result is, _{}
Hence, probability of results shown in the table above is,
_{}
Hence, solution of the equation, _{}, gives the value of _{}, i.e., the MPN value.
2. Nitrite standards for gaseous NO_{2} measurement are prepared by dissolving 2.03 g NaNO_{2 }in 1000 mL of distilled water. This primary standard is then diluted 100 times to get the secondary standard. Prove that 1 mL of this secondary standard is equivalent to 10 mL of NO_{2 }(at 298^{o}K, 1 atm.), given that 0.72 moles of NaNO_{2} produces same color as 1 mole of NO_{2}.
Solution:
Molecular Weight of NaNO_{2} = 69
69 g of NaNO_{2} has 46 g of _{}
2.03 g of NaNO2 has _{}
That is, Strength of the primary standard = 1.35 g/L _{}
That is, Strength of the secondary standard (1 to 100 dilution) is 13.5 mg/L as _{}
That is, 1 mL of the secondary standard has 13.5 mg _{} = 2.93 x 10^{7} moles of _{}
Volume of 1 mole of NO_{2} at 298^{o}K and 1 atmosphere pressure =
_{}
Therefore, 1 Liter of NO_{2} = _{}
Therefore, 10 mL of NO_{2} = _{}moles of NO_{2}
Also, 1 mole of NO_{2} = 0.72 moles of _{}
Therefore, _{}moles of NO_{2} = 2.93 x 10^{7} moles of _{}
Thus, 1 mL of the secondary standard solution is equivalent to 10 mL of NO_{2}.
3. A synthetic sample of water is prepared by dissolving 200 mg glucose, 168 mg sodium bicarbonate, 120 mg magnesium sulphate and 111 mg calcium chloride in one liter distilled deionized water. Assuming that the complete dissociation of the salts occur leading to presence of Na^{+}, Mg^{+2}, Ca^{+2}, Cl^{}, SO_{4}^{2} and HCO_{3}^{} species in addition to H^{+} and OH^{} ions, compute,
Total solids (TS), Total dissolved solids (TDS), Volatile dissolved solids (VDS) and Fixed dissolved solids (FDS).
Solution:
Total solids (TS) = 200 + 168 + 120 + 111 = 599 mg/L
Total dissolved solids (TDS) = 599 mg/L
Volatile dissolved solids (VDS) = 200 mg/L
Fixed dissolved solids (FDS) = 599200 = 399 mg/L
pH of the synthetic sample
from electroneutrality consideration (show your computations for electroneutrality).
Solution:
Concentration of _{} Added = _{} 2 millimoles
Concentration of _{} Formed = 2 milli moles
Concentration of _{} Formed = 2 millimoles
Concentration of _{}Added = _{} 1 millimoles
Concentration of Mg^{2+} Formed ^{ }= 1 millimoles
Concentration of _{} Formed = 1 millimoles
Concentration of _{} Added = _{} 1 millimoles
Concentration of Ca^{2+} Formed = 1 millimoles
Concentration of Cl^{} Formed = 2 millimoles
Electroneutrality condition states that the sums of positive and negative charges in the solution must be equal.
Solution:
Sum of positive charges:
_{} = 2.(0.001) + 2(0.001) + 1.(0.002) + _{}
Sum of negative charges:
_{} = 1.(0.002) + 2.(0.001) + 1.(0.002) + _{}
Equating: 0.006 + _{}
Also, _{}
Hence, _{} or, pH = 7
Total alkalinity (TA), Hydroxyl alkalinity (HA), Carbonate
alkalinity (CA), and Bicarbonate alkalinity (BA).
Solution:
Total Alkalinity (TA) = _{}
Hydroxyl Alkalinity (HA) = 0
Carbonate Alkalinity (CA) = 0
Bicarbonate Alkalinity (BA) = 200 mg/L as CaCO_{3}
Total hardness
(TH), Carbonate hardness (CH), Noncarbonate hardness (BH), Calcium hardness
(CaH), and Magnesium hardness (MgH),
Solution:
Total Hardness (TH) = 200 mg/L as CaCO_{3}
Carbonate Hardness (CH) = 200 mg/L as CaCO_{3}
Noncarbonate Hardness (BH) = 0
Calcium Hardness (CaH) = 100 mg/L as CaCO_{3}
Magnesium Hardness (MgH) = 100 mg/L as CaCO_{3}
(Report alkalinity and hardness values as mg/L CaCO_{3})
4. You are given an unknown sample for the determination of nitrite. You do the following:
Take 10 mL of the unknown sample. Add the required reagents for the colorimetric determination of nitrite by the NEDA method. Make up the volume to 25 mL and measure absorbance. The resulting absorbance is 0.402.
Take another 10 mL of the unknown sample. Add 10 mL of standard nitrite solution (10 mg/L concentration) to the unknown sample, add required reagents as before and make up to 25 mL, and measure absorbance. The resulting absorbance is 0.503.
Based on these readings, find the nitrite concentration in the unknown sample.
Solution:
Let the concentration of the unknown sample be ‘C’
When 10 mL of the sample is diluted to 25 mL, concentration in the diluted sample is _{}
BeerLambert law States: Absorbance (A) = _{}
Where, _{} = molar absorptivity
b = path length
c = concentration
Hence, for the first measurement, 0.402 = _{} (A)
Similarly, for the second measurement, 0.503 = _{} (B)
From equation (A) and (B)
_{}
or, C = 39.7 mg/L
5. Propionitrile has a general formula of _{}. It is completely oxidized by dichromate during the COD test.
Write a
balanced equation for reaction of _{} with
_{}. The nitrogen
endproduct of the reaction is _{}.
Solution:
Writing balanced equation:
Oxidation: _{}
Reduction: _{}
Final Equation:
_{}
What is the COD
of a 50 mg/L solution of propionitrile.
Solution:
While getting oxidized, 1 molecule of _{}releases 14 electrons.
Reduction of oxygen takes place as follows, _{}, accepting 4 electrons per molecule of oxygen reduced,
Thus reaction stoichiometry for reaction between _{} and O_{2} is as follows:
_{}
Molecular weight of _{}is 55
That is, 55 g of _{} consumes _{}
Therefore, 50 mg of _{} consumes _{}
Hence COD of a 50 mg/L _{}solution is 102 mg/L.