Quiz No. 1

Pollution Monitoring Techniques -I

### SOLUTION

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 1:105 dilution,

Probability of a negative result is,

Probability of a positive result is,

With 1:106 dilution,

Probability of a negative result is,

Probability of a positive result is,

With 1:107 dilution,

Probability of a negative result is,

Probability of a positive result is,

With 1:108 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 NO2 measurement are prepared by dissolving 2.03 g NaNO2 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 NO2 (at 298oK, 1 atm.), given that 0.72 moles of NaNO2 produces same color as 1 mole of NO2.

Solution:

Molecular Weight of NaNO2 = 69

69 g of NaNO2 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 NO2 at 298oK and 1 atmosphere pressure =

Therefore,        1 Liter of NO2 =

Therefore,        10 mL of NO2 = moles of NO2

Also,                1 mole of NO2 = 0.72 moles of

Therefore,        moles of NO2 = 2.93 x 10-7 moles of

Thus,    1 mL of the secondary standard solution is equivalent to 10 mL of NO2.

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-, SO4-2 and HCO3- 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) =             599-200 =                   399 mg/L

pH of the synthetic sample from electroneutrality consideration (show your computations for electroneutrality).

Solution:

Concentration of  Added     =                               2 milli-moles

Concentration of  Formed                         =                                  2 milli moles

Concentration of  Formed          =                                              2 milli-moles

Concentration of Added                      =                1 milli-moles

Concentration of Mg2+ Formed                             =                                  1 milli-moles

Concentration of  Formed                        =                                  1 milli-moles

Concentration of  Added                        =                 1 milli-moles

Concentration of Ca2+ Formed                          =                                  1 milli-moles

Concentration of Cl-  Formed                            =                                  2 milli-moles

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 CaCO3

Total hardness (TH), Carbonate hardness (CH), Non-carbonate hardness (BH), Calcium hardness (CaH), and Magnesium hardness (MgH),

Solution:

Total Hardness (TH)                             =          200 mg/L as CaCO3

Carbonate Hardness (CH)                    =          200 mg/L as CaCO3

Non-carbonate Hardness (BH) =          0

Calcium Hardness (CaH)                      =          100 mg/L as CaCO3

Magnesium Hardness (MgH)                =          100 mg/L as CaCO3

(Report alkalinity and hardness values as mg/L CaCO3)

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

Beer-Lambert 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 end-product 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 O2 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.