Quality Characteristics of Water


Question 1
A 100 mg of HNO_3 (strong acid) is added to water, bringing the final volume to 1.0 liter. Consider the atomic weights of H, N, and O, as 1 g/mol, 14 g/mol, and 16 g/mol, respectively. The final pH of this water is (Ignore the dissociation of water.)
A
2.8
B
6.5
C
3.8
D
8.5
GATE CE 2022 SET-2   Environmental Engineering
Question 1 Explanation: 
HNO_3\rightarrow H^++NO_3^{-}
\begin{aligned} \text{1 mole of }HNO_3&=\text{1 mole of }H^+\\ \text{No. of mole of }HNO_3&=\frac{100}{(1+14+16 \times 3)}\\ &=1.587 m\; mol\\ [H^+] \text{ in 1 lit. of water }&=1.587 m\;mol/Lit\\ &=1.587 \times 10^{-3}\;mol/Lit\\ pH&=-\log[H^+]\\ &=\log[1.587 \times 10^{-3}]\\ &=2.8 \end{aligned}
Question 2
In a water sample, the concentrations of Ca^{2+},Mg^{2+}, and HCO_3^{-} are 100 mg/L, 36 mg/L and 122 mg/L, respectively. The atomic masses of various elements are: Ca = 40, Mg = 24, H = 1, C = 12, O = 16.
The total hardness and the temporary hardness in the water sample (in mg/L as CaCO_3) will be
A
400 and 100, respectively.
B
400 and 300, respectively.
C
500 and 100, respectively.
D
800 and 200, respectively.
GATE CE 2022 SET-1   Environmental Engineering
Question 2 Explanation: 
Total Hardness,
\begin{aligned} TH&=[Ca^{+2}] \times \frac{50}{20}+[Mg^{+2}] \times \frac{50}{12}\\ &=100 \times \frac{50}{20}+36 \times \frac{50}{12}\\ &=400 mg/l \text{ as } CaCO_3 \end{aligned}
Total alkalinity,
TA=[HCO_3^-]\times \frac{50}{61}=122 \times \frac{50}{61}=100\; mg/l \text{ as } CaCO_3
Temporary hardness (carbonate hardness)
=min\left\{\begin{matrix} TA= 100\; mg/l \text{ as } CaCO_3\\ TH=400\; mg/l \text{ as } CaCO_3 \end{matrix}\right.=100 \; mg/l \text{ as } CaCO_3


Question 3
A water filtration unit is made of uniform-size sand particles of 0.4 mm diameter with a shape factor of 0.84 and specific gravity of 2.55. The depth of the filter bed is 0.70 m and the porosity is 0.35. The filter bed is to be expanded to a porosity of 0.65 by hydraulic backwash. In the terminal settling velocity of sand particles during backwash is 4.5 cm/s, the required backwash velocity is
A
5.79 \times 10^{-3} \mathrm{~m} / \mathrm{s}
B
6.35 \times 10^{-3} \mathrm{~m} / \mathrm{s}
C
0.69 cm/s
D
0.75 cm/s
GATE CE 2021 SET-2   Environmental Engineering
Question 3 Explanation: 
\begin{aligned} n^{\prime} &=\text { Porosity of expanded bed } \\ n^{\prime} &=\left(\frac{V_{B}}{V_{s}}\right)^{0.22} \\ 0.65 &=\left(\frac{V_{B}}{4.5 \mathrm{~cm} / \mathrm{s}}\right)^{0.22} \\ V_{B} &=6.35 \times 10^{-3} \mathrm{~m} / \mathrm{s} \end{aligned}
Question 4
The internal \left(d_{i}\right) and external \left(d_{o}\right) diameters of a Shelby sampler are 48 mm and 52 mm, respectively. The area ratio \left(A_{r}\right) of the sampler (in %,round off to two decimal palces) is _____________
A
12.45
B
28.56
C
47.12
D
17.36
GATE CE 2021 SET-2   Environmental Engineering
Question 4 Explanation: 
Outside diameter = 52 mm
Inside diameter= 48 mm
\begin{aligned} A_{r}&=\frac{\frac{\pi}{4}\left(D_{2}\right)^{2}-\frac{\pi}{4}\left(D_{1}\right)^{2}}{\frac{\pi}{4}\left(D_{1}\right)^{2}} \times 100 \\ A_{r}&=\frac{(52)^{2}-(48)^{2}}{(48)^{2}} \times 100=17.36 \% \end{aligned}
Question 5
A water sample is analyzed for coliform organisms by the multiple-tube fermentation method. The results of confirmed test are as follows:
\begin{array}{|l|c|c|} \hline \text { Sample size (mL) } & \begin{array}{c} \text { Number of positive } \\ \text { results out of } 5 \text { tubes } \end{array} & \begin{array}{c} \text { Number of negative } \\ \text { results out } 5 \text { tubes } \end{array} \\ \hline 0.01 & 5 & 0 \\ \hline 0.001 & 3 & 2 \\ \hline 0.0001 & 1 & 4 \\ \hline \end{array}
The most probable number (MPN) of coliform organisms for the above results is to be obtained using the following MPN Index.


The MPN of coliform organisms per 100 mL is
A
1100000
B
110000
C
1100
D
110
GATE CE 2021 SET-1   Environmental Engineering
Question 5 Explanation: 
The sample size is 0.01 ml, 0.001 ml and 0.0001 ml which is 1000 times lesser than the standard of 10 ml, 1 ml and 0.1 ml.
The positive set is 5-3-1 and w.r.t. the positive combination,
MPN/100 ml will be 110 x 1000 = 110000


There are 5 questions to complete.

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