Q21GATE 2014MCQ

For a sample of water with the ionic composition shown in the figure below, the carbonate and non-carbonate hardness concentrations (in mg/l as $CaCO_{3}$ ), respectivelyare:

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📖 Explanation

Carbonate hardness $=3.5 \times 10^{-3}$ g-eq [if NCH is present sodium alkalinity will be absent i.e. $\mathrm{NaHCO}_{3}$ absent] $=3.5 \times 10^{-3} \times \frac{50 g}{1} as \mathrm{CaCO}_{3}$ $=175 \mathrm{mg} / l\text{ as }\mathrm{CaCO}_{3}$ 2. Non carbonate hardness = total hardness carbonate hardness 3. Total hardness $=5 \times 50 \mathrm{mg} / l\text{ as }\mathrm{CaCO}_{3}$ $[total hardness is due to $\mathrm{Ca}^{2}+ and \mathrm{Mg}^{2+}$ ]$ $=250 \mathrm{mg} / l \text{ as } \mathrm{CaCO}_{3}$ $\mathrm{NCH}=250-175=75 \mathrm{mg} / l \text{ as } \mathrm{CaCO}_{3}$

Q23GATE 2011MCQ

Anaerobically treated effluent has MPN of total coliform as $10^{6}$ /100 mL. After chlorination, the MPN value declines to $10^{2}$ /100 mL. The percent removal (%R) and log removal (logR) of total coliform MPN is

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📖 Explanation

\begin{aligned} \% R &=\left(\frac{10^{6}-10^{2}}{10^{6}}\right) \times 100 \\ &=0.9999 \times 100=99.99 \% \\ \log R &=\log 10^{6}-\log 10^{2}=6-2=4 \end{aligned}

Q24GATE 2010MCQ

Ion concentrations obtained for a groundwater sample (having pH=8.1) are given below. Carbonate hardness (mg/L as $CaCO_3$ ) present in the above water sample is

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📖 Explanation

Total alkalinity in water consists of alkalinity caused by $\mathrm{CO}_{3}^{2-}, \mathrm{HCO}_{3}\text{ and }\mathrm{OH}^{-}.$ A little negative alkalinity is also caused by $\mathrm{H}^{+}$

\begin{aligned} \therefore \text { Total Alkalinity } &=\left[\mathrm{HCO}_{3}^{-}\right]\$ \times \frac{\mathrm{Eq} \cdot \mathrm{wt} . \text { of } \mathrm{CaCO}_{3}}{\mathrm{Eq}, \mathrm{wt} . \text { of } \mathrm{HCO}_{3}^{-}} \\ &=250 \times \frac{50}{61}=205 \mathrm{mg} / \mathrm{L} \end{aligned}

Carbonate hardness is equal to the total hardness or alkalinity, whichever is less. The carbonate hardness of the sample will be $205 \mathrm{mg} / \mathrm{L}$ $\therefore$ Non-carbonate hardness = Total hardness - carbonate hardness $=275-205=70 \mathrm{mg} / \mathrm{L}$

Q25GATE 2009MCQ

Following chemical species were reported for water sample form a well : Total hardness in mg/L as is $CaCO_{3}$ is

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📖 Explanation

\begin{aligned} &\text { Total Hardness \in } \mathrm{mg} / \mathrm{L} \text { as } \mathrm{CaCO}_{3} \\ &=\left[\mathrm{Ca}^{2+}\right]\$ \times \mathrm{Combining} \text { weight of } \mathrm{CaCO}_{3}+\left[\mathrm{Mg}^{2+}\right]\$\\ &\times \text { combining weight of } \mathrm{CaCO}_{3} \\ &= 12 \times 50+18 \times 50=600+900=1500 \end{aligned}

Q26GATE 2009MCQ

Following chemical species were reported for water sample form a well : Alkalinity present in the water in mg/L as $CaCO_{3}$ is

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📖 Explanation

Total alkalinity in $\mathrm{mg} / \mathrm{L} \text{ as }\mathrm{CaCO}_{3}$ $=\left[\mathrm{CO}_{3}^{2-}\right]\$ × \text {combining weight of } \mathrm{CaCO}{3}+\left[\mathrm{HCO}{3}^{-}\right]$ \times $combining weight of$ \mathrm{CaCO}_{3}=5 \times 50+30 \times 50=250+1500=1750$

Q27GATE 2008MCQ

Group I list estimation method of some of the water and wastewater quality parameter. Group-II list the indicators used in the estimation methods. Match the estimation method (Group I) with the corresponding indicator (Group II).

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Q28GATE 2008MCQ

A wastewater sample contains $10^{-5.6}$ mmol/l of $OH^{-}$ ions at 25 $^{\circ}$ C. The pH of this sample is

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📖 Explanation

\begin{aligned} \mathrm{pOH} &=-\log \left[\mathrm{OH}^{-}\right]\$ \\ \Rightarrow\quad \mathrm{pOH} &=-\log \left(10^{-5.6} \times 10^{-3}\right) \\ \Rightarrow\quad \mathrm{pOH} &=-\log \left(10^{-8.6}\right) \\ \mathrm{pOH}&=8.6\\ \text{BH}\quad \mathrm{pH}+\mathrm{pOH} &=14 \\ \Rightarrow\quad \mathrm{pH} &=14-8.6 \\ \Rightarrow\quad \mathrm{pH} &=5.4 \end{aligned}

Q29GATE 2007MCQ

The alkalinity and the hardness of a water sample are 250 mg/L and 350 mg/L as $CaCO_{3}$ , respectively. The water has

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📖 Explanation

Carbonate hardness is equal to the total hardness or alkalinity whichever is lesser. In this case alkalinity (250 mg/l )is less than total hardness (350 mg/l). carbonate hardness = 250 mg/l as $CaC0_{3}$ . Non-carbonate hardness is the total hardness in excess of the alkalinity i.e Non-carbonate hardness = Total hardness - alkalinity =350-250 $=100 \mathrm{mg} / \mathrm{L}\text{ as }\mathrm{CaCO}_{3}$

Q30GATE 2007MCQ

The presence of hardness in excess of permissible limit causes

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Q31GATE 2006MCQ

A water sample contains the following dissolved ions. $[Na^+]$ = 56mg/L; $[Ca^{2+}]$ =40 mg/L; $[Mg^{2+}]$ =30mg/L $[AI^{3+}]$ =3 mg/L; $[HCO^{-}_{3}]$ =190mg/L; $[CI^{-}]$ =165mg/L; Water pH is 7. Atomic weights: Ca:40; Mg:24; Al:27; H:1; C:12; O:16; Na:23; Cl:35.5 The non-carbonate hardness of the sample in mg/L as $CaCO_{3}$ is

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📖 Explanation

Carbonate hardness

\begin{aligned} &=\left[H C O_{3}\right] \times \frac{e q \cdot w t \cdot \text { of } C a C O_{3}}{e q \cdot w t \cdot \text { of } H C O_{3}^{-}}=190 \times \frac{50}{61} \\ &=155.74 \simeq 156 \mathrm{mg} / \mathrm{L} as \mathrm{CaCO}_{3} \end{aligned}

Noncarbonate hardness = Total hardness - Carbonate hardness $=242-156=86 \mathrm{mg} / \mathrm{L} \text { as } \mathrm{CaCO}_{3}$

Q32GATE 2006MCQ

A synthetic sample of water is prepared by adding 100 mg Kaolinite (a clay mineral), 200 mg glucose, 168 mg NaCl, 120 mg $MgSO_{4}$ , and 111 mg $CaCL_{2}$ to 1 litre of pure water. The concentrations of total solids (TS) and fixed dissolved solids (FDS) respectively in the solution in mg/L are equal to

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📖 Explanation

\begin{aligned} \text { Total Solids } &=100+200+168+120+111 \\ &=699 \mathrm{mg} / l \end{aligned}

Fixed dissolved solids

\begin{aligned} &=\text{ Total Solids }-\text{ suspended solids}\\ &=699-100-200 \\ &=399 \mathrm{mg} / l \end{aligned}

Q33GATE 2005MCQ

1 TCU is equivalent to the color produced by

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Q34GATE 2005MCQ

If tomato juice is having a pH of 4.1, the hydrogen ion concentration will be

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📖 Explanation

\begin{aligned} \mathrm{pH} &=-\log _{10}\left[\mathrm{H}^{+}\right]\$ \\ \Rightarrow\quad 4.1 &=-\log _{10}\left[\mathrm{H}^{+}\right]\$ \\ \Rightarrow\quad 10^{-4.1} &=\left[\mathrm{H}^{+}\right]\$ \\ \Rightarrow\quad \left[\mathrm{H}^{+}\right]\$ &=7.94 \times 10^{-5} \mathrm{mol} / \mathrm{L} \end{aligned}

Q35GATE 2004MCQ

A standard multiple-tube fermentation test was conducted on a sample of water from a surface stream. The results of the analysis for the confirmed test are given below. Using the above MPN index table, the Most Probable Number (MPN) of the sample is

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📖 Explanation

The MPN for the combination 4-3-1 is 33 but the standard sample sizes are 10mL, 1.0 mL and 0.1 mL. The speciman sample sizes are 1.0 mL, 0.1 mL and 0.01 mL i.e 10 times less than standard sample. Therefore MPN = $33\times 10=330$ .

Q36GATE 2004MCQ

The organism, which exhibits very nearly the characteristics of an ideal pathogenic indicator is

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Q37GATE 2004MCQ

Most of the turbidity meters work on the scattering principle. The turbidity value so obtained is expressed in

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Q38GATE 2004MCQ

Hardness of water is directly measured by titration with ethylene-di-amine-tetracetic acid (EDTA) using

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Q39GATE 2003MCQ

A water treatment plant treating 10 MLD of water requires 20 mg/L of filter Alum, $AI_2(SO_4)_3.18H_2O$ . The water has 6 mg/L of alkalinity as $CaCO_3$ (Al = 26.97, S = 32, O = 16, H = 1, Ca = 40 and C = 12). Quantity of Quick Lime required ( $10^{6}$ mg per year as CaO) shall be

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📖 Explanation

Natural alkalinity available in water =6 mg /L Total alkalinity required to be added in the form $\text { of quick lime }=9-6=3 \mathrm{mg} / \mathrm{L}$

\begin{array}{cccc} &\mathrm{CaCO}_{3} &\longrightarrow &\mathrm{CaO}+\mathrm{CO}_{2} \\ &100 \mathrm{gm} &\quad &56 \mathrm{gm} \end{array}

$\therefore 56 \mathrm{gm}$ of CaO (quick lime) is required to produce $100 \mathrm{gm} of \mathrm{CaCO}_{3}$ $\therefore$ Quantity of CaO required $=\frac{56}{100} \times 3 \mathrm{mg} / \mathrm{L}=1.68 \mathrm{mg} / \mathrm{L}$ Total amount of CaO required

\begin{aligned} &=10 \times 10^{6} \times 1.68 \times 365 \\ &=6132 \times 10^{6} \mathrm{mg} \text { per year } \end{aligned}

Q40GATE 2003MCQ

Results of a water sample analysis are as follows : (milliequivalent weight of $CaCO_3$ =50 mg/meq). Hardness of the water sample in mg/L as $CaCO_3$ is

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📖 Explanation

$\text{Total hardness} =\left[ Mg^{2+} \right]\$ × \frac{\text{eq. wt. of}; ; CaCO_{3}}{\text{eq. wt. of}; ; Mg^{+}}+\left[ Ca^{2+} \right]$\times \frac{\text{eq. wt. of}; ; CaCO_{3}}{\text{eq. wt. of}; ; Ca^{+}}$

\begin{aligned} &=\left ( 10\times \frac{50}{12.2} \right )+\left ( 55\times \frac{55}{20} \right ) \\ &= 178.48\simeq 179\; mg/l \end{aligned}

In addition to $Ca^{2+}$ and $Mg^{2+}$ , other divalent metallic icons like $Fe^{2+}$ , $Mn^{2+}$ , $Sr^{2+}$ , etc may also cause hardness but their presence is negligible in most of the waters. $Na^{2+}$ and $K^{+}$ do not contribute in hardness.