46. A water sample has a pH of 9.25. The concentration of hydroxyl ions in the water sample is

1. 1. 10^–9.25 moles/l

   
   
   

   2. 10^–4.75 moles/l

   
   
   

   3. 0.302 mg/l

   
   
   

   4. 3.020 mg/l

   
   
   

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1. View Hint View Answer Discuss in Forum

   pH = –log[H⁺]
   9.25 = –log[H⁺]
   ∴ [H⁺] = 10^–9.25 mol/l
   [H⁺ ][OH^-] = 10^-14
   

   [OH-] =	10-14	= 10-4.75mol/l
   	10-9.25

   
   = 10^–4.75 × 17 × 1000
   = 0.302 mg/l.

   Correct Option: C

   pH = –log[H⁺]
   9.25 = –log[H⁺]
   ∴ [H⁺] = 10^–9.25 mol/l
   [H⁺ ][OH^-] = 10^-14
   

   [OH-] =	10-14	= 10-4.75mol/l
   	10-9.25

   
   = 10^–4.75 × 17 × 1000
   = 0.302 mg/l.

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47. Chlorine gas (8 mg/l as Cl₂) was added to a drinking water sample. If the free chlorine residual and pH was measured to be 2 mg/l (as Cl₂) and 7.5, respectively, what is the concentration of residual OCl^- ions in the water? Assume that the chlorine gas added to the water is completely converted to HOCl and OCl^-. Atomic weight of Cl: 35.5.
    Given: OCl^- + H⁺ ⇄^K HOCl, K = 10^7.5

1. 1. 1.408 × 10^–5 moles/l

   
   
   

   2. 2.817 × 10^–5 moles/l

   
   
   

   3. 5.634 × 10^–5 moles/l

   
   
   

   4. 1.127 × 10^–4 moles/l

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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48. Anaerobically treated effluent has MPN of total coliform as 10⁶ /100 ml. After chlorination, the MPN value declines to 10²/100 ml. The per cent removal (%R) and log removal (log R) of total coliform MPN is

1. 1. %R = 99.90; log R = 4

   
   
   

   2. %R = 99.90; log R = 2

   
   
   

   3. %R = 99.99; log R = 4

   
   
   

   4. %R = 99.99; log R = 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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49. A water treatment plant is required to process 28800 m³/d of raw water (density = 1000 kg/m³, kinematic viscosity = 10–6 m²/s). The rapid mixing tank imparts a velocity gradient of 900s^–1 to blend 35 mg/l of alum with the flow for a detention time of 2 minutes. The power input (W) required for rapid mixing is

1. 1. 32.4

   
   
   

   2. 36

   
   
   

   3. 324

   
   
   

   4. 32400

   
   
   

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1. View Hint View Answer Discuss in Forum

   Velocity gradient,
   

   G = √	P
   	μv

   
   ∴ P = G² μn
   

   = 900² × (10-6 × 1000) ×	(28800 × 2)	= 32400 w
   	24 × 60

   Correct Option: D

   Velocity gradient,
   

   G = √	P
   	μv

   
   ∴ P = G² μn
   

   = 900² × (10-6 × 1000) ×	(28800 × 2)	= 32400 w
   	24 × 60

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50. The presence of hardness in excess of permissible limit causes

1. 1. cardio vascular problems

   
   
   

   2. skin discolouration

   
   
   

   3. calcium deficiency

   
   
   

   4. increased laundry expenses

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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