(1) The phenol concentration in a water solution is 20 ppm and the density of th

Question

(1) The phenol concentration in a water solution is 20 ppm and the density of this solution is 1.1 g/cm3. Please express the concentration by mg/L (2) Calculate pH for a solution that contains proton [H] of 0.02 mM (milimoleL) (3) Chlorine dioxide (C1O2) is added to the water pipeline as a predisinfection measure similar to ozonation as we discussed in the class. C1O2 reacts with organics and pathogens and follows first-order reaction kinetics. The concentration of C102 decreases to half of the initial level within 20 min. If we want to maintain the final concentration of C1O2 at 10 mg/L at the pipe's exit into the plant as shown below, what is the initial CI02 concentration that should be spiked into the water pipeline? The pipe has the dimension shown in the following graph. Flow rate, Q, is 1000 gallon/min Pipe diameter: 3 ft Pipe length: 3500 ft tt (4). If a water sample contains 10 mg/L of bicarbonate alkalinity and pH-8, what is the carbonate alkalinity in mgL as the ion, in mg/L as CaCO3, and in megL, pKal=6.35 and pka2= 10.33. pKa=-logio (Ka) and there are two Ka (K1 and K2) as shown below H2CO,

Explanation / Answer

1.

Total mass of 1 L(1000cm3) solution = 1.1g/cm3 x 1000cm3/L = 1100g/L

20 ppm of 1100g = 1100g x 10-6 = 1.1 x 10-3 g = 1.1 mg

concentration in mg/L = 1.1mg/L

Ans = 1.1 mg/L

2.

pH = -log[H+]

we have [H+] = 0.02 mM = 0.02 x 10-3 M

pH = -log[0.02 x 10-3 M] = 4.7

3.

time taken to reach half of original concentration is given = 20 min

for 1st order reaction ln[Ao]/[A] = kt

[Ao] = initial concentration

[A] = final concentration

k = rate constant

t = time

when concentration become half of orginal ,we have t = 20min

ln[Ao]/[Ao]/2 = k x 20 min

k = ln2 / 20 min = 3.465 x 10-2 min-1

assuming this to be perfect plug flow reactor

T =   -dCA/-rA

integration this we get

TK = lnCAo/CA

T = time constant = time take by reactant to reach end point of pipe

1000 gallon per minute = 133.681 ft3/min

velocity of fluid = Q/A

A = area of cross section

= pi x d2/4 = 3.14 x (3ft)2/ 4 = 7.065 ft2

(velocity)v = 133.681 ft3/min/7.065 ft2 = 18.92 ft/min

time taken = distance / velocity = 3500 ft / 18.92 ft/min = 185 min = T

Initial concentration can be calculated from = TK = lnCAo/CA

185 min x 1.465 x 10-2 min-1 = ln CAo / 10mg/L

15 = CAo / 10mg/L

CAo = 150 mg/L

Maximun initial concentation possible = 150 mg/L

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