Question

Chlorine is used to disinfect a flow of 0.20 m3/s water in a municipal drinking water treatment plant, which operates an ideal plug-flow reactor with a 185 m^3 volume. The chlorine concentration is assumed to be constant throughout the reactor. Giardia, a problematic cyst-forming pathogen, is reduced by 99.9% by 2.50 mg/L HOCl. Plant operators normally keep the pH at 6.00 for the disinfection process.
If r(N) = -k*N and k* = kC, what are the values of k* and k?
Given the dissociation reaction below with a dissociation constant Ka = 10-7.54, how much OCl⁻ will be in solution in the disinfection process above? What is the total molar concentration of chlorine?
HOCl ↔ H⁺ + OCl⁻

Answer 1

Answer:

a) Kt = 7.5x10^-3 s^-1

K* = 7.5x10^-3*2.5 = 0.018 mg/L

b) doseCl2 = 2.43 mg/L

doseCl2 in moles =  3.42x10^-5 mol/L

Explanation:

data given by the exercise:

Volume = 185^3

flow rate = 0.2 m^3/s

[HOCl] = 2.5 mg/L

pH = 6

removed pathogen = 99.9%

a) The time is equal to:

t = volume/flow rate = 185/0.2 = 925 s

The removal efficiency is equal to:

n = (1-e^-Kt*t) * 100

Clearing Kt:

-e^-Kt*t = 0.999-1

-Kt*t = ln(10^-3)

Kt = 6.91/925 = 7.5x10^-3 s^-1

K* = 7.5x10^-3*2.5 = 0.018 mg/L

b) The reaction is:

HOCl = H+ + OCl-

K = [H+]*[OCl-]/[HOCl]

Clearing [OCl-]:

[OCl-] = (K*[HOCl])/[H+]

pH = -log[H+] = 6

[H+] = 10^-6

[OCl-] = (10^-7.54*2.58)/10^-6 = 0.074 mg/L

The Cl2 dose is equal to:

doseCl2 = [HOCl-]/%HOCl

%HOCl = (HOCl/(HOCl+OCl-))*100

The reaction is:

Cl2 + H2O = HOCl + HCl

HOCl = H+ + OCl-

%HOCl = (1/(1+(OCl-/HOCl))) * 100

%HOCl = (1/(1+(0.074/2.05)))*100 = 97.1%

doseCl2 = (2.5/97.1)*100 = 2.43 mg/L

doseCl2 in moles = (concentration/(molecular weight*1000) = (2.43/(70.906*1000)) = 3.42x10^-5 mol/mL

Answer 2

Answer:

a) Kt = 0.018 mg/L

b) dose of Cl2 = 2.43 mg/L

dose Cl2 in moles =  3.42x10^-5 mol/L

Explanation

Following were provided

Volume = 185^3

flow rate = 0.2 m^3/s

[HOCl] = 2.5 mg/L

pH = 6

removed pathogen = 99.9%

Now to calculate the time we use

t = volume/flow rate,

a) time therefore is,

t = volume/flow rate

= 185/0.2 = 925 s

Next is the efficiency removal

Which is,

n = (1-e^-Kt*t) * 100

Clearing the kt,

-e^-Kt*t = 0.999-1

-Kt*t = ln(10^-3)

Kt = 6.91/925 = 7.5x10^-3 s^-1

K* = 7.5x10^-3*2.5 = 0.018 mg/L

b) now the reaction is given below,

HOCl = H+ + OCl-

K = [H+]*[OCl-]/[HOCl]

Clearing [OCl-]:

[OCl-] = (K*[HOCl])/[H+]

pH = -log[H+] = 6

[H+] = 10^-6

[OCl-] = (10^-7.54*2.58)/10^-6

= 0.074 mg/L

The Cl2 dose is,

dose of Cl2 = [HOCl-]/%HOCl

%HOCl = (HOCl/(HOCl+OCl-))*100

But the reaction is

Cl2 + H2O = HOCl + HCl

HOCl = H+ + OCl-

%HOCl = (1/(1+(OCl-/HOCl))) * 100

%HOCl = (1/(1+(0.074/2.05)))*100 = 97.1%

dose of Cl2 = (2.5/97.1)*100 = 2.43 mg/L

dose of Cl2 in moles = (con./(molar mass*1000)

= (2.43/(70.906*1000))

= 3.42x10^-5 mol/mL