Question

2. Chlorine is used as a disinfectant and is added in the form of sodium hypochlorite (NaOCl), which dissociates completely to Na+ and OCl-. The form of chlorine is then determined according to the known equilibrium reaction: HOCl ↔ H+ + OCl- with Ka = 10-7.54. A disinfection reactor treats a flow of 2,500 L/min.

a. If the pH of the treated water is 9.6, what will the ratio of HOCl to OCl- (HOCl:OCl-) be?

b. If operators decide that a concentration of 2.5 mg/L of HOCl must be applied to the treatment process, how much (mass) NaOCl should they add per second if the pH is kept at or lower than 7.3?

Answer 1

HOCl is a weak acid as its Ka value is 10-7.54 which is very low. Its conjugate base is OCl- conjugate base is determined by extracting 1 H+ from the acid. The solution which contains some amount of weak acid with its conjugate base forms an acidic buffer solution. Here the acid is HOCl and its conjugate base is OCl-

Buffer solutions are the solutions which resist a change of pH. The pH of a buffer solution is given by Henderson-Hasselbalch equation for acidic buffers which is -

pH = pKa + log([salt]/[acid])

Where pH is the pH = -log)[H+]) pKa = -log([Ka]) and log([salt]/[acid[) is the log of concentration of salt and acid which is OCl-/HOCl

Given that pH of treated water is 9.6 and Ka = 10-7.54 we have pKa = -log(10-7.54) = 7.54

So, 9.6 = 7.54 + log(OCl-/HOCl) so, log(OCl-/HOCl) = 2.06 or OCl-/HOCl = 10^2.6 = 398.1

So, HOCl/Cl- will be 1/398.1 = 0.00251

b) Per minute we are getting 2500L of water. So, water per second will be 2500/60 L per second = 41.67 L per second.

We want to keep the pH lower than 7.3 bu adding HOCl and NaOCl. Here we will again form a buffer solution becuase of the presence of weak acid HOCl and its weak base salt NaOCl.

by Henderson-Hasselbalch equation for acidic buffers which is -

pH = pKa + log([salt]/[acid])

we have 7.3 = 7.54 + log([salt]/[acid]) so, log([salt]/[acid]) = 0.24 so, [salt]/[acid] = 10^0.24 = 1.738

The concentration of HOCl added is given as 2.5mg/L the molecular mass of HOCl = molar mass of H + Molar mass O atom + molar mass of Cl = 1 + 16 + 35.46 = 52.46 g/mol

No. of moles of HOCl in 2.5 mg = weight of HOCl/molar mass of HOCl = 2.5*10^-3/52.46 = 0.04766*10^-3

So, the molar cocentration of HOCl = no. of moles of HOCl/volume in L = 0.04766*10^-3/1 = 0.04766*10^-3 M

The concentration of NaOCl = 1.738*concentration of acid = 1.738*0.04766*10^-3 = 0.0828*10^-3 M

So, each L must contatin 0.0828 moles*10^-3 of NaOCl (Molarity is no. of moles / Volume in L)

Molar mass of NaOCl = molar mass of Na + Molar mass O atom + molar mass of Cl = 22.98 + 16 + 35.46 = 74.44 g/mol

so, 0.0828 moles should contatin 0.0828*10^-3*74.44 g of NaOCl = 6.165*10^-3 g/L per second or 6.165 mg/L per second.