Question

When bromide-containing waters are disinfected with chlorine, Br– can be oxidized into HOBr and other reactive bromine species that are capable of generating toxic, brominated disinfection by-products. Consider the reaction shown below:

HOCl(aq) + Br– ⇌ HOBr(aq) + Cl–

The rate of the forward reaction can be described by the following equation:

rate=-d[Br^- ]/dt=k[HOCl][Br^-]

where the rate constant (k) equals 1.55 × 103 M–1 s–1 at 25 C. A sample of raw (untreated) drinking water (with pH 7.30 and [Br–] = 0.70 µM) is dosed with free chlorine such that TOTOCl = 28 µM (i.e., 2.0 mg/L as Cl2). Following addition of free chlorine, how much time is required for 50% of the bromide to be oxidized? Assume that the water has a high buffer intensity.

Answer 1

Given the concentration of free Chlorine, [Cl₂(g)] = 28 µM * (10^-6 M / 1µM) = 2.8*10^-5 M

Given pH = 7.30

=> pH = - log[H⁺(aq)] = 7.30

=> [H⁺(aq)] = 10^-7.30 = 5.012*10^-8 M

--HOCl(aq)  <------> OCl^-(aq) + H⁺(aq) ; Ka = 2.63*10^-8

I: 2.8*10^-5 M -------- 0 M ------- 5.012*10^-8 M

C: - X M --------------- +X M

E: (2.8*10^-5 - X) M, X M -------- 5.012*10^-8 M

Ka = 2.63*10^-8 = [OCl^-(aq)]*[H⁺(aq)] / [HOCl(aq)]

=> 2.63*10^-8 = X * 5.012*10^-8 / (2.8*10^-5 - X)

=> 2.63*10^-8 * (2.8*10^-5 - X) = 5.012*10^-8X

=> 7.364*10^-13 - 2.63*10^-8X = 5.012*10^-8X

=>7.364*10^-13 = 5.012*10^-8X + 2.63*10^-8X

=> 7.364*10^-13 = 7.642*10^-8 X

=> X = 7.364*10^-13 / 7.642*10^-8 = 9.636*10^-6 M

X = [OCl^-(aq)] = 9.636*10^-6 M

=> [HOCl(aq)] = (2.8*10^-5 - X) = (2.8*10^-5 - 9.636*10^-6)

=>  [HOCl] = 1.8364*10^-5 M

Given [Br^-] = 0.70 µM * (10^-6 M / 1µM) = 7.0*10^-7 M

k = 1.55*10³ M^-1.s^-1

Given 50% (one-half) of Br- is oxidized

=> (7.0*10^-7 M / 2) / t = k* [HOCl]*[Br^-]

=> 3.5*10^-7 M / t = 1.55*10³ M^-1.s^-1 * 1.8364*10^-5 M * 7.0*10^-7 M

=> t = 3.5*10^-7 / (1.55*10³ * 1.8364*10^-5 * 7.0*10^-7) s

=> t = 17.6 seconds (Answer)