Question

The smell often associated with public swimming pools comes from chloramines. The reaction of hypochlorous acid with ammonia from human urine will produce monochloramine (NH2Cl) as follows: HOCl(aq) + NH3(aq) ⇌ NH2Cl(aq) + H2O(l) Keq = 1.47 x 1011 One part per million (ppm) is a mg L–1. A typical public swimming pool volume (750,000 L) contains 75.0 L of urine. The concentration of ammonia in 1 L of urine is 0.200 M. If the concentration of hypochlorous acid in pool water is 1.00 ppm, determine the concentration (in ppm) of monochloramine (NH2Cl) in a typical public swimming pool. Assume the mass of 1 L of pool water is 1 kg.

Answer 1

Moles of ammonia (NH3) in the swimming pool = 75.0 L urine * (0.200 mol / 1L urine) = 15.0 mole NH3

=> [NH3] = 15.0 mole / 750,000 L = 2.00*10^-5 M

HOCl concentration in swimming pool = 1 ppm or 1 mg / L

=> Mass of HOCl in swimming pool = 750,000 L * ( 1mg / 1L) * (1g / 1000 mg) = 750 g

Molar mass of HOCl = 52.46 g/mol

=> Moles of HOCl = 750 g * (1 mol / 52.46 g) = 14.3 mol

=> [HOCl] = 14.3 mole / 750,000 L = 1.907*10^-5 M

-- HOCl(aq) + NH3(aq) ⇌ NH2Cl(aq) + H2O(l); Keq = 1.47 x 10¹¹  

I: 1.907*10^-5 M, 2.00*10^-5 M,  0 mol

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

E: (1.907*10^-5 - X), (2.00*10^-5​​-X), X

Since HOCl(aq) has smaller concentration, it is the limiting reactant and decide the amount of product formed.

Since the value of Keq is very high, almost all of HOCl and NH3 are converted to NH2Cl.

Hence [NH2Cl(aq)] = 1.907*10^-5 M or mol NH2Cl /L

=> Concentration in ppm = 1.907*10^-5 mol/L * (51.48 g / 1 mol) * (1000 mg / 1 g) = 0.982 mg/L or 0.982 ppm (Answer)