Question

Chlorination of water produces small amounts of organic chlorides. To determine the amount of organic chlorides in drinking water, drinking water is passed through charcoal filter that adsorbed organic halides. Then charcoal was burned and resulting gases containing HCl were absorbed in aqueous solution. The content of HCl was determined by coulometric titration using silver electrode. When 0.500 L of drinking water was analyzed, a current of 5.0 mA was required for 400 s to completely precipitate AgCl. Find ppb of chlorine that was present as organic chlorides in drinking water.

Answer 1

AgCl $\to$ Ag ⁺ + Cl^-

Using Faraday law

W = $\frac{Eit}{F}$

W is mass of Cl^- formed  

E = equivalent weight of Cl = 35.5 g/eqv

i = current = 5.0 mA = 5.0×10^-3 A.

t = 400 s.

W = $\frac{35.5 \times 5 \times 10^{-3}\times 400}{96500}$ = 0.000735 g = 0.735 mg

Now, 0.735 mg is dissolved in 0.500 L.

= 0.735 mg × $\frac{1}{0.500 (L)}$

= 1.47 mg/L

1 mg/L = 10³ ppb

So, 1.47 mg/L = 1.47 mg/L × 10^-3 = 1470 ppb.

So, 1417.5 ppb of chlorine was present .