Question

$500mL$ of a sample of water required $19.6mg$ of $K_{2}C{r}_2{O}_7$ for the oxidation of dissolved organic matter in it in the presence of $H_{2}S{O}_4$. The $COD$ of water sample is:

• A. $3.2ppm$
• B. $7.2ppm$
• C. $6.4ppm$
• D. $4.6ppm$

Answer 1

Answer: (C)

$6.4ppm$

The molecular weight of $K_{2}C{r}_2{O}_7$ is 294 g/mol.
$K_{2}C{r}_2{O}_7+4H_{2}S{O}_4\to K_{2}S{O}_4+C{r}_2(S{O}_4{)}_3+4H_{2}O+3[$O$]$
294 g $K_{2}C{r}_2{O}_7$ corresponds to $3\times 16=48$ g [$O$].
19.6 mg $K_{2}C{r}_2{O}_7$ corresponds to $19.6\times \frac{48}{294}=3.2$ mg [$O$]. This is for 500 mL of sample.

For 1000 mL of sample, COD $=3.2\times \frac{1000}{500}=6.4$ ppm.