Question

10 ml of water required 1.47 mg of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ (M. Wt. = 294) for the oxidation of dissolved organic matter in presence of acid. Then the C.O.D of the water sample is:
(a)- 2.44 ppm
(b)- 24 ppm
(c)- 32 ppm
(d)- 1.6 ppm

Answer 1

Hint: For calculating the C.O.D of the water sample first, calculate the number of milliequivalents by dividing the weight in mg of the sample by the equivalent weight of the sample. The equivalent weight of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is equal to the molecular weight divided by the n-factor of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$.

Complete Solution :
The C.O.D is the term used to define the chemical oxygen demand of the reaction, and it tells the amount of oxygen required or consumed during the reaction in the solution.
First, we have to calculate the number of milliequivalents by dividing the weight in mg of the sample by the equivalent weight of the sample.
$\text{Number of milliequivalent = }\dfrac{\text{weight in mg}}{\text{Equivalent weight}}$
The equivalent weight of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is equal to the molecular weight divided by the n-factor of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$. The n-factor of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is 6 because in acidic medium 6 electrons are needed for the reaction.
$Equivalent\text{ }weight =\dfrac{294}{6}=49$
So, the number of milliequivalents will be:
$\text{Number of milliequivalent = }\dfrac{1.47}{49} = 0.03\text{ }meq$

Now the COD of the sample will be:
$\dfrac{0.03}{10}\text{ x 1000 = 3 meq/L}$
So, in terms of the oxygen COD, we have to multiply the COD of the sample with $8mg{{O}_{2}}/meq$
$3\text{ meq/L x 8mg}{{\text{O}}_{\text{2}}}\text{/meq = 24 mg/L = 24 ppm}$
Therefore, the COD of the sample is 24 ppm.
So, the correct answer is “Option B”.

Note: Don’t forget to multiply the COD of the sample with 8 because in the reaction, oxygen is consumed and its number of milliequivalents is equal to $8mg{{O}_{2}}/meq$. Take the n-factor of the sample by a change in the oxidation number.