Question

For \[500mL\] 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

Hint: COD stands for chemical oxygen demand. It is the amount of oxygen required to chemically oxidize organic water waste into inorganic products.

Complete step by step answer:
The term COD refers to chemical oxygen demand. It describes a measure of water and wastewater quality present in a random sample of water solution.
The COD test is frequently used for measuring the efficacy of water treatment plants. A strong oxidizing agent under acidic conditions is used to completely oxidize the dissolved organic compound to carbon dioxide. The reaction uses oxygen which is produced under the reaction conditions.
The materials which are used as a strong oxidant are potassium dichromate or potassium permanganate under acidic conditions like sulfuric acid. The method followed for determining COD is as follows:
a. A measured excess amount of the oxidant is added to the sample for oxidation purpose.
b. After the completion of the oxidation reaction the amount of oxidant remaining in the solution indicates the amount of organics present in the sample. For this a titration is employed in the presence of an indicator solution.
COD is expressed in terms of mg/L, describing the total mass of oxygen produced and consumed per liter of solution. The chemical reaction for the formation of oxygen using \[{K_2}C{r_2}{O_7}\] and \[{H_2}S{O_4}\] is
\[{K_2}C{r_2}{O_7} + 4{H_{2}}S{O_4} \to {K_2}S{O_4} + C{r_2}{\left( {S{O_4}} \right)_3} + 4{H_2}O + 3\left[ O \right]\]
The molecular weight of \[{K_2}C{r_2}{O_7}\]​ is = \[2{\text{ }} \times \] atomic mass of \[K\] + \[2{\text{ }} \times \] atomic mass of \[Cr\] + \[7{\text{ }} \times \] atomic mass of \[O\]
=$2 \times 39 + 2 \times 52 + 7 \times 16 = 294g/mol$.
From the above balanced reaction it is clear that one mole of \[{K_2}C{r_2}{O_7}\] produces \[3\] moles of \[O\]. Hence \[294g\] of \[{K_2}C{r_2}{O_7}\]​ will produce =\[3 \times 16 = 48\;g{\text{ }}\left[ O \right]\].
The amount of \[{K_2}C{r_2}{O_7}\] supplied is \[19.6mg\] in \[500mL\] of water.
Thus \[19.6mg\] of \[{K_2}C{r_2}{O_7}\] produces =\[19.6 \times \dfrac{{48}}{{294}}mg\left[ O \right] = 3.2mg\left[ O \right]\] in \[500mL\] water.
Thus COD of water sample in \[1000mL\] is \[3.2 \times \dfrac{{1000}}{{500}}{\text{ = }}6.4ppm\].
Hence option C is the correct answer.

Note:
The COD is an important measurement for wastewater treatment. It helps in determining the amount of pollutants also called organic particles in water.