Question

A $ 100ml $ sample of water was treated to convert any iron present to $ F{e^{2 + }} $ . Addition of $ 25ml $ of 0.002M $ {K_2}C{r_2}{O_7} $ resulted in the reaction.
 $ 6F{e^{2 + }} + C{r_2}{O_7}^{2 - } + 14{H^ + } \to 6F{e^{3 + }} + 2C{r^{3 + }} + 7{H_2}O $
The excess of $ {K_2}C{r_2}{O_7} $ was back-titrated with $ 7.5ml $ of $ 0.01{\text{ M}} $ $ F{e^{2 + }} $ solution. Calculate the parts per million (ppm) of iron in the water sample.

Answer 1

Hint: A titration is a very useful technique in the laboratories and in many chemical researches. So to discuss this question and to solve it carefully, first we need to understand what the titration process is and how it helps us in the chemical reactions.

Complete answer:
A titration can be defined as a chemical reaction which is used to determine the concentration of the unknown sample using the sample with a known concentration.
In this process of titration, the titrant with a known concentration is added from the buret to the known quantity of the analyte (unknown sample) until the reaction is complete.
So, let’s get started with the solution of the above question:
First of all we will calculate the mille equivalent of the excess $ {K_2}C{r_2}{O_7} $ .
Therefore, Mili equivalent of the excess $ {K_2}C{r_2}{O_7} $ $ = 7.5 \times 0.01 \times 1 $
Here, $ 1 $ is the n- factor.
And the mili equivalent of the $ {K_2}C{r_2}{O_7} $ that is used or taken $ = 25 \times 0.002 \times 6 $
Here $ 6 $ is the n- factor.
As the oxidation state is changing from $ + 6 $ to $ + 3 $ and that too for two atoms. Hence, the n-factor will be $ 6 $ .
Now let’s calculate the mili equivalent of $ F{e^{2 + }} $ in a water sample.
So, mili equivalent of $ F{e^{2 + }} $ in a water sample $ = 25 \times 0.002 \times 6 $ $ - 7.5 \times 0.01 \times 1 $
 $ = 0.225{\text{ eq}} $
Now, mass of $ F{e^{2 + }} $ $ = 0.225{\text{ }} \times {10^{ - 3}} \times 56 $
 $ = 12.26 \times {10^{ - 3}} $ Gram.
So, in $ 100g - - - - - - - - 12.6 \times {10^{ - 3}} $
And in $ {10^6} $ gram $ - - - - - - \dfrac{{12.6 \times {{10}^{ - 3}}}}{{100}} \times {10^6} $
 $ = 126 $ Ppm
Here, the answer to our question is $ = 126 $ ppm i.e. parts per million.

Note:
ppm or parts per million can be defined as the number of units of mass of a contaminant per million units of a total mass. Parts per million units are used to calculate the concentration of very dilute solutions. It is mainly used to find out the contamination in the soils.