Question

What is the titre value of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ when $Fe$ reacts with $Fe{{(S{{O}_{4}})}_{3}}$?
a.) $25 mL$
b.) $50 mL$
c.) $75 mL$
d.) $100 mL$

Answer 1

Hint: Here we are going to use milliequivalent value to find the titre of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$. Potassium dichromate is a popular inorganic chemical reagent, the most widely used in various laboratory and industrial applications as an oxidizing agent. As with all hexavalent chromium compounds, it is toxic to health, both acute and chronic.

Complete answer:
First let us know some terms: Titres are the solution whose concentration is determined by titration and they are also known as the unknown solution and are taken pipette into the conical flask. Titrants are the solutions of known concentration and they are also known as the standard or the known solution they are taken in burette during the titration.
Standard solutions are those whose concentration is already known. Intermediate solutions are those which can react with both the known and unknown solution. Indicators are those which give information about the end point. An end point is when the completion of the reaction is observed by any means like colour change. And equivalent points are those at which the reaction between titrant and titre is complete.
As we go further there are many types of titration and we can classify them on the basis of their chemical reactions such as: Acid base titration, Oxidation Reduction Titration, Precipitation Titration and Complexometric Titration. But here we are going to talk about Oxidation Reduction titration, here we can say it as Redox Titration too.
As we know, redox reactions are due to exchange of electrons between two chemical species. The species which is oxidised when loses electrons and reduces when gain electrons and when they both proceed together so collectively they are known as Redox reactions.
 So now we are having an example of Oxidation Reduction Titration in our question so now let’s solve it first we have to find their $mEq$ values which are going to be equal for both
$Fe{{(S{{O}_{4}})}_{3}}$ and ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ so we say that,
$\begin{align}
  & mEq\text{ }of \\
 & \left( \dfrac{2\times 2.41}{964} \right)\times {{10}^{3}}=\dfrac{1}{60}\times 6\times V \\
 & V=50ml \\
 & mEq=\dfrac{mg\text{ }substance\times Valence}{Atomic\text{ }Weight} \\
\end{align}$$mEq\text{ }of$$Fe{{(S{{O}_{4}})}_{3}}$$=$$mEq\text{ }of$${{K}_{2}}C{{r}_{2}}{{O}_{7}}$
$\left( \dfrac{2\times 2.41}{964} \right)\times {{10}^{3}}=\dfrac{1}{60}\times 6\times V$
 By solving it we will get
$V=50ml$

Note: Here $mEq$ value also known as Milliequivalent is the equivalent weight of element or radical which is equal to their atomic weight or we can say formula weight divided by assumed valence in the compound.
Formula to express $mEq$ value.
$mEq=\dfrac{mg\text{ }substance\times Valence}{Atomic\text{ }Weight}$.