Question

Atomic weight of Barium is 137.34. The equivalent weight of barium in \[BaCr{O_4}\]which is used as an oxidising agent in acid medium is
\[
  A.\;\;\;\;\;137.34 \\
  B.\;\;\;\;\;45.78 \\
  C.\;\;\;\;\;114.45 \\
  D.\;\;\;\;\;68.67 \\
 \]

Answer 1

Hint: When we come across the dissociation of $BaCrO_4$, we will get \[B{a^{2 + }}\] and \[Cr{O_4}^{2 - }\]. Now use this to solve the problem.
Formula used:
${\text{Ew (Equivalent Weight) = }}\dfrac{{{\text{Mw (Molecular weight)}}}}{n}$
Where n is n factor.

Complete step by step solution:
When we add \[BaCr{O_4}\] in any acid the first thing that happens is \[BaCr{O_4}\] dissociates into two ions which are \[B{a^{2 + }}\]+ and \[Cr{O_4}^{2 - }\].
${\text{BaCrO4 }} \to {\text{ B}}{{\text{a}}^{2 + }} + {\text{ CrO}}{{\text{4}}^{2 - }}$
In this you can clearly see that \[Cr{O_4}^{2 - }\] is the one which works as an oxidising agent by changing it’s an oxidative state from +6 to +3 and gaining 3 electrons as shown below.
${\text{3}}{{\text{e}}^ - } + {\text{CrO}}{{\text{4}}^{2 - }} \to {\text{C}}{{\text{r}}^{3 + }}$
Here, it is clearly shown that the \[Cr{O_4}^{2 - }\]accepts 3 electrons and changes its state from Cr6+ to Cr3+ and hence the n factor for this is 3.
Now, using the formula ${\text{Ew (Equivalent Weight) = }}\dfrac{{{\text{Mw (Molecular weight)}}}}{n}$
We will calculate the equivalent weight of Barium.
${\text{Ew = }}\dfrac{{{\text{Mw}}}}{n}$
The Molecular weight of barium is 137.34 and the n factor is 3, put these values in the above equation we get,
${\text{Ew = }}\dfrac{{137.34}}{3}$
\[Ew = {\text{ }}45.78\]
So we conclude that the equivalent weight to barium in \[BaCr{O_4}\]is 45.78

Hence the correct option is “B”.:

Note: We must know that barium chromate $(BaCr{O_4})$ is useful for burn rate modifiers in pyrotechnics. Also barium chromate is a yellow colour powder. We can use barium chromate as a corrosion inhibitor when zinc alloy electroplating surfaces. For acids, n factors as the amount of H+ ion replaced by 1 mole of acid in a reaction. For alkali, n factors as the amount of OH- ion replaced by 1 mole of alkali in a reaction.