
Molecular weight of ${\text{KBr}}{{\text{O}}_{\text{3}}}$ is ${\text{M}}$. What is its equivalent weight, if the reaction is:
${\text{BrO}}_{\text{3}}^ - \to {\text{B}}{{\text{r}}^ - }$ (acidic medium)
(A) ${\text{M}}$
(B) $\dfrac{{\text{M}}}{4}$
(C) $\dfrac{{\text{M}}}{6}$
(D) $6{\text{M}}$
Answer
503.4k+ views
Hint: The valency factor for acids is the basicity i.e. the number of hydrogen ions released by an acid. The valency factor for bases is the acidity i.e. the number of hydroxide ions released by a base. Valency factor in a redox reaction is equal to the number of moles of electrons lost or gained per molecule.
Formula used: Equivalent weight $ = \dfrac{{{\text{molecular weight}}}}{{\text{X}}}$
Where ${\text{X}}$ is the valency factor.
Complete answer: The given compound ${\text{KBr}}{{\text{O}}_{\text{3}}}$ refers to a potassium salt and bromate salt. The molecular weight of ${\text{KBr}}{{\text{O}}_{\text{3}}}$ is ${\text{M}}$. Calculate the oxidation state of bromine in ${\text{BrO}}_3^ - $. This can be done by assuming the oxidation state of bromine as $x$. We know that the oxidation state of oxygen is $ - 2$. In a polyatomic ion, the sum of the oxidation numbers of all the atoms present in it is equal to the total charge on the ion.
So, we can write it as $x + 3( - 2) = - 1$
$x - 6 = - 1$
$x = 5$
So, the oxidation state of bromine in ${\text{BrO}}_3^ - $ is $ + 5$
The change in oxidation state in the reaction ${\text{BrO}}_{\text{3}}^ - \to {\text{B}}{{\text{r}}^ - }$ is $5 - ( - 1) = 6$
\[ \Rightarrow \] $Equivalent\;weight = \dfrac{{{\text{molecular weight}}}}{{\text{X}}}$
$Equivalent\;weight = \dfrac{{\text{M}}}{{\text{6}}}$
Therefore, option C is the correct answer.
Note:
The equivalent weight is the weight of a substance that contains a single reactive proton (or single hydrogen ion) or a single reactive hydroxide ion. Equivalent weight is the weight of a substance which has a single reactive proton in the case of acids which are proton donors and it is the weight of a substance which has a single reactive hydroxide ion in the case of bases which are proton acceptors.
Formula used: Equivalent weight $ = \dfrac{{{\text{molecular weight}}}}{{\text{X}}}$
Where ${\text{X}}$ is the valency factor.
Complete answer: The given compound ${\text{KBr}}{{\text{O}}_{\text{3}}}$ refers to a potassium salt and bromate salt. The molecular weight of ${\text{KBr}}{{\text{O}}_{\text{3}}}$ is ${\text{M}}$. Calculate the oxidation state of bromine in ${\text{BrO}}_3^ - $. This can be done by assuming the oxidation state of bromine as $x$. We know that the oxidation state of oxygen is $ - 2$. In a polyatomic ion, the sum of the oxidation numbers of all the atoms present in it is equal to the total charge on the ion.
So, we can write it as $x + 3( - 2) = - 1$
$x - 6 = - 1$
$x = 5$
So, the oxidation state of bromine in ${\text{BrO}}_3^ - $ is $ + 5$
The change in oxidation state in the reaction ${\text{BrO}}_{\text{3}}^ - \to {\text{B}}{{\text{r}}^ - }$ is $5 - ( - 1) = 6$
\[ \Rightarrow \] $Equivalent\;weight = \dfrac{{{\text{molecular weight}}}}{{\text{X}}}$
$Equivalent\;weight = \dfrac{{\text{M}}}{{\text{6}}}$
Therefore, option C is the correct answer.
Note:
The equivalent weight is the weight of a substance that contains a single reactive proton (or single hydrogen ion) or a single reactive hydroxide ion. Equivalent weight is the weight of a substance which has a single reactive proton in the case of acids which are proton donors and it is the weight of a substance which has a single reactive hydroxide ion in the case of bases which are proton acceptors.
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