Question

Calculate the equivalent weight of $NaOH$ in the following reaction:
$3B{r_2} + 6NaOH \to 5NaBr + NaBr{O_3}$

Answer 1

Equivalent weight of a substance is the number of parts by weight of it, which combines with or displaces directly or indirectly 1.008 parts by weight of hydrogen or 8 parts by weight of oxygen or 35.5 parts by weight of chlorine.

Complete step by step answer:
Here the chemical equation given is:
$3B{r_2} + 6NaOH \to 5NaBr + NaBr{O_3}$
Here we have to find the equivalent weight of sodium hydroxide i.e. $NaOH$. Since $NaOH$ is basic, and we know the equivalent weight of acid and base is given by the following formula:
${\text{Equivalent weight}} = \dfrac{{{\text{molecular weight}}}}{{\text{n}}}$
Where n= number of moles of either ${H^ + }$ or $O{H^ - }$ ions which enter the reaction or displaced.
Or we can write the above formula as:
In the case of acid, ${\text{Equivalent weight}} = \dfrac{{{\text{molecular weight}}}}{{{\text{Basicity}}}}$
In case of base, ${\text{Equivalent weight}} = \dfrac{{{\text{molecular weight}}}}{{{\text{acidity}}}}$
Now to find the molecular weight of ${NaOH}$, the atomic weight of sodium is $23u$. and the atomic weight of oxygen is 16u and the atomic weight of hydrogen atom is 1u. On adding the atomic weight of all the atom present in ${NaOH}$ we find its molecular weight, it will be
$23 + 16 + 1 = 40u$
And acidity ${NaOH}$ is one because it gives only one hydroxide ion. Hence the equivalent weight of ${NaOH}$ will be:
$\text{ }\\!\\!~\\!\\!\text{ }\\!\\!~\\!\\!\text{ Equivalent weight}=\frac{40}{1}\text{ }\Rightarrow 40\text{ }$

Note:
If we consider volume ratio instead of weight ratio then the equivalent weight of an element is defined as the number of parts by weight of it which combines with or displaced from the compound 11200 ml of hydrogen or 5600 ml of oxygen or 11200 ml of chlorine at N.T.P.
The equivalent weight of a substance is also known as chemical equivalent and is a pure number and has no unit because it represents the ratio of weight.