Question

The equivalent weight of $\text{Na}{{\text{H}}_{2}}\text{P}{{\text{O}}_{4}}$ in the following reaction is:
$\text{NaOH}+{{\text{H}}_{3}}\text{P}{{\text{O}}_{4}}\to \text{Na}{{\text{H}}_{2}}\text{P}{{\text{O}}_{4}}+{{\text{H}}_{2}}\text{O}$
A. 96
B. 100
C. 120
D. 98

Answer 1

Hint: The equivalent weight here is determined by the number of replaceable H atoms. The replaceable H atoms mean the number of hydrogen atoms lessened in that compound after the reaction. Equivalent weight is mathematically represented by Equivalent mass = $\dfrac{\text{molecular mass}}{\text{no}\text{. of replaceable H atoms}}$ .

Complete step by step solution:
Let us first find the number of replaceable H atoms of $\text{P}$ in $\text{Na}{{\text{H}}_{2}}\text{P}{{\text{O}}_{4}}$; then we will calculate the equivalent mass.

The number of replaceable H atoms can be determined by the reaction; the reaction is $\text{NaOH}+{{\text{H}}_{3}}\text{P}{{\text{O}}_{4}}\to \text{Na}{{\text{H}}_{2}}\text{P}{{\text{O}}_{4}}+{{\text{H}}_{2}}\text{O}$. This is a neutralization reaction. The compound$\text{Na}{{\text{H}}_{2}}\text{P}{{\text{O}}_{4}}$has three hydrogen atoms and the compound ${{\text{H}}_{3}}\text{P}{{\text{O}}_{4}}$ has two hydrogen atoms. So, the number of replaceable H atoms is 1; which is replaced by one atom of $\text{Na}$.
-The molecular weight of $\text{Na}{{\text{H}}_{2}}\text{P}{{\text{O}}_{4}}$is:

Atomic weight of $\text{Na}$ is: 23 grams
Atomic weight of$\text{H}$ is: 1 gram
Atomic weight of$\text{P}$ is: 31 grams
Atomic weight of$\text{O}$ is: 16 grams

The molecular mass of $\text{Na}{{\text{H}}_{2}}\text{P}{{\text{O}}_{4}}$ is $\left[ \left( 23 \right)+\left( 1\times 2 \right)+\left( 31 \right)+\left( 16\times 4 \right) \right]=120$. The molecular mass of $\text{Na}{{\text{H}}_{2}}\text{P}{{\text{O}}_{4}}$ is 120 grams.
By using the formula; Equivalent weight =$\dfrac{\text{molecular mass}}{\text{no}\text{. of replaceable H atoms}}$ the equivalent weight of $\text{Na}{{\text{H}}_{2}}\text{P}{{\text{O}}_{4}}$ is $\dfrac{120}{1}$ ; which is 120 grams. Thus, the equivalent mass of $\text{Na}{{\text{H}}_{2}}\text{P}{{\text{O}}_{4}}$ is 120 grams,

So, the correct answer is “Option C”.

Note: The number of replaceable hydrogen atoms means the basicity of compound 1 minus the basicity of compound 2. It can be represented as $\text{no}\text{. of replaceable H atoms = Basicity}{{\text{y}}_{1}}\text{- Basicit}{{\text{y}}_{2}}$; basicity is the ability to release ${{\text{H}}^{+}}$ ions. So, the basicity of ${{\text{H}}_{3}}\text{P}{{\text{O}}_{4}}$ is 3 and that of $\text{Na}{{\text{H}}_{2}}\text{P}{{\text{O}}_{4}}$ is 2. Thus, the number of replaceable H atoms are 3-2 which is 1.