Question

The equivalent mass of ${H_3}P{O_4}$ and $N{a_2}HP{O_4}$ in the reaction are respectively:
$2NaOH + {H_3}P{O_4} \to N{a_2}HP{O_4} + 2{H_2}O$
A. $49,142$
B. $49,71$
C. $98,71$
D. $98,142$

Answer 1

Hint : We know that equivalent mass is determined by the formula that Equivalent mass $=$ Molar mass / n- factor. To calculate n factor of a salt in a reaction we calculate the total moles of cationic or anionic charge replace in one mole of the salt or we can say that one mole of reactant is taken and this would will help in finding in the number of moles of elements whose oxidation state is changing.

Step-By-Step answer:
Molar mass of ${H_3}P{O_4}$ $= 3 \times$ atomic mass of hydrogen $+$ atomic mass of phosphorus $+ 4 \times$ atomic mass of oxygen $= 3 \times 1 + 31 + 4 \times 16 = 98$ gram per mole. Since here two hydrogen ions are replaced in the reaction so n factor is two. Hence equivalent weight of ${H_3}P{O_4}$ $=$ Molar mass of ${H_3}P{O_4}$ $/$ n factor $= \dfrac{{98}}{2} = 49$ gram per equivalent.
Molar mass of $N{a_2}HP{O_4}$ $=$ $2 \times$atomic mass of sodium $+$ atomic mass of hydrogen $+$ atomic mass of phosphorus $+ 4 \times$ atomic mass of oxygen $= 2 \times 23 + 1 + 31 + 4 \times 16 = 142$ gram per mole.
Here the n factor of salt $N{a_2}HP{O_4}$ is two because the total moles of cationic or anionic charge replaced in one mole of the salt is two. Hence equivalent weight $N{a_2}HP{O_4}$ $=$ Molar mass of $N{a_2}HP{O_4}$ $/$ n- factor $= \dfrac{{142}}{2} = 71$ gram per equivalent.

Hence option B is correct, that is $49,71$.

Note : We have calculated equivalent weight of ${H_3}P{O_4}$ and $N{a_2}HP{O_4}$ by calculating n- factor of them. First of all we have to calculate the molar mass of each compound. Thus we got equivalent mass. $N{a_2}HP{O_4}$ reacts with two moles of monobasic acid so n factor is two here while two hydrogen ion is replaced in order to form $N{a_2}HP{O_4}$ so n-factor of ${H_3}P{O_4}$ is also two. We know that the equivalent weight of ${H_3}P{O_4}$ and $N{a_2}HP{O_4}$ will be the ratio of their corresponding molar mass to n factor.