Question

What is the equivalent weight of ${\mathrm{H}}_3{\mathrm{PO}}_4$ in the reaction: $\mathrm{Ca}{\left(\mathrm{OH}\right)}_2+{\mathrm{H}}_3{\mathrm{PO}}_4\to {\mathrm{CaHPO}}_4+2{\mathrm{H}}_2\mathrm{O}$

Answer 1

Step 1: Equivalent weight

• Equivalent weight is equal to the molecular/atomic mass divided by the number of electrons involved in the reaction per molecule/atom/ion.

Step 2: Formula for the equivalent weight

• $\text{Equivalent weight}=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{X}}$
• Here, X represents n-factor.

Step3. Calculation of valency

In the reaction,

$\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\left(\mathrm{aq}\right)+{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{aq}\right)\to {\mathrm{CaHPO}}_4\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

we have,

${\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{aq}\right)\to {\mathrm{HPO}}_4^{2-}\left(\mathrm{aq}\right)+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)$

The basicity of acid is equivalent to the number of ${\mathrm{H}}^{+}$ produced by an acid when it gets ionized in the aqueous medium.

Thus, the basicity will be 2.

Step4. Calculation of molecular weight:

Molecular weight of ${\mathrm{H}}_3{\mathrm{PO}}_4$= $\left(3+31+64\right)\text{g=98 g}$

Step5. Calculation of equivalent weight of ${\mathrm{H}}_3{\mathrm{PO}}_4$

The equivalent weight of ${\mathrm{H}}_3{\mathrm{PO}}_4$=$\frac{98}{2}=49$

So, the equivalent weight of ${\mathrm{H}}_3{\mathrm{PO}}_4$is $49$g.