Find the equivalent weight of $\mathrm{Ca}(\mathrm{OH}{)}_2$ in the following reaction.

$\begin{array}{l}\mathrm{Ca}(\mathrm{OH}{)}_2+2{\mathrm{H}}_2{\mathrm{SO}}_4⟶\mathrm{Ca}{\left({\mathrm{HSO}}_4\right)}_2+2{\mathrm{H}}_2\mathrm{O}\\ \text{ (at. wt., }\mathrm{Ca}=40,\mathrm{O}=16,\mathrm{H}=1\text{ ) }\end{array}$

First method.

Mol. wt. of $\mathrm{Ca}(\mathrm{OH}{)}_2=40+2(16+1)=74\mathrm{g}{\mathrm{mol}}^{-}$

Reaction.

$\mathrm{Ca}(\mathrm{OH}{)}_2+2{\mathrm{H}}_2{\mathrm{SO}}_4⟼\mathrm{Ca}{\left({\mathrm{HSO}}_4\right)}_2+2{\mathrm{H}}_2\mathrm{O}$

1 molecule                               no OH-group

$= 2 \mathrm{OH}-$groups.

$\therefore$  No. of $\mathrm{OH}-$groups replaced from one molecule of

$\mathrm{Ca}(\mathrm{OH}{)}_2=2-0=2$.

$\therefore \text{ Eq. wt. of }\mathrm{Ca}(\mathrm{OH}{)}_2=\frac{\text{ Mol. wt. of }\mathrm{Ca}(\mathrm{OH}{)}_2}{\begin{array}{c}\text{ no. of OH-groups replaced }\\ \text{ from one molecule of }\mathrm{Ca}(\mathrm{OH}{)}_2\end{array}}$

                                     $=\frac{74}{2}=37$ Ans.

Second method

$\mathrm{Ca}(\mathrm{OH}{)}_2 + 2{\mathrm{H}}_2{\mathrm{SO}}_4 ⟶ \mathrm{Ca}{\left({\mathrm{HSO}}_4\right)}_2 + 2{\mathrm{H}}_2\mathrm{O}$

1 molecule     $=2\times 2{\mathrm{H}}^{+}\text{ions }$            $=2\times 1{\mathrm{H}}^{+}\text{ions }$

                      $=4{\mathrm{H}}^{+}\text{ion }$                       $=2{\mathrm{H}}^{+}\text{ion }$

$\therefore$   No. of ${\mathrm{H}}^{+}\text{ions }$that react with one molecule of  $\mathrm{Ca}(\mathrm{OH}{)}_2$

$=4-2=2$.

$\therefore \text{ Eq. wt. of }\mathrm{Ca}(\mathrm{OH}{)}_2=\frac{\text{ Mol. wt. of }\mathrm{Ca}(\mathrm{OH}{)}_2}{\begin{array}{c}\text{ No. of }{\mathrm{H}}^{+}\text{ions that react }\\ \text{ with one molecule of }\mathrm{Ca}(\mathrm{OH}{)}_2\end{array}}$

                                              $=\frac{74}{2}=37$ Ans.