# Solve this:

Your question seems to be unclear, so I am assuming that you want the detailed solution of Example 21. If you still have any doubts regarding this topic, then we request you to please repost your question with complete information regarding it. So, that our experts can help you in a better way.

Please find below the solution of example 21:

For calculating the equivalent mass of these compounds we use the following relation-

Equivalent mass = $\frac{\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{acidic}\mathrm{salt}}{\mathrm{Replaceable}{\mathrm{H}}^{+}\mathrm{ions}\mathrm{of}\mathrm{the}\mathrm{acidic}\mathrm{salt}}$ (as we know that, both the salts are acidic salts)

**NaH**_{2}PO_{4}-

Replaceable H

^{+}ions = 2

Therefore, Equivalent mass = $\frac{120}{2}$

=

**60 g eq**

^{-1}**Na**_{2}HPO_{4}-

Replaceable H

^{+}ions = 1

Therefore, Equivalent mass = $\frac{142}{1}$

=

**142 g eq**

^{-1}Hope this information clears your doubts about the topic.

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