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  4. Ca(OH)2+H3PO4 xrightarrow[]CaHPO4+2H2O the equivalent weight of H3PO4 in the above reaction is:

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Q. $ Ca{{(OH)}_{2}}+{{H}_{3}}P{{O}_{4}}\xrightarrow[{}]{{}}CaHP{{O}_{4}}+2{{H}_{2}}O $ the equivalent weight of $ {{H}_{3}}P{{O}_{4}} $ in the above reaction is:

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Solution:

The equivalent weight of $ {{H}_{3}}P{{O}_{4}}=\frac{molecular\text{ }weight}{2} $ (molecular wt. of $ {{H}_{3}}P{{O}_{4}} $ $ =3+31+64=98) $ $ =\frac{98}{2}=49 $