Calculate the equivalent weight of phosphorous acid H3PO3 in the following reaction.2NaOH+H3PO3⟶Na2HPO3+2H2O.(at. wt., H=1,P=31,O=16)

2NaOH+H3PO3⟶Na2HPO3+2H2O1 molecule              oneH-atom=3H-atoms In the above reaction, number of replaceable H-atoms in one molecule ofH3PO3=3H-atoms −1H-atom =2H-atoms ∴ Eq. wt. of H3PO3= Mol. wt. of H3PO3 no. of replaceable H-atoms                                       in one molecule of  acid(=basicity)                                                                         =(3×1)+31+(3×16)2=822=41.