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A sample of `Na_(2)CO_(3).H_(2)O` weighing `0.62 g` is added to `100 mL` of `0.1 N H_(2)SO_(4)`. The resulting solution will beA. acidicB. basicC. amphotericD. neutral |
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Answer» Correct Answer - 4 Equivalent mass of salt `(Na_(2)^(+))_(2)CO_(3)^(2-). H_(2)O` is `("Formula mass")/("Total positive charge")` `=124 u//2=62 u` Thus, Equivalents of `Na_(2)CO_(3).H_(2)O- ("Mass")/("Gram equivalent mass")` `=(0.62 g)/(62" g "eq.^(-1))=0.01` meq of `Na_(2)CO_(3).H_(2)O=(1000) eq` `=(1000) (0.01)` `=10` meq of `H_(2)SO_(4)=V_(mL)xxN` `=(100)(0.1)` `=10` Since both acid and base have equal number of meqs, they will completely neutralize each other to form salt, `Na_(2)SO_(4)`, which does not hydrolyze. Thus, the resulting aqueous solution is neutral |
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