1.

What is the normarlity and nature of a mixutre obtained by mixing `0.62 g` of `Na_(2) CO_(3). H_(2) O` to `100 mL` of `0.1 N H_(2) SO_(4)`?

Answer» mEq of `Na_(2) CO_(3) H_(2) O = (0.62)/(62) xx 1000 ((W)/(E) xx 1000 = mEq)`
mEq of `H_(2) SO_(4) = 100 xx 0.1 = 10`
`Na_(2)CO_(3) + H_(2)SO_(4) rarr Na_(2) SO_(4) _ H_(2) O + CO_(2) uarr`
mEq added 10 10 0 0 0
mEq left 0 0 10 10 10
`:. N_(Na_(2)SO_(4)) = (10)/(100) = 0.1`
Solution becomes neutral since both acid and base are used up and `Na_(2) SO_(4)` does not show hydrolysis.


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