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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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