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A mixture of `n_(1)` moles of `Na_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` is titrated separately with `H_(2)O_(2)` and `KOH`, to reach at equivalence point. Which of the following statement is/are correct?A. Moles of `H_(2)O_(2)` and `KOH` are `n_(1)+n_(2)` and `n_(2)`B. Moles of `H_(2)O_(2)` and `KOH` is `n_(1)+(n_(2))/(2)` and `n_(1)`C. n-factors of `NaHC_(2)O_(4)` with KOH and `H_(2)O_(2)`, respectively, are 1 and 2.D. n-factors of `Na_(2)C_(2)O_(4)` with `H_(2)O_(2)` and KOH, respectively, are 2 and 1. |
Answer» Correct Answer - A::C `Na_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` both reacts with `H_(2)O_(2)` as reducing agent only. (n factor for both `=2`) With `H_(2)O_(2)`: " Eq of "`H_(2)O_(2)=" Eq of "Na_(2)C_(2)O_(4)+" Eq of "NaHC_(2)O_(4)` `2xx` moles of `H_(2)O_(2)=n_(1)xx2+n_(2)xx2` Moles of `H_(2)O_(2)=(2(n_(1)+n_(2))/(2)=(n_(1)+n_(2))` With KOH: Only `NaHC_(2)O_(4)` reacts with KOH as acid base titration n factor `=1(one H^(o+)` ion) " Eq of "`KOH=" Eq of "NaHCO_(3)` `1xx` Moles of `KOH=n_(2)xx1` Moles of KOH`=n_(2)` `therefore` Moles of `H_(2)O_(2)` and KOH are: `(n_(1)+n_(2))` and `n_(2)`. n-factor of `NaHC_(2)O_(4)` with KOH and `H_(2)O_(2)=1 and 2` n-factor of `Na_(2)C_(2)O_(4)` with `H_(2)O_(2)` and `KOH=2 and 2`. |
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