InterviewSolution
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InterviewSolution
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`NaOH` and `Na_(2)CO_(3)` are dissolvedin 200 ml aqueous solution. In the presence of phenolphthalein indicator. 17.5 ml of `0.1N HCl` are used to titrate this solution. Now methyl orange is added in the same solution titrated and it requires 2.5ml of the same `HCl`. Calculate the normality of `NaOH` and `Na_(2)CO_(3)` and their mass present in the solution. |
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Answer» Milli equivalents (a) of `HCl` used in thepresence of phenolphthalein indicator. `=N xx V`(ml) `= 0.1 xx 17.5 = 1.75` `1.75` (a) = milli eq of `NaOH +1//2` milli eq of `Na_(2)CO_(3)` ..(1) Milli eq (b) of `HCl` used in the presence of methyl orange indicator `= N xx V` (ml) `= 0.1 xx 2.5 = 0.25` `0.25` (b) `= 1//2` milli equivalents of `Na_(2)CO_(3)` .......(2) For `Na_(2)CO_(3)` solution: From equation (2) Milli eq of acid used by `Na(2)CO_(3) = 2b` `= 2 xx 0.25 = 0.5` volume of `Na_(2)CO_(3)` solution =`200ml` Suppose, Normality of `Na_(2)CO_(3) = N` Milli equivalents of `Na_(2)CO_(3) = N xx V` (ml) `= 200` Putting equivalents of acid and `Na_(2)CO_(3)` equal `200N = 0.5` or (Normality of `Na_(2)CO_(3)` solution) `N = (1)/(400)` Mass of `Na_(2)CO_(3) = N xx E xx V` (litre) (E for `Na_(2)CO_(3) = 53) = 0.0265` gram For `NaOH` solution: From equation (1) and (2) Milli eq acid used by `NaOH = a-b = 1.75 - 0.25 = 1.50` Volume of `NaOH` solution `=200ml` Suppose, Normality of `NaOH` solution = N Milli eq of `NaOH = N xx V` (ml) `= 200N` Putting the mili eq of `NaOH` and acid used equal `200 N = 1.5` (Normality of `NaOH` solution) `N = (1.5)/(200)` Mass of `NaOH = N xx E xx` (V litreS) `= (1.5)/(200) xx 40 xx 0.2`(E for `NaOH = 40) = 0.06g` |
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