`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.

Explanation

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`