`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 eq (a) of `H_(2)SO_(4)` used in the presence of phenolphthalein indicator
`= N xx V` (ml) `= 0.1 xx 2 xx 2.5 = 0.5`
`a = 0.5 = 1//2` milli equivalents of `Na_(2)CO_(3)` ...(1) Milli eq of
(b) `NaHCO_(3) +1//2` milli eq of `Na_(2)CO_(3)` ...(2)
For `Na_(2)CO_(3)` solution: From equation (1)
Milli eq of acid used by `Na_(2)CO_(3) = 2xx 0.5 = 1`
Suppose, Normality of `Na_(2)CO_(3)` solution `= N`
Volume of `Na_(2)CO_(3)` solution taken `=10ml`
Milli eq of `Na_(2)CO_(3)` taken `= N xx V` (ml) `= 10N`
Putting the milli eq of `H_(2)SO_(4)` and `Na_(2)CO_(3)`
`1 = 10N` or (Normality of `Na_(2)CO_(3)) N = 0.1` Strength (S) in g/litre
`=N xx E`
`= 0.1 xx 53` (E for `Na_(2)CO_(3) = 53) = 5.3` g/litre
For `NaHCO_(3)` solution: From equations (1) and (2) milli eq of acid used by
`NaHCO_(3) = b - a = 1.0 - 0.5 = 0.5`
Suppose, Normality of `NaHCO_(3)` solution = N
Volume of `NaHCO_(3)` solution taken =10ml
Milli equivalents of `NaHCO_(3)` taken `= 10N`
Putting the milli eq of `H_(2)SO_(4)` and `NaHCO_(3)` equal, `0.5 = 10N`
or (Normality of `NaHCO_(3)` solution) `N = 0.05` Strength (S) is g/litre `=N xx E`
(E for `NaHCO_(3) = 84) = 0.05 xx 84 = 4.2` g/litre