`25 mL` of solution containing `HCl` and `H_(2) SO_(4)` is required for neutralisation of `25 mL N//2` caustic soda solution. `50 mL` of the same solution on precipitation with `BaCl_(2)` yielded `2.33 g` of `BaSO_(4)`. What weight of each acid contained in `1L` the solution? (Molecular mass of `BaSO_(4) = 233`)

Explanation

Caustic soda reacts with both `HCl` and `H_(2) SO_(4)`, whereas, `BaCl_(2)` reacts with only `H_(2) SO_(4)` to precipitate out `BaSO_(4)`. Note that in two experiments the above of sample is not the same.
Let millimoles of `HCl` and `H_(2) SO_(4)` in `25 mL` be `x` and `y`, respectively.
In the first experiment: Both `HCl` and `H_(2) SO_(4)` will react.
`implies (mEq of HCl + mEq of H_(2) SO_(4)) = mEq of NaOH`
`implies x + 1 + y xx 2 = 1//2 xx 25`
In the second experiment: Only `H_(2) SO_(4)` will react with `BaCl_(2)` to give `BaSO_(4)`.
Note that the volume is `50 mL`, which means mmoles of
`HCl = 2x` and that of `H_(2) SO_(4) = 2y`.
`underset(2 y mmol)(H_(2)SO_(4)) + BaCl_(2) rarr underset(2y mmol)(BaSO_(4)) + 2 HCl`
`2y = (2.33)/(233) xx 1000 = 10 implies y = 5`
Substituting value of `y` in equation (i)
`implies x + 2 xx 5 = 1//2 xx 25 implies x = 2.5`
`implies (W)/(36.5) xx 1000 = 2.5 implies W = 0.091 g`
This is the weight of `HCl` in `25 mL`, so the weight of `HCl` in `1 L = (1000)/(25) xx 0.091 = 3.65 g L^(-1)`
Similarly, mass of `H_(2) SO_(4)` in `1.0 L` of sample
`= (5 xx 98) xx 10^(-3) xx (1000)/(25) = 19.6 g L^(-1)`