A `100.0 mL` solution containing `HCl` and `HBr` was titrated with `0.1235 M NaOH`. The volume of base required to neutralise the acid was `47.14 mL`. Aqueous `AgNO_(3)` was then added to precipitate `Cl^(-)` and `Br^(-)` ions as `AgCl` and `AgBr`. The mass of silver halides obrained was `0.9974 g`. What were the molarities of `HCl` and `HBr` in solution?

Explanation

Let `a M HCl` and `bMHBr` be molarities.
Meq. of `HCl+Meq. of HBr=Meq. of NaOH`
`axx100+bxx100= 0.1235xx47.14=5.82`
`a+b= 5.82xx10^(-2)`…..(i)
Also Meq. of `HCl+` Meq. Of `HBr=`Meq. of halide `AgCl+Meq`. of halide `AgBr`
`=axx100+bxx100`
`:. "Total weight"=(axx100xx143.5)/(1000)+(bxx100xx188)/(1000)`
`=0.9974`
or `143.5a+188b= 9.9974`....(ii)
By eqn. (i) and (ii),
`a=0.036M, b=0.022M`