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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? |
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Answer» 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` |
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