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`0.1` mole of `CH_(3)NH_(2) (K_(b)=5xx10^(-4))` is mixed with `0.08` mole of `HCl` and diluted to one litre. The `[H^(+)]` in solution isA. `8xx 10^(-2)`B. `2xx10^(11)`C. `1.23 xx 10^(-4)`D. `8xx 10^(-11)` |
Answer» Correct Answer - D `{:(CH_(3)NH_(2),+,HCIto ,CH_(3)NH, +CI^(-)),(0.1"mol",,0.08"mol",,0),(0.02"mol",,0,,0.08):}` `pOh =pK_(b) + log .("salt")/("base")` `=- log 5 xx 10^(-4) + log .(0.08)/(0.02)` `=(4 -log 5) +log 4` `=4- 0.6989 + 0.6020 =3.903` `pOH =3.903, pH =10.0967` `[H^(+)] = 8xx 10^(-11)` |
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