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0.1 mole of `CH_(3)NH_(2)(K_(b) = 5 xx 10^(-4))` is mixed with 0.08 mole of HCl and diluted to one litre. What will be the `H^(+)` concentration in the solutionA. `8 xx 10^(-2)M`B. `8 xx 10^(-11) M`C. `1.6 xx 10^(-11) M`D. `8 xx 10^(-5) M` |
Answer» Correct Answer - B `{:(CH_(3)NH_(2),+,HCl,rarr,CH_(3)NH_(3)^(+)Cl^(-)),(0.1,,0.08,,0),(0.02,,0,,0.08):}` As it is a basic buffer solution. `pOH = pK_(b) + log.(0.08)/(0.02) = - log5 xx 10^(-4) + log 4` `= 3.30 + 0.602 = 3.902` `pH = 14 - 3.92 = 10.09`, `[H^(+)] = 7.99 xx 10^(-11) = 8 xx 10^(-11) M` |
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