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0.1 mole of `CH_(3)NH_(2) (K_(b) = 5 xx 10^(-4))` is mixed with 0.08 mole for 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):}` (Basic buffer solutions) `[OH^(-)] = K_(b) xx ("Base")/("Salt") = 5 xx 10^(-4) xx (0.02)/(0.08) = 1.25 xx 10^(-4)` `:. [H^(+)] = (10^(-14))/([OH^(-)]) = (10^(-14))/(1.25 xx 10^(-4)) = 8 xx 10^(-11) M`. |
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