1.

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 solution

Answer»

`8 xx 10^(-2)M`
`8 xx 10^(-11)M`
`1.6 xx 10^(-11)M`
`8 xx 10^(-5) M`

SOLUTION :`{:(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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