1.

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 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):}`
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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