1.

`0.1` mole of `CH_(3)CH (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):}`
(Basic buffer solution)
`[OH^(-)] = K_(b) xx ("Base")/("Salt")`
`= 5 xx 10^(-4) xx (0.02)/(0.08) x= (10^(-14))/(1.25 xx 10^(-4))`
`= 8 xx 10^(-11) M`


Discussion

No Comment Found

Related InterviewSolutions