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CH(3)NH(2)(0.1 "mole" , K(b)=5xx10^(-4))...

`CH_(3)NH_(2)(0.1 "mole" , K_(b)=5xx10^(-4))` is added to 0.08 mole of HCl and the solution is diluted to one litre, resulting hydrogen ion concentration is:

A

`1.6xx10^(-11)`

B

`8xx10^(-11)`

C

`5xx10^(-5)`

D

`8xx10^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`{:(CH_(3)NH_(2)+,HCI,to,CH_(3)NH_(3)CI,,),(0.1,0.08,,0,,),(0.02,0,,0.08,"Buffer solution",):}`
`pOH=pK+(b)+log[("salt")/(Base)]`
`K_(b)=5xx10^(-4)`
`pK_(b)=4-log_(5)`
`=4-0.7`
=3.3
`=3.3+ "log"(0.08)/(0.02)`
=3.3 + log 4
= 3.3 0.6
= 3.9
pH = 14 - 3.9 = 10.1
`[H^(+)]` = "Anilog" (-10.1)
`=8xx10^(-11)mol//L`
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