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When NH(3)(0.1 M) 50 ml mix with HCl (0....

When `NH_(3)(0.1 M) 50 ml` mix with `HCl (0.1 M) 10 ml` then what is `pH` of resultant solution (`pK_(b)=4.75)`

A

`9.25`

B

`10`

C

`9.85`

D

`4.15`

Text Solution

Verified by Experts

The correct Answer is:
C

`underset(rem.)underset()underset(Initial)(NH_(3))+underset(4mmol)underset(5mmol)underset(50xx0.1)(HCl) rarr underset(0)underset(1mmol)underset(10xx0.1)(NH_(4)Cl)`
`pOH=pK_(b)+log``("salt")/(base)`
`=4.75+log``(1)/(4)=4.15`
`pH=14-pOH=14-4.15=9.85`
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