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The pH of 0.02MNH(4)Cl solution will be...

The pH of `0.02MNH_(4)Cl` solution will be : [Given `K_(b) (NH_(4)OH) = 10^(-5) and log 2 = 0.301`]

A

`4.65`

B

`5.35`

C

`2.65`

D

`4.35`

Text Solution

Verified by Experts

The correct Answer is:
B
pH = ? `Kb = 10^(-9) " "rArr " "p^(kb) = 5 rArr 10g 2 = 0.301`
`0.02 MNH_(4)Cl`

`pH = 7 - (1)/(2) (p^(kb) + log C) = 7 - (1)/(2) (5 + log 2 xx 10^(-2)) = 7 - (1)/(2) (3.301) = 7-1.6505`
pH = 5.3495
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