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Calculate pH of 0.002 N NH(4)OH having 2...

Calculate `pH` of `0.002 N NH_(4)OH` having `2%` dissociation

A

7.6

B

8.6

C

9.6

D

10.6

Text Solution

Verified by Experts

The correct Answer is:
C
`NH_4OH` is a weak base and partially dissociated `NH_(4)OH Leftrightarrow NH_(4)^(+)+OH^(-)`
`{:(,"Concentration",1,0,0),(,"before dissociation",1-alpha,alpha,alpha),(,"after dissociation",,,):}`
`therefore [OH^(-)]=C alpha=2 xx 10^(-3) xx 2/(100)=4 xx 10^(-5) M`
`pOH=-log[OH^(-)]`
`=-log 4 xx 10^(-5)=4.4`
pH=14-4.4
=9.6
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