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The molarity of NH(3) of pH = 12 at 25^(...

The molarity of `NH_(3) of pH = 12 at 25^(@)C` is `(K_(b) = 1.8 xx 10^(-5))`

A

`11.7 M`

B

`5.5 M`

C

`0.55M`

D

`0.01M`

Text Solution

Verified by Experts

The correct Answer is:
C
`pK_(b) = 4.7447 ~~ 4.74`
`pH = 12, pOH = 2`
`pOH = (1)/(2) (pK_(b) - log C)`
`2 xx2 = pK_(b) - log C`
`4 = 4.74 - log C`
`:. Log C = 0.74`
`C = "Antilog" (0.74) = 0.55M`
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