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A buffer solution is prepared in which t...

A buffer solution is prepared in which the concentration of `NH_(3)` is `0.30 M` and the concentration of `NH_(4)^(+)` is `0.20 M`. If the equilibrium constant, `K_(b)` for `NH_(3)` equals `1.8xx10^(-5)`, what is the `pH` of this solution? (`log 2.7=0.43`)

A

`8.73`

B

`9.08`

C

`9.43`

D

`11.72`

Text Solution

Verified by Experts

The correct Answer is:
C
`pOH = pK_(b)+log. (["Salt"])/(["Base"])`
`= 4.74 + log . (0.20)/(0.30) = 4.74 +(0.301 - 0.477)`
`= 4.74 - 0.176 = 4.56`
` :. pH = 14 - 4.56 9.44 `
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