1.

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

Answer»

`9.08`
`9.43`
`11.72`
`8.72`

Solution :`pOH=pK_(a)+(["SALT"])/(["Base"])`
`pK_(b)=-LOG K_(b)=-log(1.8xx10^(-5))=4.74`
`pOH=4.74+log((0.20)/(0.30))=4.56`
`pH=14-4.56=9.44`


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