Question

How many mole of `NH_(4)CI` must be added to one liltre of `1.0M NH_(4)OH` to have a buffer of `pH=9`? `(K_(NH_(4)OH)=1.8xx10^(-5))`

Answer 1

For basic buffer solutions
`pOH= -log K_(b)+log ((["Salt"])/(["Base"]))`
Let a mole of `NH_(4)CI` are added to `(1.0xx1)` mole of `NH_(4)OH` 1 litre.
Given `pH= 9 , :. pOH =5`
`:. 5= -log 1.8xx10^(-5)+log((a)/(1.0xx1))`
or `a= 1.8`