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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» 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` |
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