For a basic buffer, `pOH = pK_(b)+log.(["Salt"])/(["Base"])`
Further, `pH + pOH = 14 ` so that `pOH = 14 - pH = 14 - 9 = 5`
`pK_(b)=-log K_(b) = - log(1.8xx10^(-5))=4.7447`
[Base] `=[NH_(4)OH]= 1 "mol" L^(-1)`
`:. 5 = 4. 7447 + log. (["Salt"])/(1) or log ["Salt"] = 0.2553 or ["Salt"]= 1.8 "mol" L^(-1)`