Calculate the amount of `NH_(3)` and `NH_(4)CI` required to prepare a buffer solution of pH `9.0` when total concentration of buffering reagents is `0.6 mol L^(-1)`. `(pK_(b)for NH_(3)=4.7,log 2=0.30)`

Explanation

`pH = 9.0 `. Hence, `pOH = 14-9 = 5 `
`pOH = pK_(b) + log. (["Salt"])/(["Base"])`
`5=4.7+ log. (["Salt"])/(["Base"]) or log. (["Salt"])/(["Base"]) = 0.3 or (["Salt"])/(["Base"]) = Antilog 0.3 = 2, i.e., ["Salt"]=2xx["Base"]`
Also. We are given : [Salt]+ [Base] = 0.6 mol `L^(-1)`
This on solving gives [Base] = 0.2 mol `L^(-1)` and [Salt]=0.4 mol `L^(-1)`