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

We know `pOH =-log K_b+log (["Salt"])/(["Base"])`
`5=4.7+log ""a/b`
`a/b=2 " "therefore a=2b`
Given a+b=0.6
2b+b=0.6
`therefore 3b=0.6`
`or b =0.2" mole or "0.2 xx 17 =3.4 g//L`
`a=0.4"mole or "0.4 xx 53.5=21.4g//L`
Thus [Salt]=0.4 M and [Base]=0.2M