`pOH = pK_(b) + log. (["Salt"])/(["Base"])=pK_(b) + log. ([NH_(4)Cl])/([NH_(4)OH])`
`[NH_(4)OH]=0.40` mol `L^(-1)`
`[NH_(4)Cl]=0.50 ` mol `L^(-1)`
`pK_(b) = - log K_(b) = - log (1.81 xx 10^(-5))=5-0.2577 = 4.7423`
`:. pOH = 4.742 + log. (0.5)/(0.4) = 4.742+ log 1.25 = 4.742 + 0.0969 = 4.8389 ~= 4.839`
`:. pH = 14-pOH = 14 - 4.839 = 9.161`.