Calculate the pOH of a solution obtained when 0.05 mol `NH_(4)Cl` is added and dissolved in 0.025 M ammonia solution. Kb for ammonia is `1.77xx10^(-5)`

Explanation

`pKb=log[Kb]`
`=log[1.77xx10^(-3)]`
`=5-0.2480=4.752`
`pOH=pK_(b)+log=(["salt"])/(["Base"])=4.752+"log"([0.05])/([0.025])`
`=4.752+log2=4.752+0.3010=5.053`