Question

If Kb for `NX_3` is `7.5x10^-6` , what is the pOH of a .175 M aqueous solution of NX3?

Answer 1

`sf(2.94)`

Explanation:

`sf(pOH=1/2[pK_b-logb])`

`sf(pK_b=-logK_b=-log(7.5xx10^(-6))=5.12)`

`sf(b=[NX_3]=0.175color(white)(x)"mol/l")`

`:.` `sf(pOH=1/2[5.12-log(0.175)])`

`sf(pOH=2.94)`