A 40 mL solution of weak base BOH is tritrated with `0.1N HCI` solution. The pH of solution is found to be `10.04` and `9.14` after the addition of `5.0mL` and `20.0` mL of acid respecctively. Find out `K_(b)` for weak base.

Explanation

Case I:
`{:(,BOH+,HCI,rarrBCI+,H_(2)O),("Millimole before reaction", a,0.1xx5=0.5,0,0),("Millimole after reaction",(a-0.5),0,0.5,0.5):}`
`{:( :.pH=10.04),( :. pOH=3.96):}:|{:( :. pOH=-log((K_(b)+log([BC]))/([BOH]))....(1)),( :. 3.96=-logK_9b+log(0.5/((a-0.05)))....(2)):}` Case II:
`{:(,BOH+,HCI,rarrBCI+,H_(2)O),("Millimole before reaction", alpha,0.1xx20=2,0,0),("Milliomole after reaction", (a-2),0,2,2):}`
`{:( :.pH=9.14),( :. pOH=4.86):}:|{:( :. pOH=-logK_(b)+log(([BC])/([BOH]))....(3)),( :. 4.86=-logK_b+log(2/((a-2)))....(4)):}`
Solving eqs. (2) and (4), `K_(b)= 1.81xx10^(-5)`