Calculate the `pH` of `500mL` Solution of `1 M BOH (K_(b)=2.5xx10^(-5))`

Explanation

`{:(,BOH,hArr,B^(+),+,OH^(-)),(t=0,1M,,0,,0),(t=eq,[1-alpha],,alpha,,alpha):}`
`:.(alpha^(2))/(1-alpha)=K_(b)` expecting `a ltlt1 :.alpha^(2)=2.5xx10^(-5)rArr alpha=5xx10^(-3)`
`:.[OH^(-)]=5xx10^(-3)MrArr pOH=3-log5rArr pOH=2.3rArr pH=11.7`
Note:
(i) Volume of the solution given in questionhas no significance.
(ii) After getting `alpha` value as negligible ,direct relation :`pOH=(1)/(2)(pK_(b)-logC)` could also be used .