Sol . As Ka of o - cresol , HC7H7O = 6.3 × 10-11
So , Kb of C7H7O-
= Kw / Ka = 10-14 / ( 6.3 × 10-11 )
= 0.1587 × 10-3
Also , pH = 11.45
So , pOH = 14 - pH = 14 - 11.45 = 2.55
Now , pOH = -1/2 log(Kb × c)
where c = initial concentration of C7H7O-
So , 2.55 = - 1/2 log ( 0.1587 × 10-3 × c )
log( 0.1587 × 10-3 × c ) = - 2.55 × 2 = - 5.10
0.1587 × 10-3 × c = 10-5.10 = 7.9432 × 10-6
c = ( 7.9432 × 10-6 ) / ( 0.1587 × 10-3 )
= 0.0500 M
Therefore , answer is 0.0500 M
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