The correct option is B 7.21
Here very dilute solution of Ca(OH)2 is given, where water dissociation needs to be considered to calculate the total [OH−] ions present in the solution.
[OH−] from Ca(OH)2=2×5×10−8 M=10−7 M.
H2O (l)⇌H+ (aq)+OH− (aq)
From H2O ,
[OH−]=[H+] = x
We know,
[H+][OH−]=10−14x(10−7+x)=10−14x2+10−7x−10−14=0x=−10−7±√10−14+4×10−142x=6.18×10−8
So, [H+]=6.18×10−8 MpH=−log[H+]pH=−log(6.18×10−8)=(8−0.79)pH=7.21