Sorry that is Incorrect
The correct answer is b. Here is why:
[OH-}= 1.58 x 10-13
-log(1.58 x 10-13)= 1.199 pOH= 1.199
14-1.199= 12.801 pH= 12.801
2nd log(-12.801)= 6.324 x 10-2 [H+]= 6.324 x 10-2
[OH-}= 1.58 x 10-13
-log(1.58 x 10-13)= 1.199 pOH= 1.199
14-1.199= 12.801 pH= 12.801
2nd log(-12.801)= 6.324 x 10-2 [H+]= 6.324 x 10-2