Sorry that is Incorrect
Here is the solution to the problem, which shows that b is correct.
[H+]= 9.77 x 10-12
-log(9.77 x 10-12)= 11.01 pH= 11.01
14-11.01= 2.99 pOH= 2.99
2nd log(-2.99)= 1.02 x 10-3 [OH-]= 1.02 x 10-3
[H+]= 9.77 x 10-12
-log(9.77 x 10-12)= 11.01 pH= 11.01
14-11.01= 2.99 pOH= 2.99
2nd log(-2.99)= 1.02 x 10-3 [OH-]= 1.02 x 10-3