We are given a solution of NaOH (a strong base) that is 0.0065 M. Because NaOH dissociates into Na+ and OH-, the [OH-] will also be 0.0065 M
If we recall that pH + pOH = 14, then if we find pOH, we can easily find pH. The pOH will be the negative log of the [OH-], so that = -log 0.0065 = 2.187
pH = 14 - pOH = 14 - 2.187 = 11.8
Another way to do this same problem is to first find the [H+] and then take the negative log to get pH. If we recall Kw = 1x10-14 = [H+][OH-] we can solve for [H+]....
[H+][OH-] = 1x10-14
[H+] = 1x10-14 / 0.0065 = 1.538x10-12
pH = -log 1.538x10-12
pH = 11.8