What is the pH of 10^-8M of KOH ?

I got #7.02#. If you get #"pH" = 6#, it won't make physical sense as you added base...

This is a problematic question... Since you have #"KOH"#, and not a known #["OH"^(-)]# AFTER having waited for equilibrium to re-establish, there is a non-negligible concentration of #"OH"^(-)# compared to #"KOH"#.

#"H"_2"O"(l) rightleftharpoons "H"^(+)(aq) + "OH"^(-)(aq)#

#"I"" "-" "" "" "0" "" "" "" "10^(-8)#
#"C"" "-" "" "+x" "" "" "+x#
#"E"" "-" "" "" "x" "" "" "10^(-8)+x#

Now, we write #K_w# at #25^@ "C"#:

#K_w = ["H"^(+)]["OH"^(-)]#

#10^(-14) = x(10^(-8) + x)#

As we said, #10^(-8)# is on the order of usual autoionization concentrations, and as such is nonnegligible...

#10^(-14) = 10^(-8)x + x^2#

#x^2 + 10^(-8)x - 10^(-14) = 0#

The exact solution is then

#x = -(10^(-8))/(2(1)) pm sqrt((10^(-8))^2 - 4(1)(-10^(-14)))/(2(1))#

#= -5 xx 10^(-9) pm sqrt(4.01 xx 10^(-16))/2#

#= -5 xx 10^(-9) pm 1.001 xx 10^(-7)#

The only physical answer is #9.51 xx 10^(-8) "M"#. This now gives:

#["H"^(+)] = 9.51 xx 10^(-8) "M"#
#["OH"^(-)] = 1.05 xx 10^(-7) "M"#

You can convince yourself that #K_w# still holds. And so, the #"pH"# is barely basic at #25^@ "C"#:

#color(blue)("pH") = -log["H"^(+)] = color(blue)(7.02)#