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)`