How do you change pOH to pH?

Remember this:

`\mathbf(color(blue)("pH" + "pOH" = 14))`

I'll tell you how to derive this below.

`"pH"` indirectly tells you the concentration of hydrogen ions (`"H"^(+)`) in solution:

`color(green)("pH" = -log["H"^(+)])`

where `["X"]` is the concentration of `"X"` in `"M"`.

`"pOH"` tells you something similar, but for hydroxide ions instead:

`color(green)("pOH" = -log["OH"^(-)])`

These can be shown to relate if you recall that water slightly ionizes according to the following equilibrium reaction:

`\mathbf("H"_2"O"(l) rightleftharpoons "H"^(+)(aq) + "OH"^(-)(aq))`

For this, we have the equilibrium constant for the autoionization of water

`K_w = ["H"^(+)]["OH"^(-)] = 10^(-14),`

which heavily favors the production of water, since `10^(-14)` is very, very small.

Now, let's try taking the negative (base 10) logarithm of both sides.

`-log(K_w) = -log(["H"^(+)]["OH"^(-)]) = 14`

But `-log(K_w) = "p"K_w = 14`, so what we have, using the additive argument properties of logarithms and our definitions of `"pH"` and `"pOH"` above, is

`\mathbf(color(blue)(14)) = -log(["H"^(+)]["OH"^(-)])`

`= -log["H"^(+)] + (-log["OH"^(-)])`

`= \mathbf(color(blue)("pH" + "pOH"))`

That tells us that:

`color(blue)("pH" = 14 - "pOH")`

`color(blue)("pOH" = 14 - "pH")`