Question

If the poH of a solution is 10, what is the pH of this solution? Is this solution acidic or basic?

Answer 1

Consider the autoionization reaction of water with itself. This is an equilibrium that is heavily favored towards water, but nevertheless, it occurs.

`2"H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "OH"^(-)(aq)`

Or, this is the same thing:

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

From this, we have the equilibrium constant known as the autoionization constant, `"K"_w`, equal to `10^(-14)`. Thus, we have the following equation (remember to not use a liquid in the expression):

`color(green)("K"_w = ["H"^(+)]["OH"^(-)] = 10^(-14))`

where `["H"^(+)]` is the concentration of hydrogen ion and `["OH"^(-)]` is the concentration of hydroxide polyatomic ion in `"M"`.

Next, let's take the base-10 negative logarithm of this. Recall that `-log("K"_w) = "pK"_w`. We then get:

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

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

Similar to what happened with `-log("K"_w) = "pK"_w`, `-log(["H"^(+)]) = "pH"` and `-log(["OH"^(-)]) = "pOH"`. Thus, we have:

`color(blue)("pK"_w = "pH" + "pOH" = 14)`

So, once you know this equation, what you get is:

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

`= 14 - 10`

`= color(blue)(4)`

Acidity is basically (pun?) when the `"pH"` is less than `7`. We know that `4 < 7`, thus the solution is acidic. The overall spectrum is:

`"pH" < 7` `->` acidic
`"pH" = 7` `->` neutral
`"pH" > 7` `->` basic