Question

Calculate the `["H"_3"O"^(+)]` and `["OH"^(-)]` for 0.00024 M NaOH. Answer in units of M?

Answer 1

Begin by recognizing that `"NaOH"` is a strong base, which dissociates completely in water.

`"NaOH"(aq) -> "Na"^(+)(aq) + "OH"^(-)(aq)`

Therefore, given `"0.00024 M NaOH"`, since `["OH"^(-)] = ["NaOH"]`, what is the concentration of `"OH"^(-)`?

At `25^@ "C"`, for the autodissociation of water,

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

`K_w = ["H"_3"O"^(+)]["OH"^(-)] = 10^(-14)`

So,

`["H"_3"O"^(+)] = K_w/(["OH"^(-)]) = ???`