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

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Answer 1 · 2018-05-15T01:01:30

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,

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

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