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"^(-)]) = ???#