Question #98abc

1 Answer
Stefan V.
Dec 11, 2017

#2.6 * 10^(-11)# #"M"#

Explanation:

The trick here is to realize that at #25^@"C"#, an aqueous solution has

#color(blue)(ul(color(black)(["H"_3"O"^(+)] * ["OH"^(-)] = 1.0 * 10^(-14)color(white)(.)"M"^2)))#

This means that for an aqueous solution at #25^@"C"#, you will have

#["OH"^(-)] = (1.0 * 10^(-14)color(white)(.)"M"^2)/(["H"_3"O"^(+)])#

So all you have to do here is to plug in your value and find the concentration of the hydroxide anions.

#["OH"^(-)] = (1.0 * 10^(-14)color(white)(.)"M"^color(red)(cancel(color(black)(2))))/(3.9 * 10^(-4)color(red)(cancel(color(black)("M")))) = 2.6 * 10^(-11)color(white)(.)"M"#

The answer is rounded to two sig figs, the number of sig figs you have for the concentration of hydronium cations, which, in your case, are given as hydrogen ions, #"H"^(+)#.

Keep in mind that you will oftentimes see this equation written without added units.

#["H"_3"O"^(+)] * ["OH"^(-)] = 1.0 * 10^(-14)#

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