Question #98abc
1 Answer
Explanation:
The trick here is to realize that at
#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
#["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,
Keep in mind that you will oftentimes see this equation written without added units.
#["H"_3"O"^(+)] * ["OH"^(-)] = 1.0 * 10^(-14)#