Question #27b47

1 Answer
Apr 13, 2016

Answer:

#["OH"^(-)] = 1.0 * 10^(-7)"M"#

Explanation:

The thing to remember about aqueous solutions at room temperature is that the concentration of hydronium cations, #"H"_3"O"^(+)#, and the concentration of hydroxide anions, #"OH"^(-)#, have the following relationship

#color(blue)(|bar(ul(color(white)(a/a)["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-14)"M"^2color(white)(a/a)|)))#

This equation is based on the self-ionization of water, which produces equal concentrations of hydronium and hydroxide ions that are equal to water's ionization constant, #K_w#.

At room temperature, you have

#K_W = 10^(-14)"M"^2#

This tells you that an aqueous solution at room temperature will be neutral when the concentrations of the hydronium and hydroxide anions are equal to #10^(-7)#.

This happens because a neutral solution must have equal concentrations of hydronium and hydroxide ions. In the case of pure water, this will get you

#x * x = 10^(-14)"M"^2 implies x= sqrt(10^(-14)"M"^2) = 10^(-7)"M"#

Here #x# represents the concentration of hydronium and hydroxide anions.

In your case, the concentration of hydronium cations is equal to

#["H"_3"O"^(+)] = 1.0 * 10^(-7)"M"#

Right from the start, this tells you that you're dealing with a neutral solution, since this would get you

#["OH"^(-)] = (10^(-14)"M"^color(red)(cancel(color(black)(2))))/(1.0 * 10^(-7)color(red)(cancel(color(black)("M")))) = color(green)(|bar(ul(color(white)(a/a)1.0 * 10^(-7)"M"color(white)(a/a)|)))#