Question

Question #27b47

Answer 1

`["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)|)))`