# Question 27b47

Apr 13, 2016

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

#### Explanation:

The thing to remember about aqueous solutions at room temperature is that the concentration of hydronium cations, ${\text{H"_3"O}}^{+}$, and the concentration of hydroxide anions, ${\text{OH}}^{-}$, 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

${K}_{W} = {10}^{- 14} {\text{M}}^{2}$

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

$x \cdot x = {10}^{- 14} \text{M"^2 implies x= sqrt(10^(-14)"M"^2) = 10^(-7)"M}$

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)|)))#