Question 5d860

Apr 13, 2016

$\text{pH} = 4.6$

Explanation:

The pH of a solution is simply of measure of how many hydronium cations, ${\text{H"_3"O}}^{+}$, it contains.

More specifically, you can find a solution's pH by taking the negative log base $10$ of the concentration of hydronium cations present in solution.

color(blue)(|bar(ul(color(white)(a/a)"pH" = - log( ["H"_3"O"^(+)])color(white)(a/a)|)))

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

["H"_3"O"^(+)] = 2.5 * 10^(-5)"M"

This means that the pH of the solution will be

$\text{pH} = - \log \left(2.5 \cdot {10}^{- 5}\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 4.6 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, is a pH of $4.6$ characteristic of an acidic, a neutral, or a basic solution?

Aqueous solutions at room temperature exhibit the following relationship between the concentration of hydronium cations and hydroxide anions, ${\text{OH}}^{-}$

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

This relationship is based on the self-ionization of water at room temperature, for which the ionization constant, ${K}_{W}$, is equal to

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

Now, a neutral solution will always have equal concentrations of hydronium cations and hydroxide anions. For pure water at room temperature, and taking $x$ to be the concentration of the two ions, you have

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

So, a neutral solution has

["H"_3"O"^(+)] = ["OH"^(-)] = 10^(-7)"M"

In your case, you have a higher concentration of hydronium cations than you would have in a neutral solution. This means that the solution will be acidic, since it contains a higher concentration of hydronium cations than of hydroxide anions.

["OH"^(-)] = (10^(-14)"M"^2)/(["H"_3"O"^(+)])

["OH"^(-)] = (10^(-14)"M"^color(red)(cancel(color(black)(2))))/(2.5 * 10^(-5)color(red)(cancel(color(black)("M")))) = 4.0 * 10^(-10)"M"#
$\left[{\text{H"_3"O"^(+)] > ["OH}}^{-}\right] \to$ the solution will be acidic