# Question cf57c

Mar 14, 2016

$\text{pOH} = 13.18$

#### Explanation:

The pOH of a solution is simply a measure of the concentration of hydroxide anions, ${\text{OH}}^{-}$.

More specifically, the pOH of a solution is calculated using the equation

color(blue)(|bar(ul(color(white)(a/a)"pOH" = - log(["OH"^(-)])color(white)(a/a)|)))

Here $\left[{\text{OH}}^{-}\right]$ represents the concentration of the hydroxide anions in solution.

So, you know that you're dealing with a solution of hydrochloric acid, $\text{HCl}$. As you know, hydrochloric acid is a strong acid, which means that it dissociates completely in aqueous solution to form hydronium cations, ${\text{H"_3"O}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$

color(red)("H")"Cl"_text((aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(color(red)(+)) + "Cl"_text((aq])^(-)

As you can see, every mole of hydrochloric acid that dissociates produces one mole of hydronium cations. This means that your $\text{0.15-M}$ solution of hydrochloric acid will contain

["H"_3"O"^(+)] = "0.15 M"

Now, to find the concentration of hydroxide anions, use the fact that at room temperature, the following relationship is valid in aqueous solutions

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

This equation, which comes from the self-ionization of water at room temperature, will allow you to find the concentration of hydroxide anions if you know the concentration of hydronium cations, and vice versa.

["OH"^(-)] = (10^(-14)"M"^color(red)(cancel(color(black)(2))))/(0.15color(red)(cancel(color(black)("M")))) = 6.67 * 10^(-14)"M"

This means that the pOH of the solution will be

$\text{pOH} = - \log \left(6.67 \cdot {10}^{- 14}\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 13.18 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

ALTERNATIVE APPROACH

Another approach to use here is to find a relationship between pH and pOH. Take the log base 10 from both sides of equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$ to get

["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-14) " "| log

$\log \left(\left[{\text{H"_3"O"^(+)] * ["OH}}^{-}\right]\right) = \log \left({10}^{- 14}\right)$

This will be equivalent to

$\log \left(\left[{\text{H"_3"O"^(+)]) + log(["OH}}^{-}\right]\right) = - 14 \cdot {\overbrace{\log \left(10\right)}}^{\textcolor{p u r p \le}{= 1}}$

$- \log \left(\left[{\text{H"_3"O"^(+)]) - log(["OH}}^{-}\right]\right) = 14$

But you know that

color(purple)(|bar(ul(color(white)(a/a)color(black)("pH" = - log(["H"_3"O"^(+)]))color(white)(a/a)|)))" " and " "color(purple)(|bar(ul(color(white)(a/a)color(black)("pOH" = - log(["OH"^(-)]))color(white)(a/a)|)))#

which means that you can write

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{pH" + "pOH} = 14 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now all you have to do is use the known concentration of hydronium cations to get the pH of the solution

$\text{pH} = - \log \left(0.15\right) = 0.82$

The pOH of the solution will once again be

$\text{pOH" = 14 - "pH}$

$\text{pOH} = 14 - 0.82 = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 13.18 \textcolor{w h i t e}{\frac{a}{a}} |}}}$