Question

Question #cf57c

Answer 1

`"pOH" = 13.18`

Explanation:

The pOH of a solution is simply a measure of the concentration of hydroxide anions, `"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 `["OH"^(-)]` represents the concentration of the hydroxide anions in solution.

So, you know that you're dealing with a solution of hydrochloric acid, `"HCl"`. As you know, hydrochloric acid is a strong acid, which means that it dissociates completely in aqueous solution to form hydronium cations, `"H"_3"O"^(+)`, and chloride anions, `"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 `"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.

In your case, you have

`["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

`"pOH" = - log(6.67 * 10^(-14)) = color(green)(|bar(ul(color(white)(a/a)13.18color(white)(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 `color(orange)("(*)")` to get

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

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

This will be equivalent to

`log(["H"_3"O"^(+)]) + log(["OH"^(-)]) = -14 * overbrace(log(10))^(color(purple)(=1))`

`-log(["H"_3"O"^(+)]) - log(["OH"^(-)]) = 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

`color(blue)(|bar(ul(color(white)(a/a)"pH" + "pOH" = 14color(white)(a/a)|)))`

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

`"pH" = - log(0.15) = 0.82`

The pOH of the solution will once again be

`"pOH" = 14 - "pH"`

`"pOH" = 14 - 0.82 = color(green)(|bar(ul(color(white)(a/a)13.18color(white)(a/a)|)))`