# Question 0209f

##### 1 Answer
Apr 14, 2016

Here's what I got.

#### Explanation:

This one is a bit tricky because, as it turns out, you can't really dissolve enough calcium hydroxide, "Ca"("OH")_2, in water at room temperature to get a $\text{0.050-M}$ solution.

Calcium hydroxide is only sparingly soluble in water; at room temperature, its solubility is equal to ${\text{1.73 g L}}^{- 1}$ at ${20}^{\circ} \text{C}$. To get its molar solubility at this temperature, use its molar mass

1.73 color(red)(cancel(color(black)("g")))/"L" * ("1 mole Ca"("OH")_2)/(74.1 color(red)(cancel(color(black)("g")))) = "0.02335 mol L"^(-1)

This means that you can only hope to dissolve $0.02335$ moles of calcium hydroxide per liter of water before the solution becomes saturated.

Any additional calcium hydroxide added to the solution will remain undissolved.

Now, calcium hydroxide is considered a strong base, which means that the amount of calcium hydroxide that does dissolve in water dissociates completely to form calcium cations, ${\text{Ca}}^{2 +}$, and hydroxide anions, ${\text{OH}}^{-}$.

${\text{Ca"("OH")_ (color(red)(2)(aq)) -> "Ca"_ ((aq))^(2+) + color(red)(2)"OH}}_{\left(a q\right)}^{-}$

Notice that every mole of calcium hydroxide that dissolved in water will produce $\textcolor{red}{2}$ moles of hydroxide anions. This means that for a saturated solution of calcium hydroxide, you will have

${\left[\text{OH"^(-)] = color(red)(2) xx "molar solubility of Ca"("OH}\right)}_{2}$

["OH"^(-)] = color(red)(2) xx "0.02335 mol L"^(-1) = "0.0467 mol L"^(-1)

As you know, you can calculate the pOH of the solution by using

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

In this case, you would have

$\text{pOH} = - \log \left(0.0467\right) = 1.33$

As you know, an aqueous solution at room temperature is characterized by the following relationship

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

This means that the pH of the solution will be

$\text{pH} = 14 - 1.33 = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 12.66 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

It's worth noting that this is the highest pH value you can get for an aqueous solution of calcium hydroxide, since it corresponds to the molar solubility of the salt at room temperature.

ALTERNATIVE SOLUTION - CHEMICALLY INACCURATE

There's always the chance that the problem wants you to ignore calcium hydroxide's molar solubility at room temperature, and simply use the dissociation of the salt to find the concentration of hydroxide anions.

In this case, you'd have

["OH"^(-)] = color(red)(2) xx "0.050 M" = "0.10 M"#

The pOH of the solution would be

$\text{pOH} = - \log \left(0.10\right) = 1$

The pH of the solution would be

$\text{pH} = 14 - 1 = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 13 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Keep in mind that this is not a chemically realistic solution!