# Question d9c3a

Jun 7, 2017

#### Answer:

Here's what I got.

#### Explanation:

The idea here is that calcium hydroxide is a strong base, which implies that it dissociates completely in aqueous solution to produce calcium cations and hydroxide anions.

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

Now, keep in mind that calcium hydroxide is not very soluble in water, but the amount that does dissolve dissociates completely.

Calcium hydroxide has a solubility of about ${\text{1.73 g L}}^{- 1}$ at ${20}^{\circ} \text{C}$.

https://en.wikipedia.org/wiki/Calcium_hydroxide

This is equivalent to

1.73color(white)(.) color(red)(cancel(color(black)("g")))/"L" * ("1 mole Ca"("OH")_2)/(74.093color(red)(cancel(color(black)("g")))) = "0.0233 mol L"^(-1)

You can thus say that a saturated solution of calcium hydroxide will be able to dissolve $0.0233$ moles of calcium hydroxide for every $\text{1 L}$ of solution.

This implies that all the calcium hydroxide that you dissolve to make your $\text{0.015-M}$ solution will be completely dissociated.

Now, notice that every $1$ mole of calcium hydroxide that dissociates produces $\textcolor{red}{2}$ moles of hydroxide anions.

This means that your solution will have

["OH"^(-)] = color(red)(2) * ["Ca"("OH")_2]

which, in your case, is equal to

["OH"^(-)] = color(red)(2) * "0.015 M" = "0.030 M"

The $\text{pOH}$ of the solution is defined as

color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))#

Plug in your value to get

$\text{pOH} = - \log \left(0.030\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.52}}}$

Finally, to get the $\text{pH}$ of the solution, use the fact that an aqueous solution at room temperature has

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{pH + pOH = 14}}}}$

You will get

$\text{pH} = 14 - 1.52 = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{12.48}}}$

The answers are rounded to two decimal places, the number of sig figs you have for the concentration of the solution.