# Question 4b7bf

Feb 14, 2016

Here's what I got.

#### Explanation:

Your starting point here is to use the percent yield of the reaction to determine how many moles of calcium hydroxide, "Ca"("OH")_2, were produced by the reaction.

Once you know how many moles of calcium hydroxide you have, you can use the volume of the solution to determine the molarity of calcium hydroxide.

Now, calcium hydroxide is not very soluble in aqueous solution, which means that in order to find the molarity of the hydroxide anions, you need to know the solubility product constant, ${K}_{s p}$, of the compound, which is listed as

${K}_{s p} = 5.5 \cdot {10}^{- 6}$

http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm

So, according to the balanced chemical equation

"CaC"_2 + 5/2 "O"_2 + "H"_2"O" -> 2 "CO"_2 + "Ca"("OH")_2

you have a $1 : 1$ mole ratio between calcium carbide, ${\text{CaC}}_{2}$, and calcium hydroxide.

This tells you that the theoretical yield, which is what you get for a reaction that has a 100% yield, is one mole of calcium hydroxide for every one mole of calcium carbide.

However, you know that the reaction has a percent yield equal to 37%. This means that for every $100$ moles of calcium carbide that react, only $37$ are converted to moles of calcium hydroxide.

This means that instead of $0.468$ moles of calcium hydroxide, which you would get for a 100% yield, your reaction will actually produce

0.468 color(red)(cancel(color(black)("moles CaC"_2))) * ("37 moles Ca"("OH")_2)/(100color(red)(cancel(color(black)("moles CaC"_2)))) = "0.1732 moles Ca"("OH")_2

Molarity is defined as number of moles per liter of solution, which means that the molarity of calcium hydroxide will be

["Ca"("OH")_2] = "0.1732 moles"/"5 L" = "0.03464 M"

Now, calcium hydroxide will partially dissociate to form calcium cations, ${\text{Ca}}^{2 +}$, and hydroxide anions, ${\text{OH}}^{-}$

${\text{Ca"("OH")_2 rightleftharpoons "Ca"^(2+) + color(red)(2)"OH}}^{-}$

Calculate the molar solubility of calcium hydroxide, $s$, at room temperature by using the ${K}_{s p}$.

${K}_{s p} = {\left[{\text{Ca"^(2+)] * ["OH}}^{-}\right]}^{\textcolor{red}{2}}$

If $s$ represents the concentration of calcium cations, you can say that

${K}_{s p} = s \cdot {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}} = 4 {s}^{3}$

This will give you

$s = \sqrt[3]{{K}_{s p} / 4} = \sqrt[3]{\frac{5.5 \cdot {10}^{- 6}}{4}} = \text{0.0111 M}$

This means that a saturated calcium hydroxide solution will contain

["Ca"^(2+)] = "0.0111 M"

["OH"^(-)] = color(red)(2) xx "0.0111 M" = "0.0222 M"#

In your case, the concentration of calcium hydroxide exceeds the compound's molar solubility. The excess calcium hydroxide will actually precipitate out of solution, leaving you with a concentration of hydroxide anions equal to

$\left[\text{OH"^(-)] = color(green)("0.022 M}\right)$

I'll leave the answer rounded to two sig figs.