Question c3f72

Oct 7, 2016

$\text{499 mL}$

Explanation:

The problem tells you that slacked lime, which is simply an aqueous solution of calcium hydroxide, "Ca"("OH")_2, has a solubility of $\text{0.185 g / 100.00 mL H"_2"O}$ at ${20}^{\circ} \text{C}$.

You can use this information to find the mass of calcium hydroxide dissolved in your $\text{10.0 mL}$ sample

10.0 color(red)(cancel(color(black)("mL"))) * ("0.185 g Ca"("OH")_2)/(100.00color(red)(cancel(color(black)("mL")))) = "0.0185 g Ca"("OH")_2

So, you know that your sample contains $\text{0.0185 g}$ of dissolved calcium hydroxide. As you know, calcium hydroxide is a strong base, which means that the amount dissolved will dissociate completely to produce calcium cations and hydroxide anions

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

As you can see, for every mole of calcium hydroxide that dissolves in solution, you get $\textcolor{red}{2}$ moles of hydroxide anions.

Use calcium hydroxide's molar mass to determine how many moles you have in your sample

0.0185color(red)(cancel(color(black)("g"))) * ("1 mole Ca"("OH")_2)/(74.093color(red)(cancel(color(black)("g")))) = "0.0002497 moles Ca"("OH")_2

This implies that the sample will contain

0.0002497color(red)(cancel(color(black)("moles Ca"("OH")_2))) * (color(red)(2)color(white)(a)"moles OH"^(-))/(color(red)(cancel(color(black)("mole Ca"("OH")_2))))#

$= {\text{0.0004994 moles OH}}^{-}$

Now, in order to have a complete neutralization, you need to use enough hydrochloric acid solution to ensure that you have equal numbers of moles of hydronium cations, ${\text{H"_3"O}}^{+}$, and hydroxide anions.

In this case, you need your acid solution to contain $0.0004994$ moles of hydrochloric acid. Keep in mind that hydrochloric acid is a strong acid that dissociates completely to form hydronium cations in a $1 : 1$ mole ratio.

${\text{HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

So, every mole of hydronium cations produced in solution comes from $1$ mole of hydrochloric acid. Since the solution is said to have a molarity of $\text{0.00100 M}$, the volume that will contain that many moles of hydrochloric acid will be

${V}_{\text{HCl" = (0.0004994 color(red)(cancel(color(black)("moles"))))/(0.00100 color(red)(cancel(color(black)("mol")))"L"^(-1)) = "0.4994 L}}$

Expressed in milliliters and rounded to three sig figs, the answer will be

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{volume of HCl " = "499 mL }} \textcolor{w h i t e}{\frac{a}{a}} |}}}$