Question

Question #c3f72

Answer 1

`"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 `"0.185 g / 100.00 mL H"_2"O"` at `20^@"C"`.

You can use this information to find the mass of calcium hydroxide dissolved in your `"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 `"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

`"Ca"("OH")_ (color(red)(2)(aq)) -> "Ca"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)`

As you can see, for every mole of calcium hydroxide that dissolves in solution, you get `color(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))))`

`= "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, `"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.

`"HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl"_ ((aq))^(-)`

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 `"0.00100 M"`, the volume that will contain that many moles of hydrochloric acid will be

`V_ "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

`color(green)(bar(ul(|color(white)(a/a)color(black)("volume of HCl " = "499 mL ")color(white)(a/a)|)))`