Question

Calculating Ksp of calcium hydroxide?

0.05M HCl was used with saturated calcium hydroxide solution, and the average volume of HCl found was 19.45cm^3. We titrated with 25cm^3 of calcium hydroxide and two drops of screened methyl orange in the conical flask.

In the calculations, I need to find:

1. the number of moles of HCl used, and
2. the number of moles of hydroxide ions present in 25cm^3 of the solution, as well as
3. the molar concentration of the hydroxide ions in the saturated solution, and from there,
4. calculate Ksp.

I have all the data presented, and all I'd like is a little nudge in the correct direction- I don't quite understand how to calculate the number of moles of HCl used, or the third question... Please help me!

Answer 1

The `"mols"` of `"HCl"` just comes from the volume and concentration you have; they react `2:1` with `"Ca"("OH")_2`, as in

`2"HCl"(aq) + "Ca"("OH")_2(aq) -> 2"H"_2"O"(l) + "CaCl"_2(aq)`

And so,

`"0.05 mols HCl"/cancel"L" xx 19.45 cancel("cm"^3) xx cancel"1 L"/(1000 cancel("cm"^3))`

`=` `"0.0009725 mols HCl"` or `"H"^(+)`

There are two `"HCl"` required for every one `"Ca"("OH")_2`, so neutralizing this completely would mean that we neutralize as much `"OH"^(-)` as we have `"H"^(+)`:

`0.0009725 cancel"mols HCl" xx cancel("1 mol Ca"("OH")_2)/(2 cancel"mols HCl") xx ("2 mols OH"^(-))/cancel("1 mol Ca"("OH")_2)`

`= ul"0.0009725 mols OH"^(-)` in `"25 cm"^3` saturated solution

But the mols scale with system size. So, to get to the molarity, we just get to the mols in `"1 L"` from the mols in `"25 mL"`.

`"0.0009725 mols OH"^(-)/("25 mL") xx (1000/25)/(1000/25) = "0.0389 mols OH"^(-)/"1000 mL"`

`=` `ul"0.0389 M"`

And from there, calcium hydroxide dissociates as:

`"Ca"("OH")_2(s) rightleftharpoons "Ca"^(2+)(aq) + 2"OH"^(-)(aq)`

Therefore, `["OH"^(-)]_(eq) = 2["Ca"^(2+)]_(eq)`, and

`["OH"^(-)]_(eq) = "0.0389 M" -= 2s`

`["Ca"^(2+)]_(eq) = "0.0195 M" -= s`

As a result, the `K_(sp)` is:

`color(blue)(K_(sp)) = ["Ca"^(2+)]["OH"^(-)]^2 = s(2s)^2 = 4s^3`

`= 4(0.0195)^3 = color(blue)(2.94 xx 10^(-5))`

The actual `K_(sp)` is about `5.5 xx 10^(-6)`.