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:
- the number of moles of HCl used, and
- the number of moles of hydroxide ions present in 25cm^3 of the solution, as well as
- the molar concentration of the hydroxide ions in the saturated solution, and from there,
- 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!
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:
- the number of moles of HCl used, and
- the number of moles of hydroxide ions present in 25cm^3 of the solution, as well as
- the molar concentration of the hydroxide ions in the saturated solution, and from there,
- 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!
1 Answer
The
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
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
"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) = "0.0389 M" -= 2s
["Ca"^(2+)]_(eq) = "0.0195 M" -= s
As a result, the
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