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!

1 Answer
Truong-Son N.
Mar 28, 2018

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)#.

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