In lab, we titrated .0525M HCl into 25ml saturated Ca(OH)2 with added CaCl2 solu

Question

In lab, we titrated .0525M HCl into 25ml saturated Ca(OH)2 with added CaCl2 solution. The volume of HCl added to reach equilibrium was 17.4ml. We have to find:

mol HCl added: .0525M*17.4ml/1000ml = .000914mol HCl

mol OH- in saturated solution: I believe this is equal to mol HCl? but I am not sure.

[OH-] at equilibrium: this depends on the above.

Molar solubility of Ca(OH2) with added CaCl2: the biggest problem I have. I believe it has to do with ICE tables, but again am not sure.

Please help. I was able to complete the first part of lab without the added CaCl2, but I don't know how to account for the added CaCl2 now in this part. Thank you.

Explanation / Answer

mol HCl added: 0.0525M*17.4ml/1000ml = 0.000914mol HCl

[HCl]=0.00091moles/0.0424L=0.0215M

mol OH- in saturated solution:

Ksp{Ca(OH)2}=Ca2+ + 2OH-

Ksp{Ca(OH)2}=x*(2x)^2

x=(Ksp/4)^1/3

this gives saturated conc of (OH-)

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