I'm trying to find the solubility of Ca(OH)2 in .0125 mol/L aqueous NaOH. Here is my data:

Vol of solution used per titration: 20mL
Conc of HCl: .1201 mol/L
Average Titre: 4.71

I need to calculte:
1. the TOTAL [OH-] in the saturated solution of Ca(OH)2 in NaOH
2. the [OH-] due to dissolved Ca(OH)2
3. [Ca2+] in the saturated solution
4. the Ksp for Ca(OH)2 for the saturated solution of Ca(OH)2 in NaOH at 21.2 degrees C
5. determine the solubility of Ca(OH)2 in the NaOH solution

I'm just not understanding how to do the calculations with NaOH included.

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  1. 1. The TOTAL OH should be determined from moles HCl used in the titration. The HCl can't tell the difference between NaOH OH and Ca(OH)2 OH and you obtain total OH with this titration.
    2. You know how much OH is there due to NaOH from its concn and volume added.
    3. Thus, the Ca^+2 is the difference between the two with an adjustment made for Ca being 1/2 of OH of Ca(OH)2.
    4.Ksp for Ca(OH)2 is then (Ca^+2)(OH^-)^2
    5. The solubility of Ca(OH)2 in NaOH is just the (Ca^+2).
    Check my thinking.

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  2. Ok, so for the first one I got mol of HCl=4.92x10^-4 and so [OH-] is 4.92x10^-4/.020L = .0246 mol/L. Is that correct?

    And then from there I have the [OH-] in NaOH to be (.0125)(.020L) = 2.5x10^-4

    [OH-] in Ca(OH)2 is .0246-2.5x10^-4 = .02435

    [Ca2+] then is .012175

    Are my calculations correct?

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  3. your calculations are correct.........

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  4. why cant we titrate ca(oh)2 with hcl insted of naoh?

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