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.

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.

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

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

You are right.

To solve these calculations, we need to consider the neutralization reaction between NaOH and HCl. Given the volume of HCl used in the titration (20 mL) and its concentration (.1201 mol/L), and the average titre (4.71), we can calculate the number of moles of HCl used.

Moles HCl = (Volume HCl) * (Concentration HCl) = (20 mL) * (.1201 mol/L) = 2.402 * 10^(-3) mol

Since HCl and NaOH react in a 1:1 ratio, the moles of NaOH present in the solution are also 2.402 * 10^(-3) mol.

Now let's proceed with the calculations:

1. To find the total [OH-] in the saturated solution of Ca(OH)2 in NaOH, we need to add the [OH-] contributed by NaOH and Ca(OH)2. [OH-] contributed by NaOH can be calculated by multiplying the moles of NaOH by its concentration.

[OH-] (NaOH) = (moles NaOH) * (Concentration NaOH)
= (2.402 * 10^(-3) mol) * (0.0125 mol/L)
= 3.003 * 10^(-5) mol/L

2. [OH-] contributed solely by dissolved Ca(OH)2 can be calculated using the stoichiometry of the reaction between Ca(OH)2 and OH- ions. The balanced equation is:

Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH-(aq)

Since 1 mole of Ca(OH)2 dissociates to produce 2 moles of OH-, the [OH-] contributed by Ca(OH)2 equals half of the moles of Ca(OH)2.

[OH-] (Ca(OH)2) = 0.5 * (moles Ca(OH)2)

3. To find [Ca2+] in the saturated solution, we need to know the concentration of Ca2+ ions. According to the stoichiometry of the reaction, 1 mole of Ca(OH)2 produces 1 mole of Ca2+ ions. Thus,

[Ca2+] = (moles Ca(OH)2) / (Volume of solution used per titration)

4. The solubility product constant (Ksp) for Ca(OH)2 can be calculated by using the concentration values of Ca2+ and OH- ions in the solution. The Ksp expression is:

Ksp = [Ca2+] * [OH-]^2

5. The solubility of Ca(OH)2 is the amount of Ca(OH)2 that can dissolve in a given volume of NaOH solution. It is determined by comparing the moles of Ca(OH)2 dissolved with the volume of the NaOH solution used.

To calculate the solubility of Ca(OH)2, we need to know the volume of the NaOH solution used.

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?