Molar Solubility Lab I am unsure of how to do the calculations Moles at stoichio

Question

Molar Solubility Lab

I am unsure of how to do the calculations

Moles at stoichiometric point for HCl is .1085

1085 m 25.0 ML Ca on Experiment 7 Report 2 t 20H Ca Molar Solubility, common-lon E Date Lab Sec. Name badeu V13 Desk No. A. Molar Solubility and Solubility Product of Calcium Hydroxide Trial 2 Trial 1 i. Volume of saturated Ca(OH) solution (ml) 25.0 25.0 50ML 085 m 3. Buret reading, initial (mL) 4. Buret reading, final (mL) 5. Volume of HCl added (mL) 6. Moles of HCI added mob 30 60qu .30 Mol 3 MDA 7. Moles of OH in saturated solution (mol) 8. [OH 1, equilibrium (molL 9. [Ca equilibrium (moll) 2+ 10. Molar solubility of Ca(OH)2 (molL) 12. Ksp of Ca (OH02 13. Average K 14. Standard deviation of Kp 15. Relative standard deviation of Ksp (%RSD) *Calculations for Trial 1.

Explanation / Answer

Trial 1

First,

moles of HCl = M*V

Vused = Vfinal - Vinitial = 11.30-0 = 11.30 mL used

M = 0.1085 M fo HCl

mol HCL = MV = 11.30*0.1085 = 1.22605 mmol of HCl

then

ratio is 1:1

mmol of OH- = 1.22605 mmol

[OH-] = mmol/Ml = 1.22605/25 = 0.049042 M

[Ca+2] = 1/2*[OH-] = 1/2*0.049042 = 0.024521 M

solubility

[Ca(OH)2] = [Ca+2] = 0.024521

assume this is the avg. solubility

so

Ksp = [Ca+2][OH-]^2

Ksp = S*(2S)^2

Ksp = (4S^3) = 4*(0.024521)^3

Ksp = 5.89758*10^-5

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