Experiment 22Report she eet Molar solubility. Common-lon Effect uDIhry and solub

Question

Experiment 22Report she eet Molar solubility. Common-lon Effect uDIhry and solubility Product of Calcium Hydroxide Trial 3 Trial 2 Trial 1 1. volume of saturated CaoH)2 solution (ml) 25.0 0,0003 M 2. Concentration of standardized HCI solution (mol/L) Buret reading, initial (mL) 222 mL 4 mL 40.2 mL Buret reading final (ml) 5. volume of HCI added (ml) eoollz mol .00008 mol .00102mo 6. Moles of HCI added (mol 00098 mol 00102 mul 0112 mo 7. Moles of OH in saturated solution (mo) 8. [OH equilibrium (mova) 9. [Ca2 equilibrium (mouL) 10. Molar solubility of Ca(OH) (moUL) 11. Average molar solubility of Ca(OH, (moll) 12. Kn of Ca(OH, 13. Average 14. Standard deviation of K Appendix B 15. Relative standard deviation of K (%RSD) Appendix B "Calculations for Trial l 8) .0012mol mL ,0222 LI M XL 3MX ,0212 L

Explanation / Answer

Question 6.

mol of HCl = Macid*Vacid = 0.0503*16.9 = 0.85007 mmol = 0.85007*10^-3 mol

Question 7.

moles of OH- related to H+

ratio is 1:1 so

0.85007 = 0.85007 mmol of OH-

mol of OH- = 0.85007*10^-3 mol of OH-

question 8

[OH-] in equilibrium = mmol of Oh- / volume in mL of base = 0.85007 / 25 = 0.0340 M

Question 9.

Molar solubility of Ca(OH)2

relate:

S = 1/2*[OH-] = 1/2*0.0340 = 0.017 M of Ca(OH)2

S =  0.017 M of Ca(OH)2

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