Question

Question

Molar Solubility and solubility Product of Calcium Hydroxide:

1. Volume of Saturated Ca(OH)2 solution (mL) = 25.0 mL

2. Concentration of Standardized HCL solution (mol/L) = 0.05 mol/L

3. Buret reading, initial (mL) = 50.0 mL

4. Buret reading, final (mL) = 32.5 mL

5. Volume of HCL added (mL) = 17.5 mL

6. Moles of HCL added (mol) = (concentration of HCl)(Volume of HCl) = (0.05)(19.1) = 0.875 mol.

7.Moles of OH- in saturated solution (mol)=8.75x10^-4

8.the [OH-] equilibrium (mol/L)=0.0205M

10.Molar solubility of Ca(OH)2 (mol/L)=9.01x10^-3

Can someone please help me with questions 9-15 please!

9.[Ca2+] equilibrium (mol/L),

11.Average molar solubility of Ca(OH)2 (mol/L),

12.Ksp of Ca(OH)2,  

13.Average Ksp

14.Satandard deviation of Ksp, and

15.The Relative Standard deviation of Ksp (%RSD)

Answer 1

9) Chemical equation:

Ca(OH)₂ <===> Ca²⁺ + 2OH¯

[Ca²⁺] = (1/2)[OH^-] = 1/2(0.0205M) = 0.01025 M

11) The Average molar solubility of Ca(OH)2 (mol/L) equals to 9) = 0.01025 mol/L

12) K_sp(ca(oh)2) = [Ca²⁺] [OH¯]² = (0.01025)(0.0205)² = 4.3E^-6

I am assuming that you have a second trial for Ca(OH)2

13) Average Ksp = (4.3E^-6 + 4E^-6) / 2 = 8.3E^-6 / 2 = 4.15E^-6

14) Standard deviation Ksp = (4.3E^-6 - 4.15E^-6)² + (4E^-6 - 4.15E^-6)² / 2-1 = 4.5E^-14/1 = 2.12E^-7

15) % Relative Standard deviation Ksp = 100 * ksp(sd) / ksp(avrg) = 100 (2.12E^-7) / (4.15E^-6) = 5.11 %