I need help filling out the rest of this chart. Please show me the calculations

Question

I need help filling out the rest of this chart. Please show me the calculations used for each step.

Experiment 22 Report Sheet Molar solubility. Common-lon Effect Date Lab Sec. Name Desk No. A. Molar solubility and solubility Product of Calcium Hydroxide Trial 1 Trial 2 Trial 3 1. Volume of saturated Ca (OH)2 solution (mL) 25.0 2. Concentration of standardized HCI solution (mol/L) 3. Buret reading, initial (mL) 4, Buret reading, final (mL) AS, 7 5. Volume of HCl added (mL) 6. Moles of HCl added (mol 7. Moles of OH in saturated solution (mol 8. [OH 1, equilibrium (moUL) 9. [Ca equilibrium (moVI) 11. Average molar solubility of Ca(oH2 (molL 13. Average Ksp Data Analysis, B 14. Standard deviation of Kep Analysis, C 15. Relative standard deviation of Ksp (%RSD) Analysis, D *Calculations for Trial 1. Experiment 22 269

Explanation / Answer

7. Reaction between Ca(OH)2 and HCl is Ca(OH)2+2HCl=CaCl2+2H2O

So moles of Ca(OH)2 is 1/2 moles of HCl=0.5*1.13*10^-3moles=0.565*10^-3 moles=5.65*10^-4 moles(for trial 1 and trial 2) and 0.5*1.12*10^-3 moles=5.6*10^-4 moles (for trial 3)

8. [OH-] in equlibrium=2*moles of Ca(OH)2=1.13*10^-3 mol/L(for trial 1 & 2), 1.12*10^-3 (for trial 3)

9.Ca2+ in equilibrium=moles of Ca(OH)2=5.65*10^-4 mol/L(for trial 1&2) and 5.6*10^-4(for trial 3).

10. Molar solubility =5.65*10^-4 mole/L(for trial 1&2) and 5.6*10^-4 mole/L(for trial 3)

11. Average molar solubility of Ca(OH)2=(2*5.65*10^-4+1*5.6*10^-4)/3 moles/L=7.5*10^-4 moles/L

12.Ksp=4S^3=4*(5.65*10^-4)^3 (for trial 1)=7.21*10^-10

13. Average Ksp=[2*7.21*10^-10+1*4*(5.6*10^-4)^3]/3=[21.44*10^-10]/3=7.11*10^-10

14.Standard deviation of Ksp=root[2*{(7.11-7.21)*10^-10}^2+{(7.11-7.02)*10^-10}^2]/(3-1)]=root of (0.5*3.23*10^-22)=1.27*10^-11

15.RSD=mean+-100*s/mean where s=standard deviation=7.11*10^-10+-(1.27*10^-9)/(7.11*10^-10)=7.11*10^-10+-12.7/7.11%=7.11*10^-10+-1.786%

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