Hi, We did the solubility lab in class and now I\'m not sure how to solve these

Question

Hi, We did the solubility lab in class and now I'm not sure how to solve these post lab questions. All my data is in the picture. I appreciate all the help and I will rate the best answers! Thank you! RP1 and 2 in my data means reaction pipet 1. Please don't answer only one question, it's due tomorrow morning and I don't have time!

The purpose of this lab was to use techniques to obtain solubility data in order to estimate the Ksp of strontium iodate, Sr(IO3)2.A centrifuge was used to separate the supernatant liquid from theprecipitate. Finally, calculations were made to compare the accepted Kspand estimated Kspof strontiumiodate

Post-Laboratory Questions

2.

a) The solubility of mercury (I) chloride, Hg2Cl2, is 0.00020 g in 100 mL of water at 25°C. What is the solubility of Hg2Cl2, in moles per liter?

b) Calculate the Ksp of Hg2Cl2. 9

3. A student did not thoroughly rinse the Sr(IO3)2 precipitate and a small amount of Sr+2 ion remained with the precipitate. How might this affect the determined value of the Ksp of Sr(IO3)2? (Will it be high or low?) Explain.

4.

a) The density of solid Sr(IO3)2 is 5.045 g/cm3 . Calculate the concentration of solid Sr(IO3)2 in moles per liter.

b) Using your average Ksp value and equation 2, calculate the value of K for Sr(IO3)2

5. Discuss your % error in the Ksp value. What errors occurred in the lab and HOW these errors affected your Ksp results (to high, too low, or cannot be determined). Be specific and detailed in your answer

Explanation / Answer

2a) The molar mass of Hg2Cl2 is 472.09 gm/mol.

The solution contains 0.00020 gm Hg2Cl2 in 100 mL water. Therefore, a liter of the same solution (= 1000 mL) will contain

(0.00020 gm Hg2Cl2)*(1000 mL/100 mL) = 0.0020 gm Hg2Cl2.

The solubility of Hg2Cl2 is 0.0020 gm/L.

Now, moles of Hg2Cl2 in 0.0020 gm of the solute = (0.0020 gm Hg2Cl2)*(1 mole/472.09 gm) = 4.236*10-6 mole 4.2*10-6 mole

Therefore, the molar solubility of Hg2Cl2 is 4.2*10-6 mole/L (ans)

b) The dissociation of Hg2Cl2 is given by the equation:

Hg2Cl2 (s) <=====> Hg22+ (aq) + 2 Cl- (aq)

The molar solubility of Hg2Cl2 is 4.2*10-6 mole/L (from above)

[Hg22+] = 4.2*10-6 mole/L

[Cl-] = 2*4.2*10-6 mole/L = 8.4*10-6 mole/L (refer to the dissociation equation, the molar ratio of Hg2Cl2 and Cl- is 1:2)

Ksp = [Hg22+][Cl-]2 = (4.2*10-6).(8.4*10-6)2 = (4.2*10-6)*(7.056*10-11) = 2.96*10-16 3.0*10-16 (ans)

3) The Ksp for strontium iodate can be written as

Ksp = [Sr2+][IO3-]2 (considering the dissociation equation is Sr(IO3)2 <=====> Sr2+ + 2 IO3-)

Now since the student did not thoroughly rinse the precipitate, some Sr2+ remained with the precipitate. Thus, some Sr2+ was lost due to improper washing. Now consider the Ksp expression above. [Sr2+] will be lower than the actual value because some Sr2+ was trapped in the precipitate and lost. Thus, the Ksp value calculated will be lower.

4a) Some other information like the weight of strontium iodate taken or the volume of the solution is required to calculate the solubility in moles/L. Please check if any other information is provided.

b) I need to know what is equation 2; you haven’t uploaded the experiment or any information related to the same. Moreover, you have noted down only the volume of the reagents added; I need to know the concentrations (or atleast some information to calculate the same) to find out Ksp.

5) I need to go through the experimental procedure to ascertain sources of errors.

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