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Chemistry Equium WCLN.ca Data and Observations: 1. Write the Dissociation equation for Lead (1) lodide. It appears in the video but you should ensure you put your charges and subscripts in the correct place. Pb 12 (6) Z P6th cag) + 2I (99) 2. Write the kip expression for Pbla. КsP = CP2L:) 3. When determining if a precipitate will form you compare a Trial Kup to the actual K. Will you see a precipitate when the Trial K. is larger than or smaller than the actual K.? If the Trial Iesp is teverger than the actual Irsp thch Percivitule will form: 4. Fill in Table 1 by watching the video Table 1: Data needed to calculate the Trial Ksp at which a precipitate begins to form 3 1 10.0 2 8,0 4 4.0 5 3.0 6 2.0 6.0 0.0 2.0 4.0 6,0 7.0 8.0 10.0 8.0 6.0 4.0 3.0 2.0 Test Tube Volume of 0.010M Pb(NO) Volume of water ml.) added Volume of 0.020M KI (mL) Volume of water (ml. added Precipitate or no precipitate at room temperature Temperature at which precipitate dissolves (°C) 0.0 2.0 4.0 6.0 7.0 8.0 Yes Yes Yes Yes NO No 63 54 44 35 N/A N/A Calculations: 1. For each test tube you will need to calculate the Trial Ky. In order to do this you will need the diluted concentration of the Pbl ion and the lion. The dilution formula is shown below and the calculations for the first test tube are done for you as an example. The old concentrations for each compound are found in Table 1. Hint: The new volume will always be 20.0 ml 2021-07-15 Page 2 of 6
Chaty Equim WCLN.ca Dilution formula New [ ] = Old [lx Old Volume New Volume Test Tube 110.0 ml. ofcach solution 10.0 ml New [Pb2+] = 0.010 Mx = 0.0050 M 20.0 mL 10.0 mL. New [1 ] = 0.020 MX = 0.010 M 20.0 mL Ksp = (0.005 M)(0.010 M)2 = 5.0x10-7 Test Tube 2 - 80 ml. of cash solution hew CPbt2] = 0010 mx 1.6th = 0.004M 20 ML (1°) - 0.020MX sonu 0.008m 2004 ksp=60.004).(0.008)² = 2-56X10? Xomil Test Tube 3 - 6,0 mL of each solution (Pb+2) = oolom x 60ML 200ML = 0.000m CSP = (0-003).60.000) 11x102 (1) = 0.020 mx bon = 0.0obom 200ML Page 3 of 6 2021-07-15
Chemistry Equi WCLN.ca Test Tube 4.4.0 ml of cach solution CPb ta] = 0.010 mx 4.0ML = 0.0020M 200ML (I -] = 0.020 Mx 40mL 0.004OM 2oom Ispa (0.002).C0.0040) - 3 Bente Test Tube 53.0 ml of cash solution CP6+2) = 0.00M X 30ml = 0.0015 2oon (I+)- 0020M X Jon = 0.0030M 200nl Ksp Cools).(0.00302 Test Tube 6 - 20 ml. of each solution ISP-60-0010).L0-0020 = 440X10-9 CPL - 0010MX 2012 = 0oolom 20 OML C1] = 0020M X 20ML - 0 OOZOM 200ML Page 4 of 6 2021-07-15
Chemistry Equilibrium WCLN.ca Questions: 1. According to the course formula sheet, what is the actual K. of Lead (II) fodide? lope 32 x 10-8 2. Based on your calculations what is the minimum value for K? 3. Based on your calculations what is the maximum value for K..? The maximun Isp value = foxlok 4. Does the actual Ki fall between your calculated min and max? 5. Between which two test tubes should the change from precipitate to no precipitate have occurred? Between test tube and so change from pot to ho ppt 6. Can you think of a reason for this discrepancy? 7. Fill in Table 2 below using your Kip calculations and the temperature numbers from the video or Table 1. The solubility is the same as the diluted [Pb?'you calculated. Test tube 1 Test tube 2 Test tube 3 Test tube 4 Page 5 of 6 2021-07-15
Chemistry Equilibrium WCLN.ca Temperature Ksp Solubility 8. What is the trend in solubility as the temperature is increased? 2021-07-15 Page 6 of 6