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Entry Task: March 5 th Tuesday

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lottie
24 diapositives · 470 Views · Jul 11, 2014

Entry Task: March 5 th Tuesday. Question: Calculate the solubility of AgI in water in moles per liter given its K sp value of 8.3 x 10 -17 .

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  1. Entry Task: March 5th Tuesday Question: Calculate the solubility of AgI in water in moles per liter given its Ksp value of 8.3 x 10-17. A saturated solution of AgI also containing NaI is found to have an iodide ion concentration of 0.020 M. What is the concentration of silver ions? You have 10 minutes

  2. Agenda: • Discuss Solubility, Precipitation and Ions ws • Major self-check on content so far & discuss it • HW: Pre-lab Determine Ksp

  3. CaC2O4(s) Ca2+(aq) + C2O42-(aq) 17.6- Precipitation and Separation of Ions 17.37 A 1.00L solution is saturated at 25°C with calcium oxalate, CaC2O4, is evaporated to dryness, giving a 0.0061 g residue of CaC2O4. Calculate the solubility-product constant for this salt. Calculate the molarity 0.0061g/127.99 = 4.77 x10-5M of CaC2O4 Ksp= [4.77 x10-5][4.77 x10-5] Ksp= 2.3 x10-9

  4. Mn(OH)2(s) Mn2+(aq) + 2OH1-(aq) 17.6- Precipitation and Separation of Ions 17.41 Calculate the solubility of Mn(OH)2 in grams per liter when buffered at (a) at pH of 7.0; pH of 7 means there are 1.0 x10-7 OH ions 1.6x10-13= [x][1.0 x10-7]2 1.6x10-13 = [x] 1.0 x10-14 The molarity is 16M (16)(1 L) = 16 moles * 89 = • 1424 or 1.4 x103g/ liter

  5. Mn(OH)2(s) Mn2+(aq) + 2OH1-(aq) 17.6- Precipitation and Separation of Ions 17.41 Calculate the solubility of Mn(OH)2 in grams per liter when buffered at (b) at pH of 9.5; pH of 9.5= pOH 4.5 means there are 3.2 x10-5 OH ions 1.6x10-13= [x][3.2 x10-5]2 1.6x10-13 = [x] 1.0 x10-9 The molarity is 1.6x10-4M (1.6x10-4)(1 L) = 1.6x10-4 moles * 89 = • 1.4x10-2g/ liter

  6. Mn(OH)2(s) Mn2+(aq) + 2OH1-(aq) 17.6- Precipitation and Separation of Ions 17.41 Calculate the solubility of Mn(OH)2 in grams per liter when buffered at (c) at pH of 11.8; pH of 11.8= pOH 2.2 means there are 6.3 x10-3 OH ions 1.6x10-13= [x][6.3 x10-3]2 1.6x10-13 = [x] 3.98 x10-5 The molarity is 4.02x10-9M (4.02x10-9)(1 L) = 4.02x10-9 moles * 89 = • 3.6x10-7g/ liter

  7. 17.6- Precipitation and Separation of Ions 17.49 (a) Will Ca(OH)2 precipitate from solution if the pH of a 0.050M solution of CaCl2 is adjusted to 8.0? Ksp= 6.5x10-6= [Ca+]OH-]2 pH of 8 = pOH of 6 = 1.0x10-6M of OH Q = [0.050]1.0x10-6]2 Q = 5.0 x 10-14 Ksp = 6.5 x 10-6 Ksp is bigger meaning no precipitate

  8. 17.6- Precipitation and Separation of Ions 17.49 (b) Will Ag2SO4 precipitate when 100 ml of 0.050M AgNO3 is mixed with 10 mls of 5.0 x10-2M NaSO4 solution? In 0.100 L of 0.050 MAgNO3 there are (0.100 L) (0.050M) = 5.0 x 10-3 moles of Ag+1 ions

  9. 17.6- Precipitation and Separation of Ions 17.49 (b) Will Ag2SO4 precipitate when 100 ml of 0.050M AgNO3 is mixed with 10 mls of 5.0 x10-2M Na2SO4solution? In 0.010 L of 5.0 x10-2MNaNO3 there are (0.010 L) (5.0 x10-2M) = 5.0 x 10-4 moles of SO4-2 ions

  10. 17.6- Precipitation and Separation of Ions 17.49 (b) Will Ag2SO4 precipitate when 100 ml of 0.050M AgNO3 is mixed with 10 mls of 5.0 x10-2M Na2SO4solution? • We have to convert the moles in to molarity but use the combined volume. 5.0 x 10-3 moles of Ag+1 ions 5.0 x 10-4 moles of SO4-2 ions

  11. 17.6- Precipitation and Separation of Ions 17.49 (b) Will Ag2SO4 precipitate when 100 ml of 0.050M AgNO3 is mixed with 10 mls of 5.0 x10-2M Na2SO4solution? • We have to convert the moles in to molarity but use the combined volume. 5.0 x10-3 moles/0.110L = • 4.55 x10-2M of Ag+1 ions 5.0 x10-4 moles/0.110L = 4.55 x10-3M of SO4-2 ions Substitute the values into the Ksp expression and solve for Q

  12. 17.6- Precipitation and Separation of Ions 17.49 (b) Will Ag2SO4 precipitate when 100 ml of 0.050M AgNO3 is mixed with 10 mls of 5.0 x10-2M Na2SO4solution? Q = [Ag+]2[SO42] • 9.4  106 (4.55 x10-2)2(4.55 x10-3) = • Q= 9.4  106 Ksp= 1.5 x 10-5 • Q is smaller than Ksp that means • No precipitate will occur

  13. 17.6- Precipitation and Separation of Ions 17. 53 A solution contains 2.0 x10-4M Ag+ and 1.5 x 10-3 M Pb+2. If NaI is added, will AgI (Ksp= 8.3x10-17) or PbI2 (Ksp= 7.9 x10-9) precipitate first? Specify the concentration of I- needed to begin precipitation.

  14. 17.6- Precipitation and Separation of Ions 17. 53 A solution contains 2.0x10-4M Ag+ and 1.5 x 10-3 M Pb+2. If NaI is added, will AgI (Ksp= 8.3x10-17) or PbI2 (Ksp= 7.9 x10-9) precipitate first? Specify the concentration of I- needed to begin precipitation. Lets look at Ag+ with I-: Ksp = [Ag+][l-] 8.3 x10-17 = (2.0 x10-4)(x) = l- ions 8.3 x10-17= 2.0 x10-4 4.2 x 10-13 l- ions

  15. 17.6- Precipitation and Separation of Ions 17. 53 A solution contains 2.0x10-4M Ag+ and 1.5 x 10-3 M Pb+2. If NaI is added, will AgI (Ksp= 8.3x10-17) or PbI2 (Ksp= 7.9 x10-9) precipitate first? Specify the concentration of I- needed to begin precipitation. Lets look at Pb+2 with I-: Ksp = [Pb+][l-]2 7.9 x10-9 = (1.5 x 10-3)(x)2 = l- ions 7.9 x10-9 = 1.5 x 10-3 x2=5.3 x 10-6 l- ions x=2.3 x 10-3 l- ions

  16. 17.6- Precipitation and Separation of Ions 17. 53 A solution contains 2.0x10-4M Ag+ and 1.5 x 10-3 M Pb+2. If NaI is added, will AgI (Ksp= 8.3x10-17) or PbI2 (Ksp= 7.9 x10-9) precipitate first? Specify the concentration of I- needed to begin precipitation. Which concentration is smaller? 4.2 x 10-13l- ions with Ag+ 2.3 x 10-3 l- ions- ions with Pb+2 This means that it will Agl precipitate as such a small concentration verses Pbl2.