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Common ion lab

The common ion effect, predicting precipitation : Read pg. 580 – 582 (common ion effect section only). Do PE 20 – 21 By the end of the period: PE 14 – 19, PE 20 – 25. Common ion lab. Cloudy / precipitate. Cl – causes shift left. PbCl 2 (s)  Pb 2 + (aq) + 2Cl – (aq). No reaction (oily) .

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Common ion lab

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  1. The common ion effect, predicting precipitation:Read pg. 580 – 582 (common ion effect section only). Do PE 20 – 21By the end of the period:PE 14 – 19, PE 20 – 25

  2. Common ion lab Cloudy / precipitate Cl– causes shift left PbCl2(s) Pb2+(aq) + 2Cl–(aq) No reaction (oily) No Cl or Pb Cloudy / precipitate Pb causes shift left Cloudy / precipitate Cl in water causes shift left Common ion: “The ion in a mixture of ionic substances that is common to the formulas of at least two.” Common ion effect: “The solubility of one salt is reduced by the presence of another having a common ion”

  3. R I C E Example 14.17 (pg. 581) Molar solubility of PbI2? Ksp = 7.9 x 10–9 Concentration of NaI is 0.10, thus [I–] = 0.10 NaI(s)  Na+(aq) + I–(aq) PbI2(s) Pb2+(aq) I –(aq) 1 2 0 0.10 x 2x x 0.10 + 2x Ksp = [Pb2+(aq)] [I–(aq)]2 Ksp = [x] [0.10 + 2x]2 = 7.9 x 10–9 x is small, thus we can ignore 2x in 0.10 + 2x Ksp = [x] [0.10]2 = 7.9 x 10–9 , x = 7.9 x 10–7M

  4. R I C E PE 20 (pg. 582) Molar solubility of AgI? Ksp = 8.3 x 10–17 Concentration of NaI is 0.20, thus [I–] = 0.20 NaI(s)  Na+(aq) + I–(aq) AgI(s) Ag+(aq) I –(aq) 1 1 0 0.20 x x x 0.20 + x Ksp = [Ag+(aq)] [I–(aq)] Ksp = [x] [0.20 + x] = 8.3 x 10–17 x is small, thus we can ignore it in 0.20 + x Ksp = [x] [0.20] = 8.3 x 10–17, x = 4.2 x 10–16

  5. R I C E PE 21 (pg. 582) Molar solubility of Fe(OH)3? Ksp = 1.6 x 10–39 Concentration of OH– is 0.050 Fe(OH)3(s)  Fe3+(aq) + 3OH–(aq) Fe(OH)3 Fe3+(aq) OH–(aq) 1 3 0 0.050 x 3x x 0.050 + 3x Ksp = [Fe3+(aq)] [OH–(aq)]3 Ksp = [x] [0.050 + 3x] = 1.6 x 10–39 x is small, thus we can ignore 3x in 0.050 + 3x Ksp = [x] [0.050]3 = 1.6 x 10–39, x = 1.3 x 10–35

  6. Predicting when precipitation occurs Read pg. 582. Do PE 22, 23 Similar to Kc vs. mass action expression to predict if equilibrium exists (and which way it will shift) E.g. in example 14.18 PbCl2(s)  Pb2+(aq) + 2Cl–(aq) (NaCl and Pb(NO3) are soluble according to the solubility rules; they will not precipitate) Ksp = 1.7 x 10-5, [Pb2+][Cl–]2 = 3.4 x 10–5 Ion product is large … to reduce, equilibrium must shift left … precipitate forms

  7. PE 22 (pg. 583) Step 1: write the balanced equilibrium: CaSO4(s)  Ca2+(aq) + SO42–(aq) Step 2: Write the Ksp equation: Ksp = [Ca2+][SO42–] Ksp = 2.4 x 10–5 [Ca2+][SO42–] = [0.0025][0.030] = 7.5 x 10–5 ion product is greater than Ksp, thus a precipitate will form

  8. PE 23 (pg. 583) Step 1: write the balanced equilibrium: Ag2CrO4(s)  2Ag+(aq) + CrO42–(aq) Step 2: Write the Ksp equation: Ksp = [Ag+]2[CrO42–] Ksp = 1.2 x 10–12 [Ag+]2[CrO42–] = [4.8 x 10–5]2[3.4 x 10–4] = 7.8 x 10–13 ion product is less than Ksp, thus no precipitate will form (more could be dissolved)

  9. Predicting when precipitation occurs • So far we have been dealing with one of two situations: • 1) dissolving a solid in a liquid (with or without initial concentrations of ions) and performing Ksp calculations • 2) given the concentrations of ions predicting if a solid (I.e. precipitate will form) • A third situation exists that is slightly different • Mixing two liquids • In this case, we need to account for both the ions and the water that is added …

  10. Predicting when precipitation occurs Q- E.g. will the addition of a NaCl solution to a saturated PbCl2 solution result in a precipitate forming? A- It depends on the concentration of the NaCl solution If the NaCl solution is very dilute, the extra water could cause more PbCl2(s) to dissolve, than the extra Cl– causes PbCl2(s) to form Read the example on pg. 583, do PE 24, 25

  11. PE 24 (pg. 583) Step 1: write the balanced equilibrium: PbSO4(s)  Pb2+(aq) + SO42–(aq) Step 2: Write the Ksp equation: Ksp = [Pb2+][SO42–] = 6.3 x 10–7 Initial concentrations: [Pb2+] = 0.0010 mol/L x 0.1 L = 1.0x10–4 mol = 1.0x10–4 mol / 0.2 L = 0.00050 M [SO42–] = 0.0020 mol/L x 0.1 L = 2.0x10–4 mol = 2.0x10–4 mol / 0.2 L = 0.0010 M [Pb2+][SO42–] = [0.0005][0.001] = 5.0 x 10–7 The ion product is smaller than Ksp, thus no precipitate will form

  12. PE 25 (pg. 583) Step 1: write the balanced equilibrium: PbCl2(s)  Pb2+(aq) + 2Cl–(aq) Step 2: Write the Ksp equation: Ksp = [Pb2+][Cl–]2 = 1.7 x 10–5 Initial concentrations: [Pb2+] = 0.10 mol/L x 0.050 L = 5.0x10–3 mol = 5.0x10–3 mol / 0.070 L = 0.0714 M [Cl–] = 0.040 mol/Lx0.020 L = 8.0x10–4 mol = 8.0x10–4 mol / 0.070 L = 0.0114 M [Pb2+][Cl–]2 = [0.0714][0.0114]2 = 9.3 x 10–6 The ion product is smaller than Ksp, thus no precipitate will form

  13. Try 14.77 and 14.78 on pg. 591 For more lessons, visit www.chalkbored.com

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