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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 –

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Presentation on theme: "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 –"— Presentation transcript:

1 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

2 Common ion lab PbCl 2 (s) Pb 2+ (aq) + 2Cl – (aq) A + BExplanation Cl in water causes shift leftCloudy / precipitate Pb causes shift leftCloudy / precipitate No Cl or PbNo reaction (oily) Cl – causes shift leftCloudy / precipitate 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 PbI 2 (s)Pb 2+ (aq)I – (aq) x2x xx R I C E Ksp = [x] [ x] 2 = 7.9 x 10 –9 Ksp = [Pb 2+ (aq)] [I – (aq)] 2 Example (pg. 581) Molar solubility of PbI 2 ? Ksp = 7.9 x 10 –9 Concentration of NaI is 0.10, thus [I – ] = 0.10 NaI(s) Na + (aq) + I – (aq) x is small, thus we can ignore 2x in x Ksp = [x] [0.10] 2 = 7.9 x 10 –9, x = 7.9 x 10 –7 M

4 AgI(s)Ag + (aq)I – (aq) xx xx R I C E Ksp = [x] [ x] = 8.3 x 10 –17 Ksp = [Ag + (aq)] [I – (aq)] 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) x is small, thus we can ignore it in x Ksp = [x] [0.20] = 8.3 x 10 –17, x = 4.2 x 10 –16

5 Fe(OH) 3 Fe 3+ (aq)OH – (aq) x3x xx R I C E Ksp = [x] [ x] = 1.6 x 10 –39 Ksp = [Fe 3+ (aq)] [OH – (aq)] 3 PE 21 (pg. 582) Molar solubility of Fe(OH) 3 ? Ksp = 1.6 x 10 –39 Concentration of OH – is Fe(OH) 3 (s) Fe 3+ (aq) + 3OH – (aq) x is small, thus we can ignore 3x in x Ksp = [x] [0.050] 3 = 1.6 x 10 –39, x = 1.3 x 10 –35

6 Predicting when precipitation occurs Read pg 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 PbCl 2 (s) Pb 2+ (aq) + 2Cl – (aq) (NaCl and Pb(NO 3 ) are soluble according to the solubility rules; they will not precipitate) Ksp = 1.7 x 10 -5, [Pb 2+ ][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: CaSO 4 (s) Ca 2+ (aq) + SO 4 2– (aq) Step 2: Write the Ksp equation: Ksp = [Ca 2+ ][SO 4 2– ] Ksp = 2.4 x 10 –5 [Ca 2+ ][SO 4 2– ] = [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: Ag 2 CrO 4 (s) 2Ag + (aq) + CrO 4 2– (aq) Step 2: Write the Ksp equation: Ksp = [Ag + ] 2 [CrO 4 2– ] Ksp = 1.2 x 10 –12 [Ag + ] 2 [CrO 4 2– ] = [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 PbCl 2 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 PbCl 2 (s) to dissolve, than the extra Cl – causes PbCl 2 (s) to form Read the example on pg. 583, do PE 24, 25

11 PE 24 (pg. 583) Step 1: write the balanced equilibrium: PbSO 4 (s) Pb 2+ (aq) + SO 4 2– (aq) Step 2: Write the Ksp equation: Ksp = [Pb 2+ ][SO 4 2– ] = 6.3 x 10 –7 Initial concentrations: [Pb 2+ ] = mol/L x 0.1 L = 1.0 x 10 –4 mol = 1.0 x 10 –4 mol / 0.2 L = M [SO 4 2– ]= mol/L x 0.1 L = 2.0 x 10 –4 mol = 2.0 x 10 –4 mol / 0.2 L = M [Pb 2+ ][SO 4 2– ] = [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: PbCl 2 (s) Pb 2+ (aq) + 2Cl – (aq) Step 2: Write the Ksp equation: Ksp = [Pb 2+ ][Cl – ] 2 = 1.7 x 10 –5 Initial concentrations: [Pb 2+ ] = 0.10 mol/L x L = 5.0 x 10 –3 mol = 5.0 x 10 –3 mol / L = M [Cl – ]= mol/L x L = 8.0 x 10 –4 mol = 8.0 x 10 –4 mol / L = M [Pb 2+ ][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 and on pg. 591 For more lessons, visit


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