# The common ion effect, predicting precipitation: Read pg

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The common ion effect, predicting precipitation: Read pg
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

Cl in water causes shift left
Common ion lab PbCl2(s) Pb2+(aq) + 2Cl–(aq) A + B Explanation 1 2 3 4 Cloudy / precipitate Cl– causes shift left 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”

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) R I C E 1 2 0.10 x 2x x x Ksp = [Pb2+(aq)] [I –(aq)]2 Ksp = [x] [ x]2 = 7.9 x 10–9 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

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) R I C E 1 1 0.20 x x x x Ksp = [Ag+(aq)] [I –(aq)] Ksp = [x] [ x] = 8.3 x 10–17 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

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) R I C E 1 3 0.050 x 3x x x Ksp = [Fe3+(aq)] [OH –(aq)]3 Ksp = [x] [ x] = 1.6 x 10–39 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

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 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

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

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)

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 …

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

The ion product is smaller than Ksp, thus no precipitate will form
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+] = mol/L x 0.1 L = 1.0 x 10–4 mol = 1.0 x 10–4 mol / 0.2 L = M [SO42–] = mol/L x 0.1 L = 2.0 x 10–4 mol = 2.0 x 10–4 mol / 0.2 L = M [Pb2+][SO42–] = [0.0005][0.001] = 5.0 x 10–7 The ion product is smaller than Ksp, thus no precipitate will form

The ion product is smaller than Ksp, thus no precipitate will form
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 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 [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

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