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chapter 17: solubility and complex ion equilibria, Exams of Chemistry

Brookdale Community CollegeChemistry

The Ksp = 1.6 × 10-72. 2. Calculate the solubility of AgCl and the concentration of ionic species present in a 2.0 M NaCl solution saturated with AgCl.

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Chapter 17 Page 1
CHAPTER 17: SOLUBILITY AND COMPLEX ION EQUILIBRIA
Part One: Solubility Equilibria
A. Ksp, the Solubility Product Constant. (Section 17.1)
1. Review the solubility rules. (Table 4.1)
2. Insoluble and slightly soluble compounds are important in nature and commercially.
a. bones and teeth - Ca3(PO4)2
b. limestone - CaCO3
c. x-ray imaging - BaSO4
3. Solubility behavior expressed mathematically by special equilib. constant = Ksp.
a. BaSO4(s) Ba2+(aq) + SO4
2-(aq)
Ksp ≡ [Ba2+][SO4
2-] in solution saturated with BaSO4
= 1.2 × 10-10 for BaSO4 in H2O at 25°C.
b. Ca3(PO4)2 3 Ca2+ + 2 PO4
3-
Ksp ≡ [Ca2+]3[PO4
3-]2
= 1.0 × 10-25
pf3
pf4
pf5
pf8

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CHAPTER 17: SOLUBILITY AND COMPLEX ION EQUILIBRIA

Part One: Solubility Equilibria A. Ksp, the Solubility Product Constant. (Section 17.1)

  1. Review the solubility rules. (Table 4.1)
  2. Insoluble and slightly soluble compounds are important in nature and commercially. a. bones and teeth - Ca 3 (PO 4 ) 2 b. limestone - CaCO 3 c. x-ray imaging - BaSO 4
  3. Solubility behavior expressed mathematically by special equilib. constant = Ksp. a. BaSO 4 ( s ) Ba 2+ ( aq ) + SO 4 2 - ( aq ) Ksp ≡ [Ba 2+ ][SO 4 2 - ] in solution saturated with BaSO 4 = 1.2 × 10 -^10 for BaSO 4 in H 2 O at 25°C. b. Ca 3 (PO 4 ) 2 3 Ca2+^ + 2 PO 43 - Ksp ≡ [Ca 2+ ] 3 [PO 4 3 - ] 2 = 1.0 × 10
  • 25
  1. Note - in all saturated aqueous solutions of BaSO 4 , no matter what other materials are present: [Ba2+][SO 42 - ] = Ksp = 1.2 × 10 -^10
  2. Table of Ksp in Table 17.1.
  3. Solubility = moles of a compound which dissolve in a liter of solution. Directly related to but not equal to Ksp. B. Finding Ksp. (Section 17.1)
  4. Usually determined by conductivity experiment.
  5. Example: Found that 6.93 × 10
  • 3 g of CaCO 3 can dissolve in 1.00 L of water. What is Ksp of CaCO 3? €

6.93 × 10

− 3 g 100.1 g/mol

= 6.92 × 10

  • 5 moles (where 100.1 g/mol is Molar mass) €

6.92 × 10

− 5 mol 1.00 L = 6.92 × 10 -^5 moles (where 1.00 L is VOL) CaCO 3 ( s ) Ca 2+ ( aq ) + CO 3 2 - ↑ ↑ ↑ 6.92 × 10 -^5 M dissolve producing 6.92 × 10 -^5 M 6.92 × 10 -^5 M Ksp = [Ca2+][CO 32 - ] = (6.92 × 10 -^5 M )^2 = 4.8 × 10 -^9

  1. Example: The molar solubility of PbBr 2 is 1.16 × 10 -^2 moles/Liter. What is Ksp? PbBr 2 ( s ) Pb2+( aq ) + 2 Br - ( aq ) ↑ ↑ ↑ 1.16 × 10 -^2 M dissolve producing 1.16 × 10 -^2 M 2 × 1.16 × 10 -^2 M Ksp = [Pb 2+ ][Br

] 2 = (1.16 × 10 -^2 M )(2 × 1.16 × 10 -^2 )^2 = 6.3 × 10 -^6 Note: If s = molar solubility Ksp = (s)(2s) 2 for a (1:2) compound

s = €

1.6 × 10

− 72 108

(^15) 5th root s = 1.7 × 10 -^15 M = molar solubility [Bi 3+ ] 2 = 2s = 3.4 × 10

  • 15 M [S^2 - ]^3 = 3s = 5.1 × 10 -^15 M D. The Common Ion Effect in Solubility Calculations. (Section 17.2)
  1. Previously showed that in saturated solution of AgCl: [Cl - ] = [Ag + ] = 1.3 × 10 -^5 M = € Ksp
  2. Calculate the solubility of AgCl and the concentration of ionic species present in a 2. M NaCl solution saturated with AgCl. Ksp of AgCl is 1.8 × 10 -^10. Here Cl - ion is present from two sources: NaCl and AgCl. AgCl( s ) Ag + ( aq ) + Cl - ( aq ) ↑ ↑ ↑ solubility ≡ s s 2.0 M + s ↑ (from NaCl) Ksp = [Ag+][Cl-] 1.8 × 10
  • 10 = (s)(2.0 + s) assume s << (2.0) 1.8 × 10 -^10 ≈ (s)(2.0) s ≈ 9.0 × 10 -^11 M (very small now) [Ag+] = s = 9.0 × 10 -^11 M (very small now) [Cl-] = 2.0 + s ≈ 2.0 M

Part Two: Uses of the Solubility Product Principle A. Predicting if Precipitation Will Occur. (Section 17.3)

  1. Precipitation will occur whenever the ion concentration product for a compound exceeds the Ksp of that compound. For example, when [Ag + ][Cl - ] > Ksp
  2. The solution is unstable, and precipitate of AgCl will occur until: [Ag + ][Cl - ] = Ksp
  3. Example: The following reagents are added to water and the final volume is adjusted to 1.00 L. What, if any, compounds will precipitate? 1.0 g NaOH ÷ 40.0 g/mol = 2.5 × 10 -^2 M 1.0 g Na 2 CO 3 ÷ 106.0 g/mol = 9.4 × 10 - 3 M 1.0 g CaCl 2 ÷ 110.9 g/mol = 9.0 × 10 -^3 M 1.0 g PbNO 3 ÷ 269.2 g/mol = 3.7 × 10 - 3 M (Hint: Don’t expect any Na + or NO 3 - salts to precipitate due to their high solubility.) Ca(OH) 2? [Ca 2+ ][OH - ] 2 = (9.0 × 10 - 3 )(2.5 × 10 - 2 ) 2 = 5.62 × 10 -^6 < Ksp = 7.9 × 10 -^6 NO! CaCO 3? [Ca 2+ ][CO 3 - ] = (9.0 × 10 - 3 )(9.4 × 10 - 3 ) = 8.46 × 10 - 5 > Ksp = 4.8 × 10 - 6 YES! Pb(OH) 2? [Pb2+][OH - ]^2 = (3.7 × 10 -^3 )(2.5 × 10 -^2 )^2 = 2.3 × 10 - 6 > Ksp = 2.8 × 10 - 16 YES! PbCO 3? [Pb 2+ ][CO 3 - ] = (3.7 × 10 - 3 )(9.4 × 10 - 3 ) = 3.5 × 10 - 5 > Ksp = 1.5× 10 - 13 YES! PbCl 2? [Pb 2+ ][Cl - ] 2 = (3.7 × 10 - 3 )(2 × 9.0 × 10 - 3 ) 2 = 1.1 × 10 - 6 < Ksp = 1.7× 10 - 5 NO! Pb(OH) 2 will precipitate first, because its ion product is the greatest relative to its Ksp.

precipitate AgCl when: [Ag

][Cl

  • ] = 1.8 × 10 - 10 [Cl

] = 1.8 × 10

  • 9 M precipitate CuCl when: [Cu

][Cl

] = 1.9 × 10

  • 7 [Cl

] = 1.9 × 10

  • 6 M
  1. What will [Au + ] be when AgCl just begins to precipitate? [Au + ][Cl - ] = 2.0 × 10 -^13 ↑ 1.8 × 10 -^9 M to precipitate AgCl [Au + ] × 1.8 × 10
  • 9 = 2.0 × 10
  • 13 [Au + ] = 1.3 × 10
  • 4 M (remember, [Au + ] was originally 0.10 M ) D. Effect of pH on Solubility (Section 17.4)
  1. Example: Limestone/Marble tends to dissolve in acidic conditions.
  2. Solubility equilibrium for CaCO 3 (s) CaCO 3 (s) Ca2+(aq) + CO 32 - (aq)
  3. This equilibrium is affected by addition of strong acid: H 3 O + (aq) + CO 3 2 - (aq) H 2 O + HCO 3 - (aq)
  4. This reaction removes carbonate ion, shifting the solubility equilibrium to the right, causing more CaCO 3 (s) to go into solution. Part Four: Complex-Ion Equilibria A. Complex-Ion Formation. (Section 17.5)
  5. Some metal ions in solution can bind varying numbers of molecules or molecular ions.
  6. Example: Ag + ion and ammonia molecules: Ag + (aq) + NH 3 (aq) Ag(NH 3 ) + (aq) Ag(NH 3 ) + (aq) + NH 3 (aq) Ag(NH 3 ) 2 + (aq)
  1. The combined complexation reaction is then: Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq)
  2. This equilibrium is governed by the formation constant , or stability constant , Kf.

Kf =

[Ag(NH 3 ) 2

]

[Ag

][NH 3 ]

2

= 1.7 × 10

7

  1. The dissociation constant Kd is simply the reciprocal of Kf.
  2. Calculations with this equilibrium. (See text Section 17.5)
  3. Formation of complex ions can have important effects on solubility of ions in solution. (See lab manual - Experiments #21-26)