Metal hydroxides in water

1
Complex Equilibria - Solubility

• Metal hydroxides in water
• Equilibria and equilibrium constants
• M(OH)2 M2 2OH-
  KspM2OH-2
• H2O H OH-
  KwHOH-
• Mass balance equation
• 2M2 H OH-
• Solve algebra

2
Complex Equilibria - Solubility

• Complex ion formation
• Example solubility of AgCl in 0.100 M NH3
• Equilibria and equlibrium constants
• AgCl Ag Cl-
  Ksp AgCl-
• Ag 2NH3 Ag(NH3)2
• NH3 H2O NH4 OH-
• H2O H OH-
  Kw HOH- 1.00 x 10-14
• Solubility s Cl-
• Mass balance equations
• Cl- Ag Ag(NH3)2
• CNH3 0.100 NH3 NH4 2Ag(NH3)2
• Charge balance equation
• Ag NH4 H Ag(NH3)2 Cl-
  OH-
• Unknowns Ag, Ag(NH3)2, NH3, NH4,
  H, Cl-, OH-
• 7 unknowns
• 7 equations

3
Complex Equilibria - Solubility

• Complex ion formation
• Example solubility of AgCl in 0.100 M NH3
• Make some assumptions
• NH3 OH-
• H ltlt OH-
• NH4 ltlt NH3 2Ag(NH3)2
• Charge balance equation becomes
• 0.100 NH3 NH4 2Ag(NH3)2
• 0.100 NH3 2Ag(NH3)2
• Ag small compared to Ag(NH3)2
• Mass balance equation becomes
• Cl- Ag Ag(NH3)2
• Cl- Ag(NH3)2

Havent reduced number of equations
4
Complex Equilibria - Solubility

• Complex ion formation
• Example solubility of AgCl in 0.100 M NH3
• Plugging into b2 and solving for Cl-
• Its left to the student to check the assumptions!

5
Complex Equilibria - Solubility

• Complex ion formation
• Solubility of a metal ion complexed by its common
  precipitating anion
• Example solubility of AgCl in solutions of Cl-
• Equilibria
• AgCl(s)
  AgCl(aq) Ks AgCl(aq)
• AgCl(aq) Ag
  Cl-
• AgCl(s)
  Ag Cl- Ksp AgCl- 1.82 x 10-10
• AgCl(s) Cl-
  AgCl2-
• AgCl2- Cl-
  AgCl32-
• Mass balance charge balance equations are the
  same
• K Ag cKCl Ag Cl-
  AgCl2- 2AgCl32- or
• Cl- CCl Ag - AgCl2- -
  2AgCl32-
• Solubility S AgCl(aq) Ag AgCl2-
  AgCl32-
• This problem is solved using Mathcad on page 59
  of Holler

KdKs Ksp
6
Complex Equilibria - Solubility

• Separation of solutions of ions by control of the
  concentration of a precipitating ion
• Reduce the concentration of one ion to a
  negligible level without initiating
  precipitation of another ion or other ions
• If the concentration of the less soluble
  substance is lt 10-4 M and a second ion does not
  precipitate, the separation is feasible
• Example Can 0.100 M I- be separated from 0.100 M
  Cl- using Cu and if so, under what conditions
  of Cu?
• Equilibria
• CuCl Cu Cl- Ksp
  CuCl- 1.2 x 10-6
• CuI Cu I- Ksp
  CuI- 1.1 x 10-12
• What is Cu when I- begins precipitating?
• What is Cu when Cl- begins precipitating?
• What is Cu when I-10-4 M? Will CuCl begin
  precipitating at this Cu?

CuI is less soluble
CuCl begins precipitating at Cu1.2x10-5
M. Keep Cu between 1.1x10-8 and 1.2x10-5M
7
Complex Equilibria - Solubility

• Separation of solutions of ions by control of the
  concentration of a
• precipitating ion
• Sulfide separations
• H2S equilibria
• H2S H HS-
• HS- H S2-
• H2S 2H S2-
• To produce a separation, find range of S2- that
  can reduce the concentration of least soluble
  cation to 10-4 M without precipitating other
  ions
• Then calculate range of H that controls the
  S2- to this range

8
Complex Equilibria - Solubility

• Separation of solutions of ions by control of the
  concentration of a
• precipitating ion
• Sulfide separations
• Example Find the conditions for the separation
  fo 0.01 M La3 from 0.10 M Mn2 using H2S.
• Equilibria

Keep pH in this range