Solubility Equilibria

Question 1

What is K_sp?

Answer 1

• Equilibrium constant K for “solubility product”
• It’s the product of the molar concentrations of the ions in the saturated solution, raised to appropriate powers
• It equals the ion product for a solution of a salt, when the solution is saturated (as evidenced by solid in solution)
  
  • So there needs to be a solid on one side of the equation. Otherwise, you have Q, ion product.

Question 2

What is the relationship between Solubility and Ksp?

Answer 2

• Solubility is the amount of the salt that dissolves in an amount of solvent to make a saturated solution (grams)
  
  • Convert solubility into molar solubility
• K_sp is the solubility product, which is the product of molar concentrations of ions in a saturated solution, raised to appropriate powers
• You can calculate K_sp from the known solubility of a salt, and vice versa
• Shortcuts:
  
  • If the salt dissociates into 2 pieces, S = square root of K_sp. K_sp=S²
  • If the salt dissociates into 3 pieces, S = cube root of K_sp/4 K_sp=4s³

Question 3

What is a common ion, and the common ion effect?

Answer 3

• Common ion: an ion that has been supplied by more than one solute.
• Effect: The common ion effect decreases the solubility of an ion in water –> precipitate
  
  • A salt is less soluble in a solution that contains one of its ions than it is in pure water
    
    • Application of Le Chatlier’s principle. When you add the common ion, there’s too much of it, so to counteract, precipitates out.

Question 4

When will a precipitate form?

Answer 4

• A precipitate will only form if a solution is supersaturated
  
  • When the ion product > K_sp
• Precipitate will NOT form if ion product ≤ K_sp
• When mixing two solutions of different salts, consider which combinations of their ions might produce a sparingly soluble salt, calculate the ion product, and compare it to the salt’s K_sp

Question 5

What is the difference between a salt’s ion product, and it’s solubility product constant?

Answer 5

• Q value and K value
• The ion product is the product of the ion molarities raised to powers from the subscripts in the salt’s formula. Not neccessarily at equilibrium. (Q value)
• The ion product gets a constant value in a saturated solution. That’s the solubility product constant (K value)
• When the ion product…
  
  • < Ksp, the solution is unsaturated
  • = Ksp, the solution is saturated
  • > Ksp, the solution is supersaturated, and precipitate will form

Question 6

What is molar solubility?

Answer 6

• The molar concentration of the salt in its saturated solution
• = # moles of salt / 1 L solution
• In the ice chart, multiply the variable for the molar solubility by the stoich coefficient

Question 7

What is K_spa , and the effect of pH on solubility?

Answer 7

• This is the constant for the solubility product of acids
  
  • ​Specifically for sulfides since without the presence of an acid, they’re insoluble.
• [H+] is in the denominator
• All K_spa values are about 10²¹ times larger than K_sp values because some metal sulfides are much more soluble in acid

Question 8

What is selective precipitation?

Answer 8

• Causing one metal ion to precipitate while holding the other in solution
• Possibile due to differences in solubility

Question 9

What is a complex ion?

Answer 9

• An ion that is composed of simpler species
  
  • The charge is whatever the original charge of the ion was, if the ligand is neutral
• Very favorable. More soluble.
• It’s an eample of metals behaving as Lewis acids – they accept electron pairs, and form coordinate covalent bonds
  • Ex: Cu(H2O)₄²⁺ or Cu(NH₃)₄^2+C
• Common ligands include neutral molecules with unshared electron pairs (like H2O or NH3) or anions (like OH- or any halide ion, like Cl-)

Question 10

What is a ligand?

Answer 10

• A Lewis base that attaches itself to a metal ion (electron donor) to form a complex ion
• Common ligands:
  
  • Neutral molecules with unshared electron pairs (like H₂O or NH₃)
  • Anions (like OH- or any halide ion, like Cl-)

Question 11

What are coordination compounds?

Answer 11

• Compounds that contain complex ions
  
  • (named this way because at least some of the bonds are coordinate covalent bonds)

Question 12

What is K_f?

Answer 12

• The formation constant
• When the product of the reaction in equilibrium is a complex ion
  
  • ​Ex: Ag+ _(aq) + 2NH_{3 (aq)} ⇄ [Ag(NH₃)₂]+_(aq) Kf = 1.6 x 10⁷
• Aka stability constant because the larger the value, the greater is the concentration of the complex at equilibrium, so the more stable is the complex
• K_sp * K_f = K_net ​
  
  • In a solubility with complex ion problem, you can multiply to get the overall equilibrium constant. (Because the net reaction is the solubility process + the complex iron formation process.)
• K_f is generally large (whereas K_sp is small)
  
  • When calculating, don’t disregard any x’s in denominator

Question 13

What is the complex ion effect?

Answer 13

• The solubility of a sparingly soluble salt increases when a substance able to form a complex with its metal ion is put into its solution
  
  • This is also based on le chatlier’s principle
• You can dissolve precipitates this way
• To get the new equilibrium constant:
  
  • K_sp * K_f = K_net

Question 14

Generally, what effect do temperature and pH have on K_sp?

Answer 14

• There is an increase in solubility with temperature and pH

Question 15

If a compound dissociates into two pieces,

what are the equations for Solubility and Ksp?

Answer 15

• S = √K_sp
  
  • _​Where S is the molar concentration of the solid
  • K_sp = S²

Question 16

If a compound dissociates into three pieces,

what are the equations for Solubility and Ksp?

Answer 16

• S = ³√(K_sp/4) the cube root of Ksp/4
• K_sp = 4s³
• The steps:
  
  • Na₂SO₄ (s) ⇄ 2Na+ (aq) + SO₄^2- (aq)
  • Ksp=[Na+]²[SO₄^2-]
  • Ksp=(2[Na₂SO₄])²[Na₂SO₄] substitute, maintain ratio

Question 17

If you’re dissolving a precipitate by forming complex ions,

how do you calculate K of the overall reaction?

Answer 17

• K_c = K_sp x K_form
• Add the equation for dissociation of the solid to the equation for the formation of the complex ion, to get the overall equilibrium equation
• Multiply the K_sp and K_f values to get the net K

Ex:

• AgCl (s) ⇄ Ag+ (aq) + Cl- (aq) Ksp = 1.8 x 10^-10
• Ag+ (aq) + 2NH₃ (aq) ⇄ [Ag(NH₃)²]⁺Kf = 1.6 x 10⁷
• ^{——————————————————————-}
• AgCl (s) + 2NH₃ (aq) ⇄ [Ag(NH₃)²]⁺ + Cl- (aq) Knet = 2.9 x 10^-3