Solubility Product and Common Ion Effect

Solubility Product

• Solubility product is discussed with reference to the solubility of ionic compounds.

Solubility Rules

• Soluble Compounds:

  • Nitrates (NO_3^-) are generally soluble with no exceptions.
  • Acetates (CH_3COO^-) are generally soluble with no exceptions.
  • Chlorides (Cl^-) are soluble except when combined with silver (Ag^+), mercury (Hg_2^{2+}), and lead (Pb^{2+}).
  • Bromides (Br^-) are soluble except when combined with silver (Ag^+), mercury (Hg_2^{2+}), and lead (Pb^{2+}).
  • Sulfates (SO4^{2-}) are soluble except for compounds of strontium (Sr^{2+}), barium (Ba^{2+}), mercury (Hg2^{2+}), and lead (Pb^{2+}).

• Insoluble Compounds:

  • Sulfides (S^{2-}) are generally insoluble except for compounds of ammonium (NH_4^+), alkali metals, calcium (Ca^{2+}), strontium (Sr^{2+}), and barium (Ba^{2+}).
  • Carbonates (CO3^{2-}) are generally insoluble except for compounds of ammonium (NH4^+), and alkali metals.
  • Phosphates (PO4^{3-}) are generally insoluble except for compounds of ammonium (NH4^+), and alkali metals.
  • Hydroxides (OH^-) are generally insoluble except for compounds of ammonium (NH_4^+), alkali metals, calcium (Ca^{2+}), strontium (Sr^{2+}), and barium (Ba^{2+}).

Solubility Equilibrium

• Example: Silver nitrate (AgNO_3) is highly soluble.
  • AgNO3(s) \rightleftharpoons Ag^+(aq) + NO3^-(aq)
  • The equilibrium lies far to the right, indicating high solubility.
• Example: Silver chloride (AgCl) is poorly soluble.
  • AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)
  • The equilibrium lies far to the left, indicating low solubility.

Solubility Product Constant (K_{sp})

• For silver chloride: K_{sp} = [Ag^+][Cl^-] = 1.6 \times 10^{-10}
• The smaller the Ksp value, the lower the solubility of the compound.

Barium Sulfate (BaSO4) and Barium Nitrate (Ba(NO3)_2)

• Barium sulfate is used in medical imaging as an X-ray absorber.
• Soluble barium is toxic. Insoluble forms are used because they are not absorbed into the body.
• Barium nitrate is soluble:
  • Ba(NO3)2(s) \rightleftharpoons Ba^{2+}(aq) + 2NO_3^-(aq)
• Barium sulfate is insoluble:
  • BaSO4(s) \rightleftharpoons Ba^{2+}(aq) + SO4^{2-}(aq)
• The K{sp} for barium sulfate is given by: K{sp} = [Ba^{2+}][SO_4^{2-}]

Toxicity of Barium Compounds

• The LD50 (lethal dose for 50% of experimental animals) of a chemical indicates its toxicity.
• Barium nitrate has an LD50 of approximately 25 g (for the person in the context).

Calculation Example: Barium Sulfate

• Problem: A person drinks 2.0 L of a barium milkshake containing 45 g of barium sulfate (K_{sp} = 1.5 \times 10^{-9}). How close did they come to death from Ba^{2+} poisoning?

• Molar masses: Ba = 137.3 amu, SO4^{2-} = 96.1 amu, NO3 = 62.0 amu.

• First, consider barium nitrate:

  • Ba(NO3)2 (s) \rightleftharpoons Ba^{2+}(aq) + 2NO_3^-(aq)
  • Calculate moles of Ba^{2+} from 25 g of Ba(NO3)2:
  • \frac{25 \text{ g } Ba(NO3)2}{25 \text{ g/mol } Ba(NO3)2} \times \frac{1 \text{ mol } Ba^{2+}}{1 \text{ mol } Ba(NO3)2} = 0.096 \text{ mols } Ba^{2+}
  • Calculate the concentration of Ba^{2+} in 2.0 L:
  • [Ba^{2+}] = \frac{0.096 \text{ mols}}{2.0 \text{ L}} = 0.048 \text{ M}

• Next, consider barium sulfate:

  • BaSO4(s) \rightleftharpoons Ba^{2+}(aq) + SO4^{2-}(aq)
  • K{sp} = [Ba^{2+}][SO4^{2-}] = 1.5 \times 10^{-9}
  • Let s be the molar solubility: K_{sp} = s^2 = 1.5 \times 10^{-9}
  • s = \sqrt{1.5 \times 10^{-9}} = 3.9 \times 10^{-5} \text{ M}

Common Ion Effect

• The common ion effect is the decrease in solubility of an ionic compound due to the addition of a soluble compound with a common ion.
• This is a consequence of Le Chatelier's principle.

Example: Lead Chloride (PbCl_2)

• PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)
• K_{sp} = [Pb^{2+}][Cl^-]^2
• If KCl is added, the concentration of Cl^- increases, shifting the equilibrium to the left, thus decreasing the concentration of Pb^{2+}.

Common Ion Effect Problem: Silver Chloride (AgCl)

• Problem 1: What mass of AgCl will dissolve in 5.0 L of water (K_{sp} = 1.6 \times 10^{-10}, MW = 143.3 amu)?

  • AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)
  • K_{sp} = [Ag^+][Cl^-] = 1.6 \times 10^{-10}
  • Let s be the molar solubility: s^2 = 1.6 \times 10^{-10}
  • s = \sqrt{1.6 \times 10^{-10}} = 1.3 \times 10^{-5} \text{ mol/L}
  • Moles of AgCl in 5.0 L: 5.0 \text{ L} \times 1.3 \times 10^{-5} \text{ mol/L} = 6.3 \times 10^{-5} \text{ mols}
  • Mass of AgCl: 6.3 \times 10^{-5} \text{ mol} \times 143.3 \text{ g/mol} = 9.0 \text{ mg}

• Problem 2: What mass of AgCl will dissolve in 5.0 L of 0.250 M NaCl(aq)?

  • AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)
  • K_{sp} = [Ag^+][Cl^-] = 1.6 \times 10^{-10}
  • Initial: [Ag^+] = s, [Cl^-] = 0.250 + s
  • K_{sp} = s(0.250 + s) = 1.6 \times 10^{-10}
  • Assume s << 0.250, then 0.250s = 1.6 \times 10^{-10}
  • s = \frac{1.6 \times 10^{-10}}{0.250} = 6.4 \times 10^{-10} \text{ M}
  • Moles of AgCl in 5.0 L: 6.4 \times 10^{-10} \text{ mol/L} \times 5.0 \text{ L} = 3.2 \times 10^{-9} \text{ mols}
  • Mass of AgCl: 3.2 \times 10^{-9} \text{ mol} \times 143.3 \text{ g/mol} = 4.6 \times 10^{-7} \text{ g} = 0.46 \mu \text{g}