Solution Equilibria

• Formal charge


  Increases by +1 when hydrogen (H+) is added
• Dissolution


  Solute (e.g. salt; NaCl(s)) dissolves in a solvent (e.g. H2O)
• NaCl


  • Ionic crystal, held together by ionic bonds (Na+ and Cl- attracted to each other by charge)

  
• H2O


  Polar molecule; oxygen is more electronegative than the hydrogens, so the oxygen gets a partial negative charge, and the hydrogens get a partial positive charge
• Ion-dipole interactions


  • The negative charge on the oxygen interacts with the positive charge on the sodium cation, 'pulling it off of' the NaCl. The H2O is the dipole, and the ion is the Na+ cation.
  • The positive charge on the hydrogens interacts with the negative charge on the chloride anion. The H2O is the dipole, and the ion is the Cl- cation.

  
• Hydration


  The water molecules break the ionic bond, pull the Cl- and Na+ ions apart and surrounding them. This allows our ions to be stabilized by a shell of our solvent molecules.

  
• Precipitation


  The opposite of dissolution. Ions come together to fold a solid, which spontaneously falls out of solution.
• When the contents of each beaker are combined, the volume increases and a precipitate (a white solid in this case) forms
• The precipitate forms because the electrostatic attractions of the ionic crystal (AgCl) must be stronger than the forces of hydration, in order to 'draw' these ions out from their interactions with water
• Spectator ions


  Ions that don't take part in the reaction and just remain in solution (e.g. Na+ and NO3+)
• Solubility


  The amount of a substance that can dissolve in a given volume of solvent at a given temperature
• 10g of solid PbCl2 added to 50 mL of water at 25oC, only 0.22 g of the solid goes into solution
• Calculating solubility


  1. Solubility in g/L = 0.22g / 0.0500 L = 4.4 g/L

  2. Solubility in mol/L = 0.00079 mol / 0.0500 L = 0.016 mol/L = 0.016 M
• Solubility product constant (Ksp)


  A constant that describes the equilibrium between a solid and its dissolved ions in a saturated solution
• Calculating Ksp


  1. Ksp = [Pb2+]1 [Cl-]2

  2. Ksp = (0.016)1 (0.032)2 = 1.6 x 10-5
• The amount of undissolved solid has no influence on molar solubility or Ksp
• Calculating solubility from Ksp

  1. Step 1: Write balanced equation

  2. Step 2: Create ICE table

  3. Step 3: Solve for x using known Ksp

  4. Step 4: Convert molar solubility to solubility in g/L
• Common-ion effect


  Occurs when there are multiple sources of a common ion, which decreases the solubility of a compound
• Calculating solubility with common-ion effect

  1. Set up ICE table with initial concentration of common ion

  2. Solve for molar solubility using Ksp equation
• pH effect on solubility

  Adding acid (H+) can increase solubility if the anion is a strong base, but not if the anion is a weak base
• Complex ion formation


  Occurs when a metal ion forms a complex with a Lewis base, which decreases the concentration of the free metal ion and increases the solubility

  
• Calculating solubility with complex ion formation


  1. Set up equilibrium with complex ion formation

  2. Multiply Ksp and Kf to get new equilibrium constant

  3. Use ICE table to solve for molar solubility
• Ksp


  Equilibrium solubility product constant
• Calculating Ksp


  1. [Ag+]1 [Cl–]1

  2. 1.8 x 10–10 = x2

  3. x = 1.3 x 10–5 M
• Molar solubility of AgCl2

  1.3 x 10–5 M
• Calculating solubility in 3.0 M NH3

  1. Ag+(aq) + 2NH3(aq) <––> Ag(NH3)2+

  2. Kf = 1.6 x 107
• Lewis acid


  Ag+ (electron pair acceptor)

  
• Lewis base


  NH3 (electron pair donor)
• Since the formation constant (Kf) is so high, equilibrium lies far to the right of this equation. Thus, this reaction 'picks up' Ag+ cations and removes them from solution.
• This decreases the concentration of one of the products (Ag+) in the original equation, and thus equilibrium in that equation will shift to the right, which increases the solubility of AgCl(s)
• Net reaction when AgCl is added to NH3

  Ag+ on both sides, so they cancel.
• When you add two reactions together, you multiply their equilibrium constants together to get the new equilibrium constant (K = Ksp x Kf)
• Recall from before: molar solubility of AgCl2 in pure water was very low (1.3 x 10–5 M). Compare this to our new molar solubility (0.14 M) in NH3. Conclusion: the formation of the complex ion Ag(NH3)2 increased the solubility of the slightly soluble compound AgCl2.
• Tools common to all types of stoichiometry

  • Dimensional analysis (allows conversion from unit to unit)
  • Avogadro's number (allows conversion between moles and the number of atoms, molecules, or ions)
  • Molecular weight (allows conversion between mass and moles)
• Calculating mole fraction


  1. Assume 100g of substance, 12g of solute, 88g of water

  2. Determine molecular weights

  3. Use g/mol and g to find mol of each

  4. Mole fraction = moles of substance of interest / (moles of substance of interest + moles of other substance)
• Calculating the number of particles

  1. Determine MW of compound

  2. Use g/mol and g to find mol

  3. Find mol in 6 mL of solution

  4. Multiply by Avogadro's number to find number of molecules or ions
• Solving across the equation using stoichiometric coefficients

  1. Find mol of H2 needed

  2. Find mol of NaOH needed

  3. Find L of NaOH needed from mol and M
• Empirical formula


  The simplest ratio between the elements in a compound
• Molecular formula


  The actual number of atoms in a compound
• Finding empirical formula


  1. Numerator is mass composition given

  2. Denominator is taken from periodic table (g/mol)

  3. Multiply to get whole number ratios

  4. Divide by anything common to find empirical formula