Mod 5 Solubility Equilibria

It is only when the solution is saturated, that the dissolution reaction is equilibrium. Water is a polar molecule, where it is slightly positive at the hydrogen ends and slightly negative on the oxygen end. These charged ends of the molecule will attract the opposite charge in the ionic compound. However, not all ionic compounds are soluble in water or are partially soluble in water. This is because the ion dipole forces between ions and the polar molecule is not strong enough to break the ionic bonds that holds the ions together in the lattice

A solution in which the maximum amount of solvent has been dissolved. Any more solute added will sit as crystals on the bottom of the container.

Max amount of solute that dissolves in a specific amount of solvent at a particular temp.

Substance that is dissolved in a solvent, forming a solution

Something that dissolves

Solubility is typically the mass of solute/volume of saturated solution, molar solubility is typically number of moles/ volume (L)

Mass of substance (m)/ volume of substance (v)

mass of substance (in mg)/ volume of substance ( in kg)

Mass of substance (in micrograms)/ volume of substance (in kg)

g/100g = %w/w

mL/100mL = %v/v

It is the equilibrium constant for the dissolution process of an ionic solid that dissociates completely in solution

AaBb (s) ⇌ aA x+ (aq)+ bB y- (aq)

[A^x+]^a [b^y-]^b

Indicates a highly insoluble solution, as there are less dissociated ions than there are solids.

Indicates a highly soluble solution, as there are more dissociated ions than there are solids.

Essentially same as Ksp but does not have to be in equilibrium

Therefore, system attempts to reach equilbrium through favouring the reaction which reduces aqueous ions –> favouring reverse reaction –> formation of precipitate

Therefore, system attempts to reach equilbrium through favouring the reaction which increases aqueous ions –> favouring forward reaction –> less precipitate, more aqueous ions –> no formation of precipitate

LCP: Equilibrium will shift to minimise the disturbance. When we add water, concentration of ions decrease, disturbing equilibrium and to counteract this change, the equilibrium will shift to favour the forward reaction to increase conc of aqueous ions –> more of solid precipitate will dissolve

Compare valus of Qsp and Ksp to predict direction of change and new equilibrium. When water is added, there is dilution of concentration of aq ions, this indicates decreased Qsp, as Qsp increases to reach Ksp, therefore the concentrations of the products have to increase –> favouring the forward reaction

When water is added, conc of ions decrease, disturbing the equilibrium. As concentrations of ions decrease, frequency of collisions for reverse reactions will decrease, thus equilibrium shifts in the right as the forward reaction is left unchanged, thus equilibrium shifts right (more precipitate dissolving)=

• Write an equation for the dissolution and the corresponding Ksp expression • Determine the ion concentrations from thee moles of the solute using the mole ratios ( • Substitute the concentrations into the equilibrium expression to solve for x

• Write equation for the dissolution and Ksp expression • Make an ICE table in units of M to set out working for calculations • Determine thee ion concentrations using the ICE tables, noting that reactants will have a ‘+x’ • Sub concentrations into equilibrium expression to solve for x