K Use (D denotes equillibrium) Name Equation In General Units/Assume Other
Kc H2(g) + I2(g) D 2HI(g) Equilibrium Constant Kc =   [HI]2     Kc = [products]Pmoles No units in this case  
             [H2][ [I2]        [reactants]Rmoles    
Kc 3In2(aq) D 6In(aq) Equilibrium Constant Kc = [In(aq)]6 Units are Units = mol3 dm-9  
      .     [In2(aq)]3 (mol dm-3)(pmoles/rmoles)    
Kc resin-H+(s) + Na+ D resin-Na+(s) + H+(aq) Partition Coefficient Kc =    [CO2(aq)]     Kc = [solvent B] No units in this case Pressure->shift to
  CO2(g) D CO2(aq) (special case of Kc)       Pressure of CO2(g)        [solvent A] Equation not required RHS, Kc increase
Kow NH3(aq) D NH3(CHCl3) (special case of Kc = [NH3(CHCl3)] Kow = [in organic solvent] No units in this case  
    partiton coefficient)        [NH3(aq)]          [in water] Assume: excess solvent  
   
Ksp CaCO3(s) D Ca2+(aq) + CO32-(aq) Solubility Product Ksp = [Ca2+(aq)][CO32-(aq)] Ksp = [anion]max[cation]max Units = mol2 dm-6  
        Units= mol2dm-6 (normally) Assume: excess solid [Ca2+]=[CO32-]
   
Ka CH3COOH(aq) D CH3COO-(aq) + H+(aq) Acid Dissociation Ka = [H+(aq)] [CH3COO-(aq)] Ka = [H+(aq)] [A-(aq)] Units = mol dm-3 or Ka =[H+(aq)] x [salt]
  ethanoic acid (weak acid HA) D sodium ethanoate (salt A-) + H+ Constant            [CH3COOH(aq)]            [HA(aq)] Assume: excess A- (from salt), [HA] constant                          [acid] .[H+] constant
Kw H2O(l) D H+(aq) + OH-(aq) Ionic Product of water Ka = [H+(aq)] [OH-(aq)] Simplified to [H+(aq)]2 Assume: excess water  @298K, Kw = 10-14
  or 2H2O(l) D H+(aq) + OH-(aq) (special case of Ka)                [H2O(l)] ([H+] = [OH-] = 10-7mol dm-3) Note pH= -log[H+]= -log Ö(Kw)  = 7