Question

It has been awhile since I have taken Chemistry and I have an equilibrium problem that I need help with.

Borax is a sparingly soluble salt and dissolves in water according to the following reaction: Na₂B₄0₇ * 10 H₂O(s) <----> 2Na+(aq)+B₄0₅(OH)₄^2-(aq)+8H₂O(l)    We can write an equilibrium expression for this dissolution as follows:   K=[Na+]²[B₄O₅(OH)₄^2-]

This is given in my lab book but I am confused as to how this reaction and equilibrium expression was determined. What I need is a good step by step explanation for how each were reached.

Answer 1

Na2B407 * 10 H2O(s) <----> 2Na+(aq)+B405(OH)42-(aq)+8H2O(l)
Na2B407 * 10 H2O is sparingly soluble in water. It dissolve in water to form ions.Here the equilibrium
exisits between sparingly soluble salt and ions. At equilibrium concentration of reactants and concentration
of products are equal. So at equilibrium rate of farward reaction and rate of back ward reaction are equl.
Rate of farward reaction [Na2B407 * 10 H2O]
rate of farward reaction =Kf [Na2B407 * 10 H2O]
Rate of backward reaction [Na+(aq)]²[B405(OH)42-(aq)][H2O(l)]⁸
rate of backward reaction =Kb[Na+(aq)]²[B405(OH)42-(aq)][H2O(l)]⁸
At equilibrium rate of farward reaction = rate of backward reaction
               Kf [Na2B407 * 10 H2O]    =Kb[Na+]^2[B405(OH)42-][H2O]^8
               Kf/Kb                    = [Na+]^2[B405(OH)42-][H2O]^8/[Na2B407 * 10 H2O]
                               Kc       = [Na+]^2[B405(OH)42-][H2O]^8/[Na2B407 * 10 H2O]
                                              Kc= Kf/Kb
                   but concentration of solids and liquids are equlto 1
                   [Na2B407 * 10 H2O] = 1
                   [H2O]               =1
                     Kc                =[Na+]²[B405(OH)42-]