equilibrium and thermodynamic study of borax; It has been awhile since I have ta
ID: 912358 • Letter: E
Question
equilibrium and thermodynamic study of borax;
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)42-(aq)+8H2O(l) We can write an equilibrium expression for this dissolution as follows: K=[Na+]2[B4O5(OH)42-]
This is given in my lab book but I am confused as to how this equilibrium expression was determined. It seems as if though just the "2Na+(aq)+B405(OH)42-(aq)" chunk was pulled out of the reaction above.... what am I missing? how is this an equilibrium expression?, where does the "Na2B407 * 10 H2O(s)" and "+8H2O(l)" from the reaction go? please explain the steps for coming to this equilibrium expression.
Explanation / Answer
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 2Na+(aq)+B405(OH)42-(aq)][8H2O(l)]
rate of backward reaction =Kb[2Na+(aq)+B405(OH)42-(aq)][8H2O(l)]
At equilibrium rate of farward reaction = rate of backward reaction
Kf [Na2B407 * 10 H2O] =Kb[Na+(aq)]^2[B405(OH)42-(aq)][H2O(l)]^8
Kf/Kb = [Na+(aq)]^2[B405(OH)42-(aq)][H2O(l)]^8/[Na2B407 * 10 H2O]
Kc = [Na+(aq)]^2[B405(OH)42-(aq)][H2O(l)]^8/[Na2B407 * 10 H2O]
Kc= Kf/Kb
but concentration of solids and liquids are equlto 1
[Na2B407 * 10 H2O] = 1
[H2O] =1
Kc =[Na+(aq)]^2[B405(OH)42-(aq)]
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