Question

thermodaynamic of borax solubility

1. delta H and Delta S may not be independent of tempreture. how this could effect the graph of borax we ploted?

Answer 1

Thermodynamics of borax solubility

In this experiment, we will investigate the effect of temperature on the solubility of a sparingly soluble salt in this case borax in water. The solubility data will permit the calculation of the equilibrium constant for the reaction, K_sp Then the experimental values of Delta H, Delta S, Delta G, will be found through graphing.

The salt chosen for this experiment is sodium tetraborate decahydrate, Na₂B₄O₇^.10H₂O, better known as borax. Borax is an example of a class of compounds named berates, which contain polyanions composed of trigonal BO₃ and/or BO₄ units linked by bridging oxygen atoms to form chain or ring structures.

Borax is mildly basic, as shown below.

B₄O₅(OH)₄^2- + 5H₂O -> 4H₃BO₃ + 2OH^-

This property provides the means to determine the solubility of borax. We simply titrate a known volume of the saturated borax solution with a standard acid solution and calculate the concentration of the tetraborate anions. The reaction of interest is as follows:

B₄O₅(OH)₄^2- + 3H₂O + 2H⁺ -> 4H₃BO₃

The formation of a saturated borax solution is represented by Equation.

Na₂[B₄O₅(OH)₄] 8H₂O (s) -> 2Na⁺ (aq) + B₄O5(OH)₄^2-(aq)

The equilibrium law expression is therefore as follows:

K_sp=[Na+]²[B₄O₅(OH)₄^2-]

If we let x represent the concentration of tetraborate ions found by titration with acid, then above Equation shows that the concentration of Na⁺ ions must equal 2x. We may substitute these values into the equilibrium law expression as shown in Equation.

K_sp = [2x]²[x]=4x³

If our solutions are ideal and if Delta H is reasonably constant, we can calculate important thermodynamic quantities from the dependence of K_sp on temperature. We begin with the fundamental expressions:

Delta G = Delta H - T Delta S

Delta G= -RT In K_sp

These may be combined and rearranged as follows:

-RT ln K_sp = Delta H - T Delta S

ln K_sp = Delta H/RT + Delta S/R

Note that the above Equation is of the form

y = mx+b

where y = ln K_sp, x = 1/T, m = - Delta H/R, and b= Delta S/R

A plot of ln K_sp versus 1/T should therefore be linear, with a slope

equal to -Delta H/R and an intercept of Delta S/R.