In this experiment, you will determine the values of ∆H° and ∆S° for the reaction which occurs when borax (sodium tetraborate octahydrate) dissolves in water.

In previous experiments, you have determined ∆H° values directly, by measuring temperature changes when the reaction occurred. However, in many cases, this technique is not practicable. For example, the reaction may not go to completion, or it may give off such a small amount of heat that the temperature change is too small to measure. In addition, there is no direct method for measuring ∆S° for a reaction. It is therefore useful to be able to determine ∆H° and ∆S° indirectly, by using their relationship to the equilibrium constant of a reaction.

The equilibrium constant of any reaction can be related to the free energy change of the reaction:

∆G° = –RT ln K

The free energy change is also related to the enthalpy and entropy changes during the reaction:

∆G° = ∆H° – T∆S°

Combining these two equations gives the general relationship between K, ∆H°, and ∆S°:

–RT ln K = ∆H° – T∆S°

Dividing both sides by –RT gives a particularly useful form of this relationship:

This represents a linear equation of the form y = mx + b. In this case, y = ln K and x = 1/T; a plot of ln K against 1/T will therefore be linear. In addition, the slope of this line (m) will equal –(∆H°/R), and its y-intercept (b) will equal (∆S°/R). It is therefore possible to determine ∆H° and ∆S° by simply measuring the equilibrium constant at two different temperatures, graphing ln K against 1/T, and measuring the slope and intercept of the resulting line. In practice, K is measured at several temperatures, so that the effect of any experimental errors in one measurement will be minimized.

The reaction you will study is the dissolution of borax (sodium tetraborate octahydrate) in water. “Borax” is a naturally occurring compound; it is in fact the most important source of the element boron, and it has been used for many years as a water softening agent. Borax is a rather complicated ionic salt which has the chemical formula Na2B4O5(OH)4⋅8 H2O. When it dissolves, it dissociates as follows:

Na2B4O5(OH)4⋅8 H2O (s) → 2 Na+ (aq) + B4O5(OH)4 2– (aq) + 8 H2O (l) (1)

Notice that the products of this reaction are two sodium ions and one other ion (this ion is called “tetraborate”), along with the eight molecules of water. Since water does not appear in equilibrium constant expressions, the K expression for this reaction is:

K = [Na+]2[ B4O5(OH)4 2– ]

You will measure K by analyzing a saturated solution of borax (i.e. a solution in which Reaction (1) has come to equilibrium!) for the tetraborate ion. Tetraborate is a weak base, so it can be titrated with a strong acid. It may surprise you that tetraborate can react with only two hydrogen ions -- not four! -- and that in this reaction, the tetraborate ion “falls apart”, producing four molecules of boric acid:

B4O5(OH)4 2– (aq) + 2 H3O+ (aq) + H2O (l) → 4 H3BO3 (aq) (2)

Once you know the number of moles of tetraborate in the solution, you can calculate the number of moles of sodium ion by using the stoichiometry of Reaction (1). Then, you can calculate the molar concentrations of the two ions and, finally, the value of K.

Procedure Handout

Please read carefully the lab procedure before moving on to the pre-lab quiz. The pre-lab quiz is specifically focused around a detail of the lab procedure. The handout is found at the bottom of this page as an attachment.