The thermodynamics of the solubility of borax Determination of ΔH° and ΔS°

Borax Borax has the formula Na 2 [B 4 O 5 (OH) 4 ].8H 2 O [B 4 O 5 (OH) 4 ] 2- is the tetra-borate anion Borax is a weak base and reacts with acid Na 2 B 4 O 7 ·10H 2 O + 2 HCl → 4 H 3 BO NaCl + 5 H 2 O H 3 BO 3 is the conjugate acid (Boric acid) which as a pH around 4 or so depending on the molarity It is a good water softener Ca 2+ (aq) + Na 2 B 4 O 7 (aq) → CaB 4 O 7 (s)↓ + 2 Na + (aq) Mg 2+ (aq) + Na 2 B 4 O 7 (aq) → MgB 4 O 7 (s)↓ + 2 Na + (aq)

Purpose To determine the thermodynamic quantities ΔH° and ΔS°,for the solvation reaction of borax in water Na 2 [B 4 O 5 (OH) 4 ].8H 2 O(s) ↔ 2 Na + (aq) + [B 4 O 5 (OH) 4 ] 2- (aq) + 8 H 2 O (l) by measuring the solubility product constant, K sp, over the temperature range 50−15°C The temperature dependence of the equilibrium constant K sp depends on the enthalpy of solvation ΔH° and the entropy of solvation ΔS°

What is ΔS° ? Imagine the solvation is an elementary step reaction Na 2 [B 4 O 5 (OH) 4 ].8H 2 O (s) 2 Na + (aq) + [B 4 O 5 (OH) 4 ] 2- (aq) + 8 H 2 O (l)

What is ΔS° ? ΔS° is called the entropy of solvation we can see it is somehow related to the ratio of the collision factors for reaction in the forward and backward direction If A 1 > A 2 ΔS° > 0 If ΔS° > 0 we say that process is spontaneous, it means that it is more probable for the reactants to come together to react, than for the products to come together and react to make reactants

What is ΔS° ? Consider what happens when the borax solid dissolves in water? When the orange particles dissolve in water two things can happen can hang around near the crystal and potentially re-attach themselves they can move off further away from the crystal It’s like reaching a crossroad where the road is going 4 ways. You randomly choose a road – 3 take you further away and 1 takes you back

What is ΔS° ? You are 3 times more likely to leave than return home In the same way when borax dissolves there are more choices which take it into solution than back to solid In this case A 1 > A 2 ΔS° > 0 Entropy is measuring the number of choices available to the system

Objective: knowing K sp (T) If we know the quantities ΔH° and ΔS° then we now how the equilibrium constant changes with T Make saturated solutions of borax in water at different temperatures Measure the concentration of the tetraborate x=[B 4 O 5 (OH) 4 ] 2- in the solution (by titration) Determine K sp at that temperature T using K sp = 4x 3 (ICE table)

Objective: knowing K sp (T) Plot the ln(K sp ) vs 1/T (where T is in Kelvin) Should give a straight line graph Where

Preparing a Saturated Borax Solution 20 g Borax in 80 mL of deionized H 2 O Stirrer and hotplate thermometer Heat to 52 o C-55 o C DO NOT LET IT GET ABOVE 55 o C Leave it at 52 o C-55 o C for 30 mins While it is heating use a pipette to measure precisely 5.00 mL of water into each of six small test tubes and mark the levels with a wax pencil Label the test tubes ~ 50°, ~ 45°, ~ 35°, ~ 30°, ~ 20°, and ~15 °C After 30 mins remove the beaker from hot plate As it cools decant 5 mL into test tubes at ~ 50°, ~ 45°, ~ 35°, ~ 30°, ~ 20°, and ~15 °C Record the actual temps to nearest 0.1 o C

Standardized HCl Solution While 2 students are making the saturated solutions and marking test tubes, the other student(s) will make a standardized HCl solution for later titrations In a fume hood, add 8 mL of concentrated HCl to about 400mL of distilled Water in a 500mL Erlenmeyer Flask. Stir well. This gives a solution of approximately 0.2M HCl. To determine the exact concentration of the HCl in the solution, we will titrate it against a base Na2CO3 whose mass can be accurately measured and whose endpoint is pH=4 Na 2 CO 3 (aq) + 2 HCl(aq)  2 NaCl(aq) + H 2 CO 3 (aq) Since the endpoint is at pH = 4 we use bromocresol green

Standardized HCl Solution HCl to Standardize Erlenmeyer Flask 0.15 g anhydrous Na 2 CO 3 50 mL deionized water 12 drops of bromocresol green indicator endpoint

Standardization

Calculation of K sp

Getting ΔH° and ΔS° Plot ln(K sp ) vs 1/T where T is in Kelvin ΔH° = -slope x J/mol/K ΔS° = intercept x J/mol/K