Question

Thermodynamics of Dissolution of Substance

Reaction Eqn of borax and HCL

B4O5(OH)4 2-(aq) + 2HCl + 3H2O <_> 4H3BO3 (aq) + 2Cl-

Temp	TRIAL ONE Molarity of HCL	Volume Borax 1	Initial Buret Reading	Final Buret Reading	TRIAL TWO Molarity of HCL	Volume Borax 2	Initial Buret Reading	Final Buret Reading
(°C)	(M)	(mL)	(mL)	(mL)	(M)	(mL)	(mL)	(mL)

25

0.25

5

18

20.45

0.25

5

21

23.5

calculate

Moles of HCL	Moles of Borate	Concentration of Borate	Ksp	Average Ksp	ΔG	ΔH	ΔS
(moles)	(moles)	(M)			(Kj/mol)	(Kj/mol)	(J/mol k)

** I understand that for ΔH ΔS, you need to plot ln(Ksp) vs 1/T and determine their value from that graph.

Values for enthalpy and entropy of the dissolution of borax in water are 110 kJ/mol and 380 J/mol*K respectively. Do your values agree? Calculate your percent error.

You assumed that H and S are constant over the temperature range studied. Are they?

Answer 1

trial 1

voulme of Hcl = 2.45

moles of acid = 2.45 *.25 =6.125* 10^-4 moles

by stoichiometry moles of borax = 6.125* 10^-4 / 2 = 3.06 * 10^-4 moles

moles of boric acid = 2* 6.125* 10^-4 =1.225* 10^-3 moles

volume of water = .08 ml

so Conc of boric acid = 1.225* /7.45 = .164

Ksp = .164^3/.25*.125

= .141

trial 2

voulme of Hcl = 2.5

moles of acid = 2.5 *.25 =6.25* 10^-4 moles

by stoichiometry moles of borax = 6.25* 10^-4 / 2 = 3.125 * 10^-4 moles

moles of boric acid = 2* 6.25* 10^-4 =1.25* 10^-3 moles

volume of water = .08 ml

so Conc of boric acid = 1.25/7.45 = .167

Ksp = .167^3/.25*.125

= .149

Average Ksp = (.149+.147)/2 = .148

from graph we gett slope = -12.204 = - H/R

H= 101.4 KJ /mol

percentage error = (110 - 101.4)*100/110 = 7.81 %

intercept = S/R= 43.8

S= 43.8 * 8.314 = 364.15

Error = (380 - 364.5 )* 100 /380 = 4.07 %