Consider the following data for questions 1-6: Two solution of an unknown slight

Question

Consider the following data for questions 1-6: Two solution of an unknown slightly soluble salt, A(OH)2, were allowed to equilibrate- one at 25 degrees Celsius and the other at 80 degrees celsius. A 15.00 mL aliquot of each solution is titrated with 0.200 M HCL. 8.00 mL of the acid is required to reach the endpoint of the tiration at 25 degress celsius, while 69.15 mL are required for the 80 degrees Celsius solution.

1. Calculate the Ksp^a+ at 25 degrees celsius.

2. Calculate the Ksp at 80 degrees celsius.

3. Calculate the gibbs free energy ( in kJ/mol) at 25 degrees celsius

4. Calculate the Gibbs free energy (in kJ/mol) at 80 degrees celsius.

5. Assuming that the change in enthalpy is negligible over this temperature range, calculate delta H.

6. Assuming that the change in entropy in negligible over this temperauter range, calculate delta S.

Explanation / Answer

Reaction A(OH)2 ----> A+ + 2OH-

at equilibrium x 2x

so at 25C 8ml of 0.2 M HCl was required so moles of HCl

= 8*0.2 = 1.6m mol

so 2x *15= 1.6m mol

x=0.053M

ksp = [A+][OH-]^2

ksp = x*(2x)^2

ksp = 5.955*10^-4....................................ANS

so at 80C 69.15ml of 0.2 M HCl was required so moles of HCl

= 69.15*0.2=13.83m mol

so 2x *15= 13.83m mol

x=0.461M

ksp = [A+][OH-]^2

ksp = x*(2x)^2

ksp = 0.392.................................ANS

For T=25C

dG = -RTlnKsp

T = 298K = 20C

R=8.314 j/mol-k

dG = -8.314*298*ln(5.955*10^-4) = 18398.72 j

dG = 18.399 KJ

For T=80C

dG = -RTlnKsp

T = 353K

R=8.314 j/mol-k

dG = -8.314*353*ln(0.392) = 2748.46j

dG = 2.748KJ

For T = 298K = 20C

dG = dH -TdS

18.399 = dH - 298dS

for T = 353K

2.784 = dH-353dS

Solving above equations

dH = 103.0039KJ

dS = 0.2839 KJ/K

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