Question

Consider the following data: Two solutions 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 0.250 M HCl. 6.49 mL of the acid is required to reach the endpoint of the titration at 25 degrees Celsius, while 63.07 mL are required for the 80 C solution.

1. Assuming that the change in enthalpy is negligible over this temperature range, calculate deltaH.

2. Assuming that the change in entropy is negligible over this temperature range, calculate deltaS.

Answer 1

Step 1: Calculate Ksp for the reaction at both the given temperatures:

                                               A(OH)_{2 (s)}$\rightleftharpoons$ 2 OH^-_(aq) + A ⁺_(aq)

concentration at equilibrium                                          2x                   x

and Ksp = (2x)² (x)   = 4x³

At 25 °C , mmoles of HCl used = 0.250 M x 63.07 mL =15.77 = mmoles of OH^- present.

total volume (in ml) = 15 + 63.07 = 78.07

[OH^-] = 15.77/78.07 = 0.202 M   = 2x , $\Rightarrow$ x = 0.101 M

$\Rightarrow$Ksp = 2.19 x 10^-4

At 80 °C , mmoles of HCl used = 0.250 M x 6.49 mL =1.62 = mmoles of OH^- present.

total volume (in ml) = 15 + 6.49 = 21.49

[OH^-] = 1.62/21.49 = 0.075 M   = 2x , $\Rightarrow$ x = 0.038 M

$\Rightarrow$Ksp = 4.12 x 10^-3

Step 2: Calculate
G T
at both temperatures, using the formula:

G T
= -RT ln (K)

At 25 °C, T= 298 K and R = 8.314 J/mol. K

= -(8.314 J/mol. K )x 298 K ln ( 2.19 x 10^-4) = 20. 9 kJ/mol

G T

Similarly,

At 80 °C, T= 353 K and R = 8.314 J/mol. K

= -(8.314 J/mol. K )x 353 K ln ( 4.12 x 10^-3) = 16.12 kJ/mol

G T

Step 3: Using Gibbs Helmholtz equation, that is:

            

а) ΔΗ т? dT

And, Assuming that $\Delta H$ is independent of Temperature in the given range, we can simplify the above equation as:

$d\frac{\Delta G}{T} =- \frac{\Delta H}{T^{2}} dT$

and integrating the above from T₁ to T₂,

$\frac{\Delta G_{2}}{T_{2}} - \frac{\Delta G_{1}}{T_{1}} = \Delta H (\frac{1}{T_{2}} -\frac{1}{T_{1}})$

Putting the values of obtained above for T₁ = 298K and T₂ = 353 K, we have

G T

$\frac{16.12}{353} - \frac{20.9}{298} = \Delta H (\frac{1}{353} -\frac{1}{298})$

which gives, $\Delta H = 46.80 kJ/mol$

Step 4: To calculate entropy change,

Use the equation;

$\Delta G ( T_{2})= \Delta H - T_{2}\Delta S$               ...... Note that H and S are assumed constant for the given Temperature range.

$\Delta G ( T_{1})= \Delta H - T_{1}\Delta S$

Subtract the second equation from the first one,

$\Delta G ( T_{2}) -\Delta G ( T_{1})= T_{1}\Delta S-T_{2}\Delta S$

or,

$\frac{\Delta G ( T_{2}) -\Delta G ( T_{1})}{(T_{1}-T_{2})}= \Delta S$

Putting the values in above equation,

$\Delta S = \frac{16.12-20.9}{(298-353)}$

$\Rightarrow \Delta S = 86.9 J/K$

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