Question

Consider the following data for questions A–F: Two solutions of an unknown slightly soluble salt, A(OH)2, were allowed to equilibrate—one at 25 °C and the other at 80 °C. A 15.00 mL-aliquot of each solution is titrated with 0.320 M HCl. 6.19 mL of the acid is required to reach the endpoint of the titration at 25 °C, while 60.14 mL mL are required for the 80 °C solution.

QUESTION: Calculate the Gibbs free energy (in kJ/mol) at 80 degrees Celsius.

I keep getting -0.171 kJ/mol when using the Gibbs free energy equations and entering the Ksp=1.06. For some reason my answer is wrong and I can't figure out where I'm going wrong. Help is much appreciated :)

Answer 1

Gibbs free energy at 80^oC = -0.160 kJ/mol

Explanation

At 80^oC, volume of HCl = 60.14 mL

moles H⁺ = (concentration HCl) * (volume of HCl)

moles H⁺ = (0.320 M) * (60.14 mL)

moles H⁺ = 19.2448 mmol

moles OH^- = moles H⁺

moles OH^- = 19.2448 mmol

[OH^-] = (moles OH^-) / (volume of aliquot)

[OH^-] = (19.2448 mmol) / (15.00 mL)

[OH^-] = 1.283 M

[A^-] = [OH^-] / 2

[A^-] = 0.6415 M

K_sp = [A^-][OH^-]²

K_sp = (0.6415 M) * (1.283 M)²

K_sp = 1.05593

Gibbs free energy, $\Delta$ G^o = -(R) * (T) * ln(K_sp)

$\Delta$G^o = -(8.314 J/mol-K) * (353 K) * ln(1.05593)

$\Delta$G^o = -159.7 J/mol

$\Delta$G^o = -0.160 kJ/mol