Below are some thermochemical data for the dissolutionequilibrium of solid aluminum hydroxide in water:

	Al(OH)3(s)	→	Al3+(aq)	+	3OH−(aq)
ΔfH∘/kJ⋅mol−1	−1284		−538.4		−230.02
S∘/J⋅mol−1⋅K−1	85.4		−325		−10.9

Part A Compute ΔrH^∘ (Express youranswer in kilojoules per mole as an integer.)

Part B Compute ΔrS^∘ (Express youranswer in joules per mole per kelvin to one decimal place.)

Part C Compute ΔrG^∘ at 298 K(Express your answer in kilojoules per mole as an integer.)

Part E Determine the equilibrium constant forthe dissolution of Al(OH)₃ at 298 K

Part F Determine the Gibbs energy change forthe dissolution of Al(OH)₃ when the concentrations ofAl³⁺ and OH^− are both1.20×10^−7molâ‹…L^−1 (Express your answer inkilojoules per mole as an integer.)

Part H At what temperature willAl(OH)₃(s) be at equilibrium with Al³⁺ andOH^−, each at 1.20×10^−7molâ‹…L^−1?

Part I Explain why the entropy of dissolutionfor this reaction, solid Al(OH)₃(s) dissolving to formAl³⁺ and 3OH^−, is so negative. Match thewords in the left column to the appropriate blanks in the sentenceson the right.

lose, holds on to, low, is releasedfrom, gain, high

The Al ³⁺ ion has a quite __ charge so that it __water molecules, which causes the water molecules to __translational and rotational degrees of freedom so that the entropyassociated with Al ³⁺ in the aqueous phase is verynegative.