Question

sodium hydroxide dissolves very easily, and exothermically, in water. Using values from the appendix in your text, calculate the Ksp for this salt, as well as the temperature at which this reaction becomes spontaneous. Does the temperature need to be above or below this value in order for the process to proceed without outside intervention? Na0H(s): -379.4 kJ/mol

Answer 1

(Note: please check values from your text. In this solution, the values used may not be exactly same as that in your text book but would be very close to it).

The equation for dissolution of NaOH in water can be given as follows:

NaOH (s) <===> Na⁺ (aq) + OH^- (aq)

G for this process can be given as follows:

              = [(1 mol)(-261.90 kJ/mol)+(1 mol)(-157.24 kJ/mol)] - [(1 mol)(-379.4 kJ/mol)]

              = -39.74 kJ.

similarly,

                            = [(1 mol)(-240.12 kJ/mol)+(1 mol)(-229.99 kJ/mol)] - [(1 mol)(-425.93 kJ/mol)]

                             = -44.18 kJ

Also,

                       = [(1 mol)(59 J/mol)+(1 mol)(-10.75 J/mol)] - [(1 mol)(64.44 J/mol)] = -16.19 J