Consider the ionisation of a monoprotic acid (HX) in water and in the gas phase at 298 K.

HX(aq)  ⇌  H⁺(aq) + X^-(aq)       ΔG°_water = +15.40 kJ mol⁻¹

HX(g)    ⇌  H⁺(g)   + X^-(g)        ΔG°_gas = +1050.50 kJ mol⁻¹

Please give your numerical answers to 2 decimal places.

1. Calculate the pK_a of HX in water. [1 mark]
   pK_a =

2. The degree of ionisation refers to the percentage of acid that exists in the ionised form. Calculate the degree of ionisation of a 0.8 mol L-1 aqueous solution of HX. [1 mark]
   % ionisation = 

3. A thermodynamic cycle can be constructed to relate the gas phase and aqueous phase Gibbs free energies of ionisation as shown below.

   

   According to this cycle, ΔG

   ^o_water can be expressed as some linear combination of ΔG^o_gas and the hydration free energies (ΔG°_hyd) of the reactant and products. [1 mark]
   The correct expression for ΔG°_water 

   is

   ΔG°_gas - ΔG°_hyd(H⁺) - ΔG°_hyd(X^-) + ΔG°_hyd(HX)

   ΔG°_gas + ΔG°_hyd(H⁺) + ΔG°_hyd(X^-) - ΔG°_hyd(HX)

   ΔG°_gas - ΔG°_hyd(H⁺) - ΔG°_hyd(X^-) - ΔG°_hyd(HX)

   ΔG°_gas + ΔG°_hyd(H⁺) + ΔG°_hyd(X^-) + ΔG°_hyd(HX)

4. Using the relationship in part (b), and given that the ΔG°_hyd(X^-) and ΔG°_hyd(HX) are -400 and -20 kJ mol^-1 respectively, determine the value of ΔG°_hyd(H⁺) [1 mark]
   kJ mol^-1.

5. The reason why HX is a stronger acid in water than in the gas phase is because [1 mark]
   the acid is more stable in the gas phase

   the ionic products have large negative hydration free energies

   The H-X bond is weaker in water

   the entropy of the reaction is very positive in water