Question

Reaction 1: NH4+(aq) + H2O(l)NH3 + H3O+(aq) pKA = 9.2

the system described above, it has been found that the activity coefficients for the species in solution are: (NH4+) = 0.80, (H3O+) = 0.90, (OH-) = 0.88 – the activity coefficients for neutral molecules are 1.0.

a.) Write down the equilibrium expression for Reaction 1 in terms of activity coefficients and concentrations – i.e. the correct equilibrium expression which is NOT just in terms of concentrations.

b.) Now algebraically manipulate this expression so the right-hand side of it is only in terms of concentrations (i.e. algebraically move the activity coefficients to the other side of the equation) and calculate the new “effective” KA’ and pKA’ for this reaction, in terms of concentrations, under these specific conditions.

c.) Utilize the ICE Table Method to calculate the equilibrium concentrations of each species (as well as the pH – recall that pH = -log10 AH3O+) presented in the chemical equation.

Answer 1

a) NH⁺₄ [ aq] + H₂O [l] NH₃[g] + H₃O⁺ [ aq]

Equilibrium expression ( concentration ) = Products in gaseous or aqueous form/ Reactants in gaseous or aqueous form

that gives =( [ H₃O⁺] [ NH₃] )/ [ NH⁺₄]

Equilibrium expression ( activity coefficient ) =

For this, activity coefficient tells us the deviation from the ideal state. For solid and liquid state, it is 1

[ A_H3O⁺] [ A _NH3] / [ A_NH4+] [ A_H2O] = [0.90] [ A _NH3] / 0.80             - eq1

NH₃ + H₂O NH₄ + OH-                    - eq2                                             

Activity coefficient of NH₃ = [1/2] C_X Z²_x        - eq3

Where C_x = activity coefficient of component X at RHS & Z_x= Ionic charge of component X at RHS

So from eq 2 and eq 3

Since NH₄ and H₂O has zero ionic charge ,

Activity coefficient of NH₃ = [1/2] [ A_OH^- * ( -1)² ]= 0.88/2= 0.44       - eq4

Putting value from eq4 in eq1

K ( activity coefficient)= ([0.90] [ 0.44])/ [ 0.80] = 0.495 M , where M= Molar concentration