##### J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 17.84

Use thermodynamic data from Appendix C to determine Keq for the following reaction

Mg(OH)2(s) + 2 NH4+(aq) Mg2+(aq) + 2 NH3(aq) + 2 H2O(l)

Then obtain Keq from other tabulated equilibrium constants in Appendix C, and compare the results.

Assume 298 K.

Gfo (Mg2+(aq)) = –454.8 kJ/mol

Gfo (NH3(aq)) = –26.57 kJ/mol

Gfo (H2O(l)) = –237.2 kJ/mol

Gfo (NH4+(aq)) = –79.31 kJ/mol

Gfo (Mg(OH)2(s)) = –833.9 kJ/mol

Go = [(–454.8) + 2(–26.57) +2(–237.2)] – [(–833.9) + 2(–79.31)] = 10.2 kJ

Go = –RTlnKeq

10200 = –(8.314)(298)lnKeq

Keq = e–4.12 = 1.6×10–2

To use published equilibrium constants, a series of reactions must be used:

Mg(OH)2(s)Mg2+(aq) + 2 OH(aq)    Ksp = 1.8×10–11

NH4+(aq) + H2O(l)H3O+(aq) + NH3(aq)    Ka = 5.6×10–10

NH4+(aq) + H2O(l)H3O+(aq) + NH3(aq)    Ka = 5.6×10–10

H3O+(aq) + OH(aq)2 H2O(l)    Kc = 1/Kw = 1/1.0×10–14 = 1.0×1014

H3O+(aq) + OH(aq)2 H2O(l)    Kc = 1/Kw = 1/1.0×10–14 = 1.0×1014

Summing the reactions gives the desired reaction with

Kc = Ksp×Ka×Ka×1/Kw×1/Kw = (1.8×10–11)(5.6×10–10)(5.6×10–10)(1.0×1014)(1.0×1014) = 5.6×10–2

The two equilibrium constants are the same order of magnitude, but differ by about a factor of 3.5. This is not too bad.