Chemistry Labs

Problem 3

Two binding centers — competition or cooperation? A host molecule H has two binding centers aa and bb with different affinities for a guest molecule G: H+G⇌HGa\ce{H + G <=> HGa} (KaK_a) and H+G⇌HGb\ce{H + G <=> HGb} (KbK_b), where HGa and HGb denote complexes with G bound at center aa and bb respectively. Attachment of one G can change the second center's affinity, described by the interaction factor β\beta: the second binding step HGa+G⇌HGX2\ce{HGa + G <=> HG2} (G now binds center bb) has equilibrium constant βKb\beta K_b (HG2 is the doubly bound complex). (a) Give the range of β\beta values corresponding to cooperation, competition, and independence of the centers. (b) Find the equilibrium constant of HGb+G⇌HGX2\ce{HGb + G <=> HG2} in terms of KaK_a, KbK_b, β\beta. (c) A solution was prepared with [H]0=1[\ce{H}]_0 = 1 and [G]0=2[\ce{G}]_0 = 2 mol dm−3^{-3}. At equilibrium [H][\ce{H}] decreased tenfold and [G][\ce{G}] fourfold. Given Kb=2KaK_b = 2K_a, find the concentrations of all species and the values of KaK_a and β\beta. (d) 1 mol of H and some amount of G were dissolved to give 1 dm3^3 of solution in which [HGX2]=[HGa]+[HGb][\ce{HG2}] = [\ce{HGa}] + [\ce{HGb}]. With KaK_a, KbK_b and β\beta from part (c), find the initial amount of G.
Step 2 of 5: Constant for the second binding at center aa
HGb+G⇌HGX2:K=βKa\ce{HGb + G <=> HG2}: \quad K = \beta K_a
Analysis

The second guest now binds center aa; by the definition of β\beta its constant is β\beta times the unperturbed constant of center aa, i.e. K=βKaK = \beta K_a. This is consistent with the thermodynamic cycle Ka⋅βKb=Kb⋅βKaK_a \cdot \beta K_b = K_b \cdot \beta K_a.