Problem 3

Two binding centers — competition or cooperation? A host molecule H has two binding centers

a

and

b

with different affinities for a guest molecule G:

\ce{H + G <=> HGa}

(

K_a

) and

\ce{H + G <=> HGb}

(

K_b

), where HGa and HGb denote complexes with G bound at center

a

and

b

respectively. Attachment of one G can change the second center's affinity, described by the interaction factor

\beta

: the second binding step

\ce{HGa + G <=> HG2}

(G now binds center

b

) has equilibrium constant

\beta K_b

(HG2 is the doubly bound complex). (a) Give the range of

\beta

values corresponding to cooperation, competition, and independence of the centers. (b) Find the equilibrium constant of

\ce{HGb + G <=> HG2}

in terms of

K_a

K_b

\beta

. (c) A solution was prepared with

[\ce{H}]_0 = 1

and

[\ce{G}]_0 = 2

mol dm

^{-3}

. At equilibrium

[\ce{H}]

decreased tenfold and

[\ce{G}]

fourfold. Given

K_b = 2K_a

, find the concentrations of all species and the values of

K_a

and

\beta

. (d) 1 mol of H and some amount of G were dissolved to give 1 dm

^3

of solution in which

[\ce{HG2}] = [\ce{HGa}] + [\ce{HGb}]

. With

K_a

K_b

and

\beta

from part (c), find the initial amount of G.

Step 4 of 5: Extract

K_a

K_b

\beta

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K_a = \frac{[\ce{HGa}]}{[\ce{H}][\ce{G}]} = \frac{0.1}{0.1 \times 0.5} = 2;\quad K_b = 4;\quad \beta K_b = \frac{[\ce{HG2}]}{[\ce{HGa}][\ce{G}]} = \frac{0.6}{0.05} = 12 \Rightarrow \beta = 3

Analysis

Reading off the constants: $K_a = [\ce{HGa}]/([\ce{H}][\ce{G}]) = 0.1/(0.1\times0.5) = 2$ dm $^3$ mol $^{-1}$ , $K_b = 2K_a = 4$ , and $\beta K_b = [\ce{HG2}]/([\ce{HGa}][\ce{G}]) = 0.6/0.05 = 12$ , so $\beta = 3$ — a cooperative system.

Common pitfall. Count G atoms correctly: each HG2 consumes two guests, so the G balance contains

2[\ce{HG2}]

, not

[\ce{HG2}]