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 3 of 5: Solve the mass balances

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[\ce{HGb}] = 2[\ce{HGa}];\; 0.1 + 3[\ce{HGa}] + [\ce{HG2}] = 1;\; 0.5 + 3[\ce{HGa}] + 2[\ce{HG2}] = 2 \Rightarrow [\ce{HGa}] = 0.1,\; [\ce{HGb}] = 0.2,\; [\ce{HG2}] = 0.6

Analysis

Since $K_b = 2K_a$ , $[\ce{HGb}] = 2[\ce{HGa}]$ at equal $[\ce{H}]$ and $[\ce{G}]$ . With $[\ce{H}] = 0.1$ and $[\ce{G}] = 0.5$ , the H and G balances give two linear equations whose solution is $[\ce{HGa}] = 0.1$ , $[\ce{HGb}] = 0.2$ , $[\ce{HG2}] = 0.6$ mol dm $^{-3}$ .