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 5 of 5: Initial amount of G

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[\ce{HG2}] = 3[\ce{HGa}];\; [\ce{H}] = 0.25,\; [\ce{G}] = 0.25,\; [\ce{HGa}] = 0.125 \Rightarrow n_0(\ce{G}) = 0.25 + 3(0.125) + 2(0.375) = 1.375\ \text{mol}

Analysis

The condition $[\ce{HG2}] = [\ce{HGa}] + [\ce{HGb}] = 3[\ce{HGa}]$ plus the H balance gives $[\ce{H}] = 0.25$ ; with $K_a = 2$ , $[\ce{HGa}] = K_a[\ce{H}][\ce{G}]$ and $[\ce{HG2}] = 12[\ce{HGa}][\ce{G}]$ solved simultaneously give $[\ce{G}] = 0.25$ , $[\ce{HGa}] = 0.125$ , $[\ce{HG2}] = 0.375$ ; the G balance then yields $n_0(\ce{G}) = 1.375$ mol.