Iron-enterobactin [Fe (ent)]3- complex has a dissociation content of 10-49 at pH = 7. Calculate the number of uncomplexed Fe3+ ions in a litre of 1 M solution in water at pH = 7.
Less than 1
Enterobactin is the siderophore that E. coli secretes to scavenge iron, and it is the strongest Fe3+ chelator known. The question is really asking what a dissociation constant of 10-49 means in terms of actual free ions.
Write the dissociation: \([\mathrm{Fe(ent)}]^{3-} \rightleftharpoons \mathrm{Fe}^{3+} + \mathrm{ent}^{6-}\), with \(K_{d} = \frac{[\mathrm{Fe}^{3+}][\mathrm{ent}]}{[\mathrm{Fe(ent)}]} = 10^{-49}\).
Starting from a 1 M solution of the complex, let \(x\) be the concentration that dissociates. Then \([\mathrm{Fe}^{3+}] = [\mathrm{ent}] = x\) and, since \(x\) is tiny, \([\mathrm{Fe(ent)}] \approx 1\).
So \(x^{2} = 10^{-49}\), giving \(x = 10^{-24.5} \approx 3.2 \times 10^{-25}\) mol dm-3.
Now convert to a number of ions in one litre by multiplying by the Avogadro constant:
\(N = 3.2 \times 10^{-25} \times 6.022 \times 10^{23} \approx 0.19\).
The result is a fraction of one ion — which physically means that in a whole litre of 1 M complex you would, on average, not find even a single free Fe3+ ion at any instant. That is the point of the question: the binding is so tight that free iron is effectively absent, which is exactly how the bacterium wins iron away from the host's transferrin.
Hence the number of uncomplexed Fe3+ ions is less than 1.
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Enzymolysis with the following mechanism :
\[ E + S \xrightleftharpoons[k_{-1}]{k_1} ES \] \[ ES \xrightarrow{k_2} E + P \] \[ E + I \rightleftharpoons EI \] \[ \therefore\ (I \equiv \text{inhibitor}) \quad I_1 < I_2 < I_3 \]
The following behaviors are noted.

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II. Sweets
III. Pasta
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pKa 3 = 10.54 (ε - amino group)
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