Question

Show that the maximum enzyme catalysed reaction rate will occur for $[H^{+}{]}_{opt}=\sqrt{k_1{k}_2}$.

Answer 1

$[H^{+}{]}_{opt}=\sqrt{k_1{k}_2}$
The general pH control enzyme catalysed reaction is
$E{H}_2^{2+}\rightleftharpoons E{H}^{+}\rightleftharpoons E$
Only $E{H}^{+}$ is capable for binding substrate and catalysing the desired reaction;
$k_1=\frac{\left[E{H}^{+}\right]\left[H^{+}\right]}{\left[E{H}_2^{2+}\right]};k_2=\frac{\left[E\right]\left[H^{+}\right]}{\left[E{H}^{+}\right]}$
[ Hint: $k_1{k}_2=\frac{\left[E\right][H^{+}{]}^2}{\left[E{H}_2^{2+}\right]}$
In pH control it may be considered that $\left[E\right]\approx \left[E{H}_2^{2+}\right]$, i.e., enzyme combines almost completely with $H^{+}$ ions.
Thus, $k_1{k}_2\approx [H^{+}{]}^2;[H^{+}]=\sqrt{k_1{k}_2}$ ]