Question

A histidine is involved in an interaction with a glutamic acid that stabilizes the charged form of the histidine, such that the value of ∆Go for deprotonation is 15 kJ•mol –1 at pH 7.0 and 293 K (calculated using the biochemical standard state). What is the pKa of this histidine

Answer 1

Answer:

pKa of the histidine = 9.67

Explanation:

The relation between standard Gibbs energy and equilibrium constant is shown below as:

$\Delta{G^0} =-RT \ln \frac{[His]}{[His+]}$

R is Gas constant having value = 0.008314 kJ / K mol  

Given temperature, T = 293 K

Given, $\Delta{G^0}=15\ kJ/mol$

So,  Applying in the equation as:-

$15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}$

Thus,

$15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}$

$\frac{[His]}{[His+]}=e^{\frac{15}{-0.008314\times 293}$

$\frac{[His]}{[His+]}=0.00211$

Also, considering:-

$pH=pKa+log\frac{[His]}{[His+]}$

Given that:- pH = 7.0

So, $7.0=pKa+log0.00211$

pKa of the histidine = 9.67