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

pKa of the histidine = 9.67

Step-by-step explanation:

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

`\Delta{G^0} =-RT \ln ([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* 293\ K* \ln ([His])/([His+])`

Thus,

`15\ kJ/mol=-0.008314\ kJ/Kmol* 293\ K* \ln ([His])/([His+])`

`([His])/([His+])=e^{(15)/(-0.008314* 293)`

`([His])/([His+])=0.00211`

Also, considering:-

`pH=pKa+log([His])/([His+])`

Given that:- pH = 7.0

So,
`7.0=pKa+log0.00211`

pKa of the histidine = 9.67