Respuesta :
Answer :
The correct answer is 212.77mM
The reaction between glucose and glucose-6-phosphate is given as :
Glucose + pi → Glucose-6-phosphate + H₂O
The equilibrium constant for this reaction (Keq) is 0.0047
The expression for equilibrium constant is given as :
[tex]Keq = \frac{[Glucose-6-phosphate]}{[Glucose][pi]}[/tex]
Where all the concentration are
Given : Concentration of Glucose 12.6mM
Concentration of Glucose-6-phosphate = 12.6 mM
Asked : Concentration of phosphate (pi) = ?
Plugging above values in equilibrium constant expression :
[tex]0.0047 = \frac{[12.6mM]}{[12.6mM][pi]}[/tex]
[tex]0.0047 = \frac{[1]}{[pi]}[/tex]
By Cross multiplying :
[tex][pi]= \frac{[1]}{0.0047}[/tex]
[pi] = 212.77 mM
The concentration of phosphate (pi) = 212.77mM .
