Question

If glucose and glucose-6-phosphate are both at 12.6 mm, what is the equilibrium concentration of phosphate (pi)?

Answer 1

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 :

`Keq = ([Glucose-6-phosphate])/([Glucose][pi])`

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 :

`0.0047 = ([12.6mM])/([12.6mM][pi])`

`0.0047 = ([1])/([pi])`

By Cross multiplying :

`[pi]= ([1])/(0.0047)`

[pi] = 212.77 mM

The concentration of phosphate (pi) = 212.77mM .