Question

What is the concentration of B expressed in terms of A if the Kd is 37.0UM, and the concentration of AB is 16.0 uM?

A+B≪m>AB
O B=430/A
O B=592/A
O B=0.592/A
O B=0.430/A
O B=2.31/A

Answer 1

To find the concentration of B expressed in terms of A given the dissociation constant Kd and concentration of AB, we use the equilibrium expression Kd = [A][B] / [AB]. Plugging in Kd = 37.0 µM and [AB] = 16.0 µM and solving for [B], we find that [B] = 592 µM / [A]. The correct option is 2 therefore O B=592/A.

Step-by-step explanation:

To determine the concentration of B expressed in terms of A given the dissociation constant (Kd) and the concentration of AB, we can use the expression of the equilibrium constant for the reaction A + B ⇌ AB. Since Kd is the reverse of the equilibrium constant for the formation of AB, we have Kd = [A][B] / [AB]. Given that Kd = 37.0 µM and [AB] = 16.0 µM, we can express [B] as a function of [A].

Let's solve the equation for [B]: Kd = [A][B] / [AB]

Multiplying both sides by [AB]:
[A][B] = Kd * [AB]

Dividing both sides by [A]:
[B] = (Kd * [AB]) / [A]

Substituting the values:
[B] = (37.0 µM * 16.0 µM) / [A] = 592 µM / [A]

Therefore, B = 592/A and the correct option is O B=592/A.