Question

4. A Beer’s Law plot was prepared for the reaction A(aq) + B(aq)  AB(aq), plotting absorption over AB(aq) concentration. The linear equation for this plot was y = 78.3x. A solution was prepared by mixing 10.0 mL of 0.100M A with 5.00 mL of 0.100 M B and adding enough water to bring the total volume to 50.0 mL. The absorption of this solution was measured as 0.646. Given this information, calculate the following:

a) The initial concentrations of each reactant.

b) The equilibrium concentration of AB

c) The equilibrium concentrations of each reactant

d) The equilibrium constant with respect to concentration

Answer 1

a) Use the dilution equation to calculate the initial concentrations for A and B. The dilution equation is

M₁*V₁ = M₂*V₂ where M₁ = concentration of stock solution; M₂ = initial concentration in the dilute solution; V₁ = volume of stock solution taken and V₂ = volume of final solution.

A: (10.0 mL)*(0.100 M) = (50.0 mL)*[A]_i

===> [A]_i = 0.02 M.

B: (5.0 mL)*(0.100 M) = (50.0 mL)*[B]_i

===> [B]_i = 0.01 M (ans).

b) Use the linear equation to find out the equilibrium concentration of AB. Plug y = 0.646 and obtain [AB]_eq.

0.646 = 78.3*[AB]_eq

===> [AB]_eq = 0.00825 M (ans).

c) Subtract the equilibrium concentration of AB from the initial concentrations of A and B to obtain the equilibrium concentrations.

[A]_eq = [A]_i – [AB]_eq = (0.02 M) – (0.00825 M) = 0.01175 M.

[B]_eq = [B]_i – [AB]_eq = (0.01 M) – (0.00825 M) = 0.00175 M (ans).

d) Use the expression for the equilibrium constant to find out the equilibrium constant, K_eq.

K_eq = [AB]_eq/[A]_eq[B]_eq = (0.00825 M)/(0.01175 M).(0.00175 M) = 401.2158 M^-1 ≈ 401.2 M^-1 (ans).