Question

Under certain conditions, the equilibrium constant of the reaction below is Kc=7.76. If the reaction begins with a concentration of 0.0599 M for each of K+ and Br− and a concentration of 0 M for KBr, what is the equilibrium concentration of KBr?

KBr(aq)⇌K+(aq)+Br−(aq)

Answer 1

ICE Table:

Equilibrium constant expression is

Kc = [K+]*[Br-]/[KBr]

7.76 = (5.99*10^-2-1*x)(5.99*10^-2-1*x)/((1*x))

7.76 = (3.588*10^-3-0.1198*x + 1*x^2)/(1*x)

7.76*x = 3.588*10^-3-0.1198*x + 1*x^2

-3.588*10^-3 + 7.88*x-1*x^2 = 0

This is quadratic equation (ax^2+bx+c=0)

a = -1

b = 7.88

c = -3.588*10^-3

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 62.08

roots are :

x = 4.554*10^-4 and x = 7.879

x can't be 7.879 as this will make the concentration negative.so,

x = 4.554*10^-4

At equilibrium:

[KBr] = x = 4.554*10^-4 M

Answer: 4.55*10^-4 M