ICE Method for Equilibrium calculations

For an equilibrium process,

`AearrB+C`

The equilibrium constant can be expressed,

`K=([B][C])/([A])`

We construct an Initial, Change and Equilibrium (ICE) table,

	`A`	`stackrel(K) earr B`	`+ C`
I	`A_0`	0	0
C	`-x`	`+x`	`+x`
E	`A_0-x`	`x`	`x`

Using these equilibrium concentrations, we can re-construct the equilibrium expression,

`K=([B][C])/([A])=x^2/(A_0-x)`

To find the equilibrium concentrations, we must find the roots of the quadratic. We could use the quadratic equation or we could complete the square - we'll use the latter method here. Thus we first rearrange it to be of quadratic standard form. Expand the brackets,

`K(A_0-x)=x^2=KA_0-Kx`

Move the x terms to one side,

`x^2+Kx=KA_0`

Add one-half the coefficient of x squared to both sides,

`x^2+Kx+K^2/4=K^2/4+KA_0`

The square is now complete, such that

`(x+K/2)^2=K^2/4+KA_0`

Hence,

`x+K/2=+-sqrt(K^2/4+KA_0)=+-sqrt(K^2+4KA_0)/2`

Finally,

`x=-K/2+-sqrt(K^2+4KA_0)/2`

Since there are two roots to a quadratic, one positive and one negative. However, concentrations can only be positive, so only one root is a valid solution,

`x=-K/2+-sqrt(K^2+4KA_0)/2`

Example I:

Calculate the equilibrium proton concentration when a solution of 0.1moldm^-3 acetic acid is made up in water. For acetic acid, the acid dissociation constant is 1.77x10^-5moldm^-3.

`x=-K/2+-sqrt(K^2+4KA_0)/2`

`=-(1.77xx10^(-5))/2+-sqrt((1.77xx10^(-5))^2+4xx1.77xx10^(-5)xx0.1)/2`

`=0.0013moldm^(-3)`