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	HA(aq)	H3O+(aq)	A-(aq)
Initial Conc. (M)	0.500	0	0
Change in Conc. (M)	- x	+ x	+ x
Equilibirum Conc. (M)	0.500 - x	x	x

Substitute into the equilibrium expression.  Assume that 0.500 - x ~ 0.500.  Simplify equation and solve for the change.

Check answer to see if it is within limits set by your instructor. (Here we use 5%.)

(0015/0.500) x 100 = 0.03%

The change is only 0.03% of the initial value and is negligible.

Determine the equilibrium concentrations of each species

[H₃O⁺] = [A^-] = x = 1.5 x 10^-4 M

[HA] = 0.500 - 1.5 x 10^-4 = 0.500 M

Check work.

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K and Q Lie on Opposite Sides of One (K>>Q or K<<Q)

When K is much larger than Q, or K is much smaller than Q, the change in amount of each species will be very large as the system moves towards a state of equilibirum.  When this occurs, finding the change in concentration can often be facilitated by doing the following:

• When K>>Q and K > 1, assume 100% conversion into products, followed by the back reaction to establish equilibrium.  When K<<Q and K < 1, assume 100% conversion into reactants, followed by the forward reaction to establish equilibrium.
• Make an ICE chart to determine change and equilibrium quantities starting with those resulting from the 100% conversion.
• Substitute quantities into the equilibrium expression.
• Assume the change is near zero such that "[A] - x" is equal to "[A]."
• Solve for the variable.
• Check to see if the change is less than 5% of the maximum amount, or within the limits set by your instructor. If not, use the method of approximations, a programmable calculator, or other method to solve.
• Solve for the equilibrium concentrations if asked to do so.
• Check your work.

Example: An evacuated flask is filled with sufficient H₂ and I₂ gases so that the concentration of each gas is 0.620 M. It is then heated to 298 K. What is the concentration of each species when equilbrium is established?

H₂(g) + I₂(g)  2 HI(g) K_c = 794 @ 298 K

• Initially the [HI] = 0, so K >>Q  and K is > 1. The change in the concentration of each species will be large so we calculate the quantity of product formed assuming 100% conversion.

100% conversion will result in the formation of 1.24 M HI (1 to 1 to 2 proporation) with neither reactant remaining.

• Make an ICE chart starting with the concentrations after the 100% conversion.

	H2	I2	HI
Initial Concentration (M)	0	0	1.24
Change in Concentration (M)	+ x	+ x	- 2 x
Equilibrium Concentration (M)	0 + x	0 + x	1.24 - 2 x