Techniques for Solving Equilibrium Problems
If Possible, Take the Square Root of Both Sides
Sometimes the mathematical expression used in solving an equilibrium problem can be solved by
taking the square root of both sides of the equation. This can be done when the variables in the
numerator and in the denominator of the equilibrium expression are multiplied by themselves
(squared).
Example:
If you can take the square root of both sides you will spend less time in solving an equation!
Using the Quadratic Equation
If you cannot take the square root of both sides of the equation, you can use the quadratic
equation for an equation of the form:
For example:
Rearrange to
the form: ax2 + bx + c = 0.
x2 + 33.3x -
166.5 = 0
Substitute the coefficients into the quadratic equation and solve for x.
Solving Equations Containing x3, x4, etc.
At times, you will have problems involving equilibrium expressions with the variable "x" raised to
a power "n".
xn = a number
With the aid of a calculator there are two ways you can go about solving for x.
 Using logarithms
 Finding the nth root
The following example will be used to illustrate each method.
Example:
(2x)2(3x)3 = 2.4 x 10-18
Using either method, the first step is to simplify the equation.
108 x5 = 2.4 x 10-18
x5 = 2.2 x 10-20
Using logarithms
Take the log of both sides of the equation.
log x5 = log (2.2 x 10-20)
Factor out the exponent (log xn = n log x) and find the log of the number.
log x5 = -19.66
5 log x = -19.66
Solve for "log x".
log x = -3.93
Solve for x (if log x = y, then x = 10y).
x = 10-3.93 = 1.2 x 10-4
Finding the nth root.
Most calculators will have a button labeled:
Often this button is in conjunction with the "yx" button. You may need to push "2ND" or "INV"
before the button to find the nth root. Consult the manual that came with the calculator for the
exact procedure.
Enter in the number for which you want the nth root.
2.2 x 10-20
Press the button for finding the nth root followed by the exponent, n, and the "=" or "enter"
button to obtain the result.
1.2 x 10-4
The Method of Successive Approximations
One method of solving what appears at first to be very daunting equations is to:
1. assume an approximate value for the variable that will simplify the equation
2. solve for the variable
3. use the answer as the second approximate value and solve the equation again repeat this
process until a constant value for the variable is obtained
Example:
Approximate a value for the variable that will simplify the equation.
Since 8.4 x 10-4 is a small number, the value of x must be small. We will make the assumption
that:
0.200 - x ~ 0.200
Simplify the equation and solve for the variable.
Using the result, make a second approximation.
0.200 - x = 0.200 - 0.013 = 0.187
Using the second approximation, simplify the equation and solve for the variable
Repeat the process until a constant value is obtained.
In this example, a consistent value has been obtained after making only two approximations.
This method will work with most polynomials. A consistent value is often obtained in less than
five successive approximations. With the aid of a calculator, the method of successive
approximations can be done quickly.
Assuming That the Change is Small
There are two special cases where we can solve an equation by assuming the variable is small.
1. K and Q Are Very Close in Size
2. K and Q Lie on Opposite Sides of One (K>>Q or K<<Q)
K and Q Are Very Close in Size
When K and Q are close to being the same value and are on the same side of 1, the change in
amount of each species will be very small 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:
1. Calculate Q, the reaction quotient and compare to K.
2. Make an ICE chart.
3. Substitute concentrations into the equilibrium expression. Assume that [A] - x = [A],
simplify the equation, and solve for the change.
4. Check to see if the change is less than 5% of the starting quantity.
5. Calculate the equilibrium amounts if asked to do so.
6. Check your work.
Example: Determine the concentration of each species present in a 0.500 M solution of a weak
acid HA. HA reacts with water according to the equation:
HA(aq) + H2O(aq)
H3O+(aq) + A-(aq) Ka = 4.6 x 10-8
1. Calculate Q and compare to K.
In this example, initially there are no products so Q = 0. K > Q so the reaction will proceed in
the forward direction. However, K and Q are < 1. The change in the concentration will be small.
2. Make an ICE chart.
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
3. 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 5%.
(0.00015/0.500) 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.
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 equilibrium. 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) Kc = 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
-2x
Equilibrium Concentration (M)
0+x
0+x
1.24 - 2 x
Substitute equilibrium amounts into the equilibrium expression.
Assume the change in the concentration of the product is 0. Substitute into the equation and
solve for "x."
Check to see if the change is within the limits set by your instructor. (Here we use 5%.)
The change in the HI is "2x" or 2(0.044) = 0.088 M
(0.088/1.24) x 100 = 7.1 %
This is greater than 5%. Using one of the other methods of solution (quadratic, successive
approximations, or programmable calculator) we arrive at:
x = 0.041
Calculate the equilibrium concentrations.
[H2] = [I2] = x = 0.041 M
[HI] = 1.24 - 2x = 1.24 -(2)(0.041) = 1.16 M
Check work.