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!
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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: ax^{2}
+ bx + c = 0.
x^{2} + 33.3x  166.5 = 0

Substitute the coefficients into the quadratic equation and solve for x.
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Solving
Equations Containing x^{3}, x^{4}, etc.
At times, you will have problems involving equilibrium expressions with
the variable "x" raised to a power "n".
x^{n} = a number
With the aid of a calculator there are two ways you can go about solving
for x.
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 x^{5} = 2.4 x 10^{18}
x^{5} = 2.2 x 10^{20}

Using logarithms
x = 10^{3.93} = 1.2 x 10^{4}

Finding the n^{th} root.

Most calculators will have a button labeled:
Often this button is in conjunction with the "y^{x}" button.
You may need to push "2ND" or "INV" before the button to find the n^{th} root.
Consult the manual that came with the calculator for the exact procedure.

Enter in the number for which you want the n^{th} root.
2.2 x 10^{20}

Press the button for finding the n^{th} root followed by the exponent,
n, and the "=" or "enter" button to obtain the result.
1.2 x 10^{4}
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The
Method of Successive Approximations
One method of solving what appears at first to be very daunting equations
is to:

assume an approximate value for the variable that will simplify the equation

solve for the variable

use the answer as the second apporximate 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

Simpify 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 consistant value has been obtained after making
only two approximations.
This method will work with most polynomials. A consistant 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.
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Assuming
That the Change is Small
There are two special cases where we can solve an equation by assuming the variable is small.
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:

Calculate Q, the reaction quotient and compare to K.

Make an ICE chart.

Subsitute concentrations into the equilibrium expression. Assume that [A]  x = [A], simplify the equation, and solve for the change.

Check to see if the change is less than 5% of the starting quantity, or within the
limits set by your instructor.

Calculate the equilibrium amounts if asked to do so.

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) + H_{2}O(aq)
H_{3}O^{+}(aq) + A^{}(aq) K_{a}
= 4.6 x 10^{8}

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.

Make an ICE chart.

HA(aq)

H_{3}O^{+}(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%.)
(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
[H_{3}O^{+}] = [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 H_{2}
and I_{2} 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?
H_{2}(g) + I_{2}(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.

H_{2}

I_{2}

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


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 programable calculator) we arrive
at:
x = 0.041

Calculate the equilibrium concentrations.
[H_{2}] = [I_{2}] = x = 0.041 M
[HI] = 1.24  2x = 1.24 (2)(0.041) = 1.16 M

Check work.
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