# Determining the Equilibrium Constant

"An equilibrium constant expression describes the relationship among the concentrations (or partial pressures) of the substances present in a system at equilibrium" (General Chemistry).

For a given chemical reaction:

\[aA + bB \rightleftharpoons cC + dD\]

the equilibrium constant expression is as follows:

\[K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]

To calculate the equilibrium constant (also known as the dissociation constant), the concentrations of each species in the reaction at equilibrium must be measured. Consider the general acid dissociation equation:

\[HA + H_2O \rightleftharpoons A^- + H_3O^+\]

Where HA is the acid, H_{2}O is water, A^{-} is the conjugate base of the acid, and H_{3}O^{+} is the hydronium ion, a protonated water molecule.

Any equilibrium problem given provides some needed information, including a variety of numbers and/or units. Either K_{a} or K_{b}, the dissociation constants for the acid or the base in the equation, respectively, will be provided. For this example, K_{a} = 2.2 x 10^{-4} and the initial concentration of HA, the acid in solution is 2 mol L^{-1}.

The equilibrium concentrations are calculated using an ICE Table. An ICE Table is a way to keep all your information organized throughout the process. ICE stands for 'Initial', 'Concentration', and 'Equilibrium'.

\[HA + H_2O \rightleftharpoons A^- + H_3O^+\]

HA | + | H_{2}O | ⇌ | A^{-} | + | H_{3}O^{+} | |
---|---|---|---|---|---|---|---|

I | 2 mols/L | n/a | 0 | 0 | |||

C | -x | n/a | +x | +x | |||

E | 2-x | n/a | x | x |

In this case,

\[K_a = \frac{[A^-][H_3O^+]}{[HA]}\]

H_{2}O is ignored in the ICE Table and when calculating the dissociation constant because it is a pure liquid. Any hypothetical solids would also be ignored.

Substituting the values from the "E" row of the table,

\[\begin{eqnarray} K_a &=& \frac{(x)(x)}{2-x} \\ &=& \frac{x^2}{2-x} \end{eqnarray} \]

A simple way to solve this equation is to first estimate the value for x. Assume x is very small, so the quantity 2-x is essentially 2. Keep in mind that the value for K_{a }is known. The equation simplifies to the following:

\[x^2 = 2K_a\]

The calculated value for x is very close to 0. It can be plugged into the 2-x term in the original equation to find a more accurate value for x. A few iterations of this process will converge on the correct concentration of H_{3}O^{+} and, equivalently, A^{-} at equilibrium in this specific case. In this example, the concentration of H_{3}O^{+} and A^{-} is 0.02086 or 2.09 x 10^{-2 }mol L^{-1}

The relationship between K_{a} and K _{b} is the following:

\[K_W = K_a \times K_b\]

K_{w} is equal to 10^{-14}. If the problem provides a value for K_{a} but the reaction involves a base, K_{b} can be calculated by dividing K_{w} by K_{a}. The same holds true if K_{b} is given for an acid dissociation problem.

### Resources

- General Chemistry: Principles & Modern Applications 9th edition.

### Contributors

- Gregory Arch