"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'. \[HA + H_2O \rightleftharpoons A^- + H_3O^+\] Unnamed: 0 HA + H2O ⇌ A- +.1 H3O+ I 2 mols/L NaN NaN NaN 0 NaN 0 C -x NaN NaN NaN +x NaN +x E 2-x NaN NaN NaN x NaN 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.