Question

How do you use ice tables to solve for acid base reactions?

Answer 1

ICE tables are most useful for cases where you have a weak acid or weak base mixed with its conjugate base or acid respectively to form a buffer solution since these will involve an equilibrium. Before worrying about the ICE table, make sure to deal with any neutralization reactions (if applicable, e.g. addition of a strong acid to a weak base) to determine how much of the weak acid/base remains after the neutralization reaction is complete (using standard stoichiometry and limiting reagent strategies).

Now focus on the weak acid/base itself. Write out the equilibrium reaction for that compound (e.g. for a weak acid, HA <--> H+ + A-). I'll use the example of a weak acid mixed with its conjugate base for the rest of the explanation, but the weak base/conjugate acid situation is identical (just realize you may be solving for pOH, not pH). Under this reaction you will have three rows to form the ICE chart: I, for initial concentrations; C, for change in concentration; and E, for equilibrium concentrations.

In the "I" row, under each compound in your equilibrium reaction, write out the initial concentration of that compound as given in the question. Very often you will be given concentrations of the weak acid and its conjugate base that form the buffer solution, but not a value for the H+ concentration, so your first row will often have a number, then 0, then another number.

In the "C" row, write the change in concentration that will occur as the system shifts to equilibrium. This involves adding/subtracting "x" amount of concentration to each compound in the reaction based on the stoichiometric ratios and the direction in which the reaction proceeds. In the example above, there would be a change of -x for HA, +x for H+, and +x for A-. This is because we know the reaction needs to shift towards products, because it can't shift to reactants without having any H+ already present. We therefore lose some of our reactant (hence the -x for HA) and gain the corresponding amount of products (hence the +x for H+ and A-).

In the "E" row, write the sum of the "I" and "C" rows. So for HA, you would have number-x, for H+ you would have 0+x, and for A- you would have number+x. Remember the "number" here just represents whatever the starting concentrations are. These expressions (number-x, 0+x, number+x) are what you plug into the Ka equation (you will often be given a value for Ka), which will enable you to solve for x. If the acid is extremely weak (e.g. Ka is more than 1000 less than your starting concentrations), you may be able to ignore the change in HA and A- concentrations because they will effectively be negligible. This will simplify your equation so that you only need to use number, 0+x, and number, making it much easier to solve for x. You can then find the -log of x (which is the concentration of H+) to determine the pH of the solution.

I know this explanation is long but hopefully I was somewhat clear. If you aren't sure of something let me know!