- A series of two ICE tables is a uesful tool when determining the pH after mixing an acid and a base.
In the previous section on salts, we saw that in addition to acids and bases reacting with water, they can also react with each other. In this section, we will determine the pH of a solution after an acid (or base) is mixed with a base (or acid) other than water. To solve such problems, we’ll use a series of two ICE tables: our first ICE table will consider the reaction between the mixed acid and base, and the second ICE table will consider the reaction between the products and/or remaining reactants of the first reaction with the water that acts as the solvent.
Reaction between a strong acid and strong base
If a strong acid and a strong base react with each other, they will react as much as possible until one of them is fully neutralized. Whichever species (acid or base) is in excess will then react with water to determine the final pH.
Example: What would the pH be if 0.54 mol HCl and 0.26 mol NaOH were mixed in 1.0 L of water?
In the first ICE table, consider the reaction directly between HCl (a strong acid) and NaOH (a strong base). Strong acids/bases react essentially completely, so the NaOH (limiting reagent) will react with HCl until it is completely consumed:
That is, after this reaction, we have 0.26 M NaCl, 0.28 M HCl, and no remaining NaOH. NaCl is a neutral salt so will not affect the pH. HCl, however, is a strong acid and will continue to react with water. We can assess this reaction using a second ICE table, inputting the initial amount of HCl as the final amount from the previous reaction
The pH is given by
Reaction between a strong acid (or base) with a weak base (or acid)
Strong acids (or bases) react essentially completely with weak bases (or acids). Here, we’ll discuss a strong acid reacting with a weak base, but the analysis would be parallel for a strong base reacting with a weak acid.
Example: What is the pH of a solution after 12 mL of 1.2 M HCl are added to 23 mL of 2.3 M CH3COO−? The Ka of CH3COOH is 1.8 x 10−5 at 298 K.
Our first ICE table will analyze the reaction directly between HCl and CH3COO−. In this question, the volume is changing. In an ICE table, we must use either moles or concentrations based on total volume. Thus, as a first step in this solution, let’s use to calculate the concentration of each species based on the final total volume of the solution, where
is the final concentration,
is the initial concentration,
is the final total volume and
is the initial volume:
Now we can construct an ICE table. Strong acids like HCl react completely with weak bases like CH3COO−, so the strong acid (limiting reagent) will react completely here:
Since Cl− does not act as an acid nor a base, it will not affect the pH and can be ignored for the rest of the analysis. CH3COO−, a weak base, and CH3COOH, its weak conjugate acid, however, will both react with water and thus affect the pH of the final solution. Since CH3COOH and CH3COO− are a conjugate acid-base pair that exist in equilibrium with each other, we will be able to assess their effects on the pH at the same time in a second ICE table by considering either the acid or the base reacting with water (either approach will give the same final pH). We’ll consider the acid in water here since we were given the Ka value in the question. We could equally consider the base in water instead to get the final pH, but this would require the extra step of converting Ka to Kb. The second ICE table shows how the CH3COOH produced in the first reaction goes on to react with water:
Notice that we included initial amounts of both CH3COOH and CH3COO−, since both were still present after the reaction in the first ICE table. The equilibrium constant of this reaction, Ka, is equal to the concentrations of the products divided by the reactants:
Since CH3COOH is a weak acid, we can make the simplifying assumption that x is small such that and
. This simplifies our equation to
And thus x is given by
This value of x is less than 5% of either 0.4114 M or 1.100 M, so our assumption that x is small was valid. Since x represents the [H3O+] at equilibrium, the pH is given by
Note that mixtures of weak acids and weak bases will be discussed in the next section.
Interactive: No practice available for this section yet.