When a strong acid or a strong base is added to water, it nearly completely dissociates into its ion constituents because it has a pK a or pK b less than zero. For example, a solution of H2SO4 in water contains mostly H+ and SO4 2-, and almost no H2SO4 is left undissolved. This makes calculating the pH of a strong acid or strong base solution exceedingly simple--the concentration of acid equals the concentration of H+. Recall that pH is computed by taking the negative log of of [H+]. Common strong acids that should be memorized include HCl (hydrochloric), HNO3 (nitric), HClO4 (perchloric), and H2SO4 (sulfuric). Strong bases include Group I hydroxides (LiOH, NaOH, KOH, etc.) and Group II hydroxides except for Be(OH)2 and Ba(OH)2.
Calculating the pH of weak acid and weak base solutions is much more complicated than the above case--weak acids and bases do not completely dissociate in aqueous solution but are in equilibrium with their dissociated forms. Therefore, we must apply what we know about equilibria to solve these types of problems. For example, let's calculate the pH of a 0.10 M solution of acetic acid in water. To do this, we first write down the equilibrium involved and the expression for the equilibrium constant:
Next, you should compile a table of values for the concentration of all species involved in the equilibrium. We already know that the initial concentration, [ ]o, of acetic acid is 0.10 M and that the initial concentration of H+ is 10-7 (since the solvent is neutral water). Even though there is an initial concentration of H+ in solution, it is so small compared to the amount produced by the acid that it is usually ignored. From the stoichiometry of the reaction, one mole of H+ and one mole of acetate (Ac-) are produced for each mole of acetic acid dissociated. Therefore, if we denote the amount of acetic acid dissociated as x, the final concentrations of H+ and Ac- are both x, and the final concentration of HAc in solution is 0.10 M - x. This data is summarized in :
After you have compiled the table of values, you can substitute the equilibrium concentration values for each species into your expression for K a as shown below:
You should note that the above equation is a quadratic equation in x and therefore requires use of the quadratic equation to solve for x. If, however, we can make the assumption that [HAc]o - x = [HAc]o, the equation becomes much easier to solve. We can do this if HAc is a weak enough acid that it dissociates very little, and the change in [HAc] is negligible.
Is this assumption valid? Solving the quadratic equation using the quadratic formula gives a pH of 2.88. The difference of 0.01 pH unit is small enough to be insignificant, so the assumption is valid in this case and will certainly save you some time on a test. We can make the approximation [HA]o - x = [HA]o so long as x is less than 5% of the initial concentration of HA. X will be greater than 5% of [HA]o with stronger weak acids at low concentrations. Consider these guidelines when you decide whether or not to make the approximation: you can simplify the quadratic equation if the solution is more concentrated than 0.01 M and the pK a of the acid is greater than 3.
The pH of a weak base solution is calculated in the same manner as that of a weak acid solution, using K b instead of a K a.
To calculate the pH of a mixture of acids in aqueous solution, first decide which acid has the lowest pK a. Calculate the pH as if the strongest acid were the only one in solution. We can ignore the contributions to the pH of the weaker acids because they will be minor in comparison to that of the strongest acid in the group. The exact solution to such a problem requires more complex mathematics and will not be covered in this SparkNote.
A salt of a strong acid and a strong base (such as NaCl from HCl and NaOH) produces a neutral solution when dissolved in water. However, when a salt of a weak acid and a strong base (e.g. NaAc from acetic acid and NaOH), a strong acid and a weak base (e.g. NH4Cl from ammonia and HCl), or a weak acid and a weak base (NH4Ac) is dissolved in water, the solution does not have a neutral pH. These phenomena are explained by the reaction of the salts of weak acids and weak bases with water in hydrolysis reactions. As shown in , these hydrolysis reactions produce the parent weak acids and weak bases of the salts:
As you can see in the figure above, the salt of a weak acid, such as acetate ion, acts as a base in water, and the salt of a weak base, such as ammonium ion, acts as a weak acid. From our previous discussion on the reactions of acids and bases with water in Disassociations, you should know that the K b of acetate ion can be calculated from the K a of acetic acid, and that the K a of ammonium ion can be calculated from the K b of ammonia, as shown in .
Another type of hydrolysis reaction comes from the reaction of metal ions with high charges. Such ions act as Lewis acids to water molecules, as shown in . A metal ion can bond to water by accepting an lone pair from the oxygen of a water molecule, and this increases the acidity of water molecule. Like other acids, we can calculate the K a and calculate the pH of a solution containing such ions.
To calculate the pH of a solution containing the salt of a weak acid or a weak base, treat the problem exactly as you did when calculating the pH of weak acid and base solutions above in Calculating pH's, Heading . Mixtures of salts of weak acids and weak bases present a challenging mathematical problem that we will not cover in our treatment of acid-base chemistry.
So far we have dealt with acids that donate only one proton per molecule. However, this is not the case for polyprotic acids--acids that can donate more than one proton per molecule. Two key features of polyprotic acids are that they lose their protons in a stepwise manner and that each proton is characterized by a different pK a. The factors contributing to the pK a of each acidic proton in a polyprotic species are the same factors that determine the relative acidity of monoprotic acids--the dominant factor is strength of the acid-H bond. Consider, for example, the triprotic acid H3PO4 shown in :
As each proton is lost from phosphoric acid, the phosphorous becomes more electron rich, and less electron withdrawing. Therefore, the loss of each proton strengthens the O-H bond and increases the pK a of the phosphate species. This trend is evident in the pK a data given in . In general, it is true that K a1, K a2, K a3, and so on, for polyprotic acids.
As you may have guessed, calculating the pH of a polyprotic acid solution is not as simple as it is for monoprotic acids. In fact, it is quite a messy problem. However, that mess can be quickly cleaned up by making the assumption, as we did for a mixture of acids, that only the strongest acid (i.e. only the first dissociation) has a significant effect on the pH. Making that assumption turns the problem into one you already know how to solve--calculating the pH of a weak acid solution.