Our discussion of acid-base ionization and salt hydrolysis was limited to solutions containing a single solute. In this section, we will consider the acid-base properties of a solution with two dissolved solutes that contain the same ion (cation or anion), called the common ion.
The presence of a common ion suppresses the ionization of a weak acid or a weak base. If sodium acetate and acetic acid are dissolved in the same solution, for example, they both dissociate and ionize to produce CH3COO- ions:
CH3COONa is a strong electrolyte, so it dissociates completely in solution, but CH3COOH, a weak acid, ionizes only slightly. According to Le Châtelier’s principle, the addition of CH3COO- ions from CH3COONa to a solution of CH3COOH will suppress the ionization of CH3COOH (that is, shift the equilibrium from right to left), thereby decreasing the hydrogen ion concentration. Thus, a solution containing both CH3COOH and CH3COONa will be less acidic than a solution containing only CH3COOH at the same concentration. The shift in equilibrium of the acetic acidionization is caused by the acetate ions from the salt. CH3COO- is the common ion because it is supplied by both CH3COOH and CH3COONa.
The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The common ion effect plays an important role in determining the pH of a solution and the solubility of a slightly soluble salt (to be discussed later in Section 12.7). Here we will study the common ion effect as it relates to the pH of a solution. Keep in mind that despite its distinctive name, the common ion effect is simply a special case of Le Châtelier’s principle.
Let us consider the pH of a solution containing a weak acid, HA, and a soluble salt of the weak acid, such as NaA. We start by writing
The ionization constant Ka was defined in Equation 11.10:
If the concentrations are sufficiently low, then the activities of solutes in Equation 11.10 can be replaced with molarities (without units) and the activity of pure water can be assumed to be unity, giving (Equation 11.11)
Rearranging this equation gives
Taking the negative logarithm (base 10) of both sides, we obtain
Using the definitions of pH (Equation 11.5) and pKa (Equation 11.15) gives (assuming dilute solution behavior)
Equation 12.1 is called the Henderson-Hasselbalch equation. A more general form of this expression is
In our example, HA is the acid and A- is the conjugate base. Thus, if we know Ka and the concentrations of the acid and the salt of the acid, we can calculate the pH of the solution.
It is important to remember that the Henderson-Hasselbalch equation is derived from the equilibrium constant expression. It is valid regardless of the source of the conjugate base (that is, whether it comes from the acid alone or is supplied by both the acid and its salt).
In problems that involve the common ion effect, we are usually given the starting concentrations of a weak acid HA and its salt, such as NaA. If the concentrations of these species are reasonably high (> 0.1 M), we can neglect the ionization of the acid and the hydrolysis of the salt. This is a valid approximation because HA is a weak acid and the extent of the hydrolysis of the A- ion is generally very small. Moreover, the presence of A- (from NaA) further suppresses the ionization of HA and the presence of HA further suppresses the hydrolysis of A-. Thus, in such cases, we can use the starting concentrations as the equilibrium concentrations in Equation 12.1.
In Example 12.1 we calculate the pH of a solution containing a common ion.
The common ion effect also operates in a solution containing a weak base, such as NH3, and a salt of the base, say NH4Cl. At equilibrium
We can derive the Henderson-Hasselbalch equation for this system as follows: Rear- ranging the preceding equation we obtain
Taking the negative logarithm of both sides gives
A solution containing both NH3 and its salt NH4Cl is less basic than a solution containing only NH3 at the same concentration. The common ion NH4+ suppresses the ionization of NH3 in the solution containing both the base and the salt.
(a) Calculate the pH of a solution containing 0.20 M CH3COOH and 0.30 M CH3COONa.
(b) What would the pH of a 0.20 M CH3COOH solution be if no salt were present?
Strategy (a) CH3COOH is a weak acid and CH3COONa is a soluble salt that is completely dissociated in solution The common ion here is the acetate ion (CH3COO-). At equilibrium, the major species in solution are CH3COOH, CH3COO-, Na+, H+, and H2O. The Na+ ion has no acid or base properties, and we ignore the ionization of water. Because Ka is an equilibrium constant, its value is the same whether we have just the acid or a mixture of the acid and its salt in solution. Therefore, we can calculate [H+] at equilibrium and hence pH if we know the equilibrium concentrations of both [CH3COOH] and [CH3COO]. For CH3COOH, Ka is 1.8 * 10-5 (see Table 11.3) giving a pKa of 4.74. (b) For a solution of CH3COOH, a weak acid, we calculate the [H+] at equilibrium, and hence the pH, using the initial concentration of CH3COOH. This problem can be solved by following the same procedure as in Example 11.11.
Solution (a) Sodium acetate is a strong electrolyte, so it dissociates completely in solution:
The initial concentrations, changes, and final concentrations of the species involved in the equilibrium are
Assuming that x is small, then 0.30 + x ≈ 0.30 and 0.20 - x ≈ 0.20. The Henderson-Hasselbalch equation (Equation 12.1) then gives
The value of x = [H+] corresponding to this pH is between 10-4 M and 10-5 M, which is significantly less than 5 percent of the initial concentrations of the acid or salt, so the approximation is valid.
(b) As in Example 11.11, x = [H+] is determined from
In this case, x is less than 5 percent of the initial acid concentration, so the approximation is valid.
Check Comparing the results in parts (a) and (b), we see that when the common ion (CH3COO-) is present, according to Le Châtelier’s principle, the equilibrium shifts from right to left. This action decreases the extent of ionization of the weak acid. Consequently, fewer H+ ions are produced in (a) and the pH of the solution is higher than that in part (b). As always, you should check the validity of the assumptions.
Comment Note that, if the initial concentrations were small enough that the approximation used in part (a) were not valid, then the problem must be solved by substituting the concentrations [CH3COOH] = 0.20 -x, [CH3COO-] = 0.30 - x, and [H+] = x into the expression for Ka and solving the resulting quadratic equation for x.