Acid-Base Equilibria:

pH is the Measure of Solution Acidity

The effective strength of an acid or base in solution, or acidity, will vary with both the intrinsic strength of the acid or base itself and the amount of the acid or base which is present in the aqueous solution.

The qualitative ranking of solutions as "more acidic" or "more basic" is possible on an empirical basis of qualitative tests, but such an approach is not useful for quantitative discussion.  Quantitatively, the acidity of a solution is measured by and is equal to the concentration of hydronium ion in that solution.

pH of Aqueous Solutions
The acidity of a solution was originally viewed as the molar concentration of H3O+ present in it.  This idea was used by the Swedish chemist S. P. L. Sorenson when in 1909 he defined the acidity of a solution expressed in logarithmic notation as:

pH = -log[H3O+]

The value of the solution pH is a quantitative measurement of the acidity of a solution.  Although the analogous concept of basicity is much less commonly used, the pOH notation is sometimes convenient.  Only one scale need be used, since they are related through the ion product of water:

[H3O+][OH-] = Kw = 1 x 10-14

in molar concentration units. Taking the logarithm of both sides;

log[H3O+] + log[OH-] = logKw = -14

Then

-log[H3O+] + (-log[OH-]) = -logKw = +14

or

pH + pOH = pKw = 14

In water we find an actual range of pH from about 0 to 14, although in solutions containing high concentrations of strong acids or bases this range can be exceeded somewhat.  It is convenient to classify aqueous solutions according to their pH.  If the pH of a solution is less than 7, the solution is called acidic; if the pH is about 7, the solution is neutral; if the pH is greater than 7, the solution is is called basic.  In an acidic solution, then, the concentration of hydrogen ions is greater than the concentration of hydroxide ions. In a neutral solution, the concentrations of hydrogen ions and hydroxide ions are roughly equal.  In a basic solution, the concentration of hydroxide ions is greater than the concentration of hydrogen ions.

The ionization constant of an acid or base is a measure of the intrinsic strength of the acid or base itself in a particular solvent. Equal acidity can be obtained from a higher concentration of a weaker acid or a lower concentration of a stronger acid.
 

Acidity of Solutions of Strong Acids

A strong acid, by definition, is an acid that is completely ionized in aqueous solution, that is, an acid which produces one or more protons for every acid molecule originally present. In the case of hydrochloric acid, the ionization equilibrium reaction is:

HCl + H2O  H3O+ + Cl-

[H3O+] = [HCl]

Example. To calculate the pH of 0.0010 M aqueous HCl solution;

[H3O+] = 0.0010
pH = -log[H3O+] = 3.00

Is there any other source of protons other than the acid itself?  The autoionization of water could give 1 x 10-7 M H3O+.  But this additional source of protons is usually neglegibly small compared to the amount produced by the acid. In the presence of any significant concentration of strong acid, the ionization of water will be repressed by the strong acid in accordance with Le Chatelier's principle. When the molar concentration of strong acid is 0.001, the concentration of hydroxide ion is only 1 x 10-11 M and the concentration of H3O+ produced by water ionization has the neglegible value of 1 x 10-11 M also. Therefore, in an aqueous solution of a strong acid, acidity is essentially determined by the concentration of strong acid alone.

Only when the concentration of strong acid present is very low does the autoionization of water become a comparable or dominant contributor to the solution pH.

Example. At an added [H3O+] of 1 x 10-8 M, [H3O+] = [H3O+](added) + [H3O+](ionization).  Then

[H3O+] = 1 x 10-8 + 1 x 10-7 = 1.1 x 10-7 M

The pH will be about 6.96 after addition of the acid, a negligible change from 7.00 obtained in water without the acid. The assumption that [H3O+] is actually equal to 10-8 M would give an altogether unrealistic pH of 8.  Addition of a strong acid cannot make any solution more basic!

Acidity of Solutions of Strong Bases

A strong base, like a strong acid, is virtually totally dissociated in water; thus 0.01 molar NaOH is actually an aqueous solution 0.01 M in Na+ and also 0.01 M in OH-. Sodium ion cannot act as an acid (no protons) or as a base (positive charge will repel a proton). The sodium ions have a negligible effect on the pH.  So the pH is determined by the hydroxide ion concentration.  It is often more convenient to calculate the pH of basic solutions by first calculating the pOH as is done in the following example.

Example. The pH of 0.010 M aqueous NaOH solution is calculated as follows:
[OH-] = 0.010 M
pOH = 2.00
pH + pOH = pKw = 14.00, so
pH = 12.00

As in the case of an aqueous solution of a strong acid, the contribution from the autoionization of water is almost always neglegible. In an aqueous solution of a strong base, acidity is essentially determined by the concentration of the strong base alone.

Weak acids and bases

Equilibria involving weak acids and weak bases can be treated in the same manner as other equilibrium problems:  Identify initial concentrations, define a variable that indicates how much the system must shift to establish equilibrium, and use the resulting (final) equilibrium concentrations in the expression for the equilibrium constant to obtain an algebraic equation that can be solved.

Let c equal the initial concentration of a weak acid, HA, in aqueous solution.

HA(aq)   +   H2O(l)    H3O+(aq)   +   A-(aq)

HA H3O+ A-
initial c 0 0
change -x x x
equilibrium c-x x x

The final concentrations at equilibrium result in the following expression:

To solve the above equation exactly, the quadratic formula is required. The equation above can be recast in the following manner:

Working with the quadratic formula is a bit tedious and in some cases, may not be needed. An approximation can be made when x << c. When x is much smaller than c, the denominator c-x can be reduced to simply c:

One can, of course, check the validity of this approximation by using the value obtained for x by this approximate method in the exact equilibrium expression and checking if the calculated Kaagrees with the given value. A useful rule of thumb is that if x is no larger than 5% of c, the approximation is valid. The degree of how small x is compared to c is, without doubt, dictated partly by the value of Ka. (From the equation above, one can see that x depends as well on the square root of c. Thus, at very low values of c, the approximation will also fail.)

For this class, we will follow this rule:

"If Ka is less than 0.001 (10-3) and c is at least 0.01 M, the approximation can be made. Otherwise, one should use the exact solution via the quadratic formula."

One additional thing that a student should be able to solve after obtaining the equilibrium concentrations is the fraction or percent ionized for a particular acid or base:

Example 1:

What is the pH of a 0.20 M solution of acetic acid (Ka = 1.8 x 10-5)?

HC2H3O2(aq)   +   H2O(l)    H3O+(aq)   +   C2H3O2-(aq)

HC2H3O2 H3O+ C2H3O2-
initial 0.20 0 0
change -x x x
equilibrium 0.20-x x x

Approximate solution:

Ka = (x)(x) / (0.20 - x)  @  x2 / 0.20 = 1.8 x 10-5

x = [(1.8 x 10-5 )(0.20)]1/2

x = 1.9 x 10-3 M = [H3O+] ; pH = -log(1.9 x 10-3 ) = 2.72

Exact solution (using quadratic formula): x = 0.001888 or 0.0019 M.
 

More examples (The following values pertain to a weak acid, Ka=1.8 x 10-5)
 
Initial Conc.

 HA (c)

 x (approx) x(exact) [H+](exact) pH %ionized pH if strong acid
1.00000 0.004243 0.004234 0.004234 2.373 0.42% 0.000
0.90000 0.004025 0.004016 0.004016 2.396 0.45% 0.046
0.80000 0.003795 0.003786 0.003786 2.422 0.47% 0.097
0.50000 0.003000 0.002991 0.002991 2.524 0.60% 0.301
0.20000 0.001897 0.001888 0.001888 2.727 0.94% 0.699
0.10000 0.001342 0.001333 0.001333 2.875 1.33% 1.000
0.05000 0.000949 0.000940 0.000940 3.027 1.88% 1.301
0.01000 0.000424 0.000415 0.000415 3.382 4.15% 2.000
0.00500 0.000300 0.000291 0.000291 3.536 5.82% 2.301
0.00100 0.000134 0.000125 0.000125 3.901 12.55% 3.000
0.00010 0.000042 0.000034 0.000034 4.464 34.37% 4.000
0.00001 0.000013 0.000007 0.000007 5.145 71.55% 5.000

Example 2:

What is the pH of a 0.10 M solution of hydrofluoric acid?

HF(aq)   +   H2O(l)    H3O+(aq)   +   F-(aq)

HF H3O+ F-
initial 0.10 0 0
shift -x x x
final 0.10-x x x

Approximate solution:

Ka = (x)(x) / (0.10 - x)  @  x2 / 0.10 = 6.8 x 10-4

x = [(6.8 x 10-4 )(0.10)]1/2

x = 8.2 x 10-3 M = [H3O+] ; pH = -log(8.2 x 10-3 ) = 2.08

Exact solution (using quadratic formula): x = 0.007913 or 0.0080 M; pH = 2.10.