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 = -log[H_{3}O^{+}]
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:
[H_{3}O^{+}][OH^{-}] = K_{w} = 1 x 10^{-14}
in molar concentration units. Taking the logarithm of both sides;
log[H_{3}O^{+}] + log[OH^{-}] = logK_{w} = -14
Then
-log[H_{3}O^{+}] + (-log[OH^{-}]) = -logK_{w} = +14
or
pH + pOH = pK_{w} = 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.
HCl + H_{2}O H_{3}O^{+} + Cl^{-}
[H_{3}O^{+}] = [HCl]
[H_{3}O^{+}] = 0.0010
pH = -log[H_{3}O^{+}] = 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 H_{3}O^{+}. 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 H_{3}O^{+} 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.
[H_{3}O^{+}] = 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
[H_{3}O^{+}] 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!
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) + H_{2}O(l) H_{3}O^{+}(aq) + A^{-}(aq)
HA | H_{3}O^{+} | 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 K_{a}agrees 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 K_{a}. (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 K_{a} 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:
What is the pH of a 0.20 M solution of acetic acid (K_{a} = 1.8 x 10^{-5})?
HC_{2}H_{3}O_{2}(aq) + H_{2}O(l) H_{3}O^{+}(aq) + C_{2}H_{3}O_{2}^{-}(aq)
HC_{2}H_{3}O_{2} | H_{3}O^{+} | C_{2}H_{3}O_{2}^{-} | |
initial | 0.20 | 0 | 0 |
change | -x | x | x |
equilibrium | 0.20-x | x | x |
Approximate solution:
K_{a} = (x)(x) / (0.20 - x) @ x^{2} / 0.20 = 1.8 x 10^{-5}
x = [(1.8 x 10^{-5} )(0.20)]^{1/2}
x = 1.9 x 10^{-3} M = [H_{3}O^{+}] ; 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, K_{a}=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) + H_{2}O(l) H_{3}O^{+}(aq) + F^{-}(aq)
HF | H_{3}O^{+} | F^{-} | |
initial | 0.10 | 0 | 0 |
shift | -x | x | x |
final | 0.10-x | x | x |
Approximate solution:
K_{a} = (x)(x) / (0.10 - x) @ x^{2} / 0.10 = 6.8 x 10^{-4}
x = [(6.8 x 10^{-4} )(0.10)]^{1/2}
x = 8.2 x 10^{-3} M = [H_{3}O^{+}] ; pH = -log(8.2 x 10^{-3} ) = 2.08
Exact solution (using quadratic formula): x = 0.007913 or 0.0080 M; pH = 2.10.