Most substances that are acidic in water are actually weak acids. Because weak acids dissociate only partially in aqueous solution, an equilibrium is formed between the acid and its ions. The ionization equilibrium is given by:

HX(aq) $H+(aq)$ + $X-(aq)$

However, keep in mind that the "Before Dissociation" state never actually exists in solution--the solution is always at equilibrium. The left hand state is theoretical.

The equilibrium constant is then:

Click here or on the button on the menu bar for a table of K_a values for common weak acids.

The smaller the value for $K$_a, the weaker the acid. Weaker acids dissociate less ($[H+]$ is smaller compared to [HX]) and therefore have a less drastic effect on pH.

POINT OF EMPHASIS: Do not confuse a weak acid with a dilute acid. A weak acid has a small $K$_a, and a dilute acid has a low concentration. It is possible to have a dilute, strong acid or a concentrated, weak acid.

Calculating the pH for Solutions of Weak Acids

EXAMPLE:

STEP 1: Write the ionization equilibrium for acetic acid:

$HC$₂H₃C₂ (aq) H⁺ (aq) + C₂H₃O₂^- (aq)

STEP 2: Create an I.C.E. grid, and determine the concentration from information provided in the problem:

$HC$₂H₃C₂ (aq) H⁺ (aq) + C₂H₃O₂^- (aq)

STEP 3: Substitute the equilibrium concentrations into the equilibrium constant expression:
The equilibrium constant expression is:

This equation has only one unknown and can be solved using the quadratic formula. However, we can make things easier:

STEP 4: Since the value of $K$_a is quite small, we can guess that the value of $x$ will be very small (ie. only a small portion of the $HC$₂H₃O₂ actually dissociates). If we assume that $x$ is much smaller (less than 5%) than the initial concentration of the acid, we can avoid using the quadratic formula, because when a small number is subtracted from a much larger one, the answer will be approximately the larger one. The equation becomes:

and $x\; =\; 1.3$x $10-3M\; =\; [H+]$, which is less than 5% of 0.10M, the initial concentration of the acid. Tip

STEP 5: Now we find pH:

pH = 2.89

Learning Goal 11
Calculate the percentage ionization for an acid or a base, knowing its concentration in solution, and the value of $K$a or $K$b.

Percent ionization or percent dissociation is defined as:

The percent dissociation of an acid varies with the concentration of the acid. The more dilute an acid is, the greater the percent ionization. Why?

Let's look at the equilibrium of a weak acid again:

HX(aq) + $H$₂O $H$₃O⁺(aq) + X^-(aq)

From Le Chatelier's Principle, adding water to the equilibrium would cause the equilibrium to shift to the right. A shift to the right implies that more acid would be in dissociated form, and thus the percent ionization increases accordingly.

EXAMPLE:

Calculate the percent of HF molecules ionized in a 0.10 M HF solution. The $K$_a for HF is 6.8 x $10-4$.

STEP 1: Ionization equation:

$HF\; (aq)$ H⁺ (aq) + F^- (aq)

STEP 2: I.C.E. grid:

$HF\; (aq)$ H⁺ (aq) + F^- (aq)

	HF	$H+$	$F-$
INITIAL	0.10M	0	0
CHANGE	-$x$M	+$x$M	+$x$M
EQUILIBRIUM	(0.10 - $x$)M	$x$M	$x$M

STEP 3: Equilibrium constant expression:

$K$_a = { [H⁺][F^-] / [HF] } = { (x)(x) / (0.10 - x) } = 6.8 x $10-4$
STEP 4: Solve equation, assuming $x$ << 0.10:

$6.8$x $10-4=\; \{\; x2/\; 0.10\}$

$x\; =\; 8.2$x $10-3M$

However,

$$

{ 8.2 x 10^-3M / 0.1M } x 100 = 8.2%,

which is greater than 5% of the initial concentration of the acid, so x is not x << 0.10M!

STEP 5: Use quadratic:

$6.8$x $10-4=\; \{\; x2/\; (0.10\; -\; x)\; \}$

$x2+\; (6.8$x $10-4)\; x\; -\; 6.8$x $10-5=\; 0$

$x\; =\; 7.9$x $10-3M\; =\; [H+]$

Therefore, $percent\; ionization\; =\; \{\; [\; ionized\; ]\; /\; [\; initial]\; \}$ x $100$:

$\{\; 7.9\; x\; 10-3M\; /\; 0.10M\; \}$ x $100\; =\; 7.9\%$

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