How Do Weak Acids Differ from Strong Acids?

Calculating the pH of a weak acid is a significantly more involved process than what you encountered with strong acids. A quick review of strong acids illustrates why.

Consider a 0.1 M solution of HCl. Because HCl is a strong acid, it dissociates 100% into H⁺ and Cl⁻. That complete dissociation means the H⁺ concentration—in a one-to-one ratio—is simply 0.1 M. Take the negative log: pH = 1. Done in about 15 seconds.

Now consider 0.1 M acetic acid (CH₃COOH). Acetic acid is a weak acid, and weak acids do not dissociate completely—they only dissociate partially. So how much H⁺ do you actually get? The honest answer: we cannot determine that immediately. What we can say with certainty is:

• The H⁺ concentration does not equal 0.1 M
• The H⁺ concentration is less than 0.1 M—and probably considerably less, since typical weak acids dissociate less than 5%
• The pH is therefore greater than 1 (remember the inverse relationship: as H⁺ goes down, pH goes up)

Notice that so far we have only established what the pH is not. To determine the actual pH, we need additional information—specifically, the K_a expression and K_a value for the acid.

What Is K_a? The Acid Dissociation Constant

K_a is simply an equilibrium constant—specifically, the acid dissociation constant for the dissociation of a weak acid in water. For acetic acid, K_a = 1.76 × 10⁻⁵. Rather than writing this for one specific acid, we can express it generically for any monoprotic weak acid:

HA → H⁺ + A⁻

Some instructors prefer to show the water explicitly, reflecting the fact that "naked" H⁺ ions (bare protons) are so reactive they almost certainly exist as hydronium (H₃O⁺) in aqueous solution:

HA + H₂O → H₃O⁺ + A⁻

The equilibrium constant expression for this generic dissociation is:

K_a = [H⁺][A⁻] / [HA]

If you write the reaction with H₂O, then H₃O⁺ replaces H⁺ on top. The water term drops out because liquids do not appear in equilibrium expressions. The two forms are completely equivalent—H⁺ and H₃O⁺ are treated as synonymous.

Comparing Acid Strengths Using K_a

An acid's strength is measured by how much it dissociates. Greater dissociation means more products, which translates to a larger equilibrium constant. The K_a value therefore provides a single number that quantifies relative acid strength.

Earlier in the chapter, you learned to compare acidities conceptually—using binary acid trends, oxyacid trends, and polyprotic acid sequences. K_a values now give you a numerical tool for comparing any two weak acids: whichever has the higher K_a is the stronger acid, no conceptual analysis required.

How to Calculate pH of a Weak Acid: The ICE Table Method

Let's work through the first actual calculation: find the pH of a 0.1 M solution of acetic acid (K_a = 1.76 × 10⁻⁵). We will use the generic weak acid notation HA rather than writing out acetic acid's specific formula, since the mathematics works identically for any weak monoprotic acid.

Setting Up the ICE Table

Start with the generic dissociation reaction and set up the ICE (Initial, Change, Equilibrium) table:

Technically, pure water already contains 10⁻⁷ M H⁺, but this is negligible compared to the H⁺ produced by the weak acid, so we round the initial H⁺ and A⁻ concentrations to zero. The variable x represents the equilibrium concentration of H⁺—the very number we need to calculate pH.

Substituting into the K_a expression:

1.76 × 10⁻⁵ = (x)(x) / (0.1 − x) = x² / (0.1 − x)

The 5% Approximation: Avoiding the Quadratic

Solving this equation rigorously requires the quadratic formula—and that gets tedious quickly. Fortunately, there is a powerful shortcut. Because K_a values for weak acids are small (reflecting less than 5% dissociation in most cases), the value of x is correspondingly small. When you subtract a very small number from 0.1, the result is essentially still 0.1.

The logic: a small equilibrium constant means the reaction heavily favors reactants. Starting with only reactants and no products, the system barely shifts to the right before reaching equilibrium. The amount that reacts (x) is therefore small. Multiplying by a small number changes the equation significantly, but subtracting a small number from 0.1 does not.

Solving for x and Calculating pH

Applying the 5% approximation (dropping the −x in the denominator):

x² / 0.1 = 1.76 × 10⁻⁵

x² = 1.76 × 10⁻⁵ × 0.1 = 1.76 × 10⁻⁶

x = √(1.76 × 10⁻⁶) = 0.0013 M

This x value is the equilibrium H⁺ concentration. To verify our approximation: 0.0013 / 0.1 × 100 = 1.3%, which is well below the 5% threshold. The approximation was valid.

Finally, the pH:

pH = −log(0.0013) = 2.9

The Weak Acid pH Shortcut Formula

Now that you understand the full ICE table derivation, here is the payoff. When you apply the 5% approximation (which works for the vast majority of weak acid problems), the entire process simplifies to a single formula:

[H⁺] = √(K_a × [HA]_initial)

Multiply the K_a by the initial acid concentration, take the square root, and you have the H⁺ concentration. Then take the negative log to obtain the pH. That's it.

You need two pieces of information: the K_a and the initial acid concentration. Given both, the calculation takes seconds rather than minutes.

How to Find K_a from pH and Calculate Percent Dissociation

Not every weak acid problem asks for pH. Consider this: what is the K_a of a weak acid HA if the pH of a 0.1 M solution is 3?

The critical insight: if pH = 3, then [H⁺] = 10⁻³ M. Recall that knowing any one of the four values—[H⁺], pH, [OH⁻], or pOH—allows you to calculate the other three. By providing the pH, the problem effectively hands you the equilibrium H⁺ concentration.

From the ICE table, [H⁺] = x = 10⁻³ M. Substitute directly into the K_a expression:

K_a = x² / (0.1 − x) = (10⁻³)² / (0.1 − 10⁻³)

K_a = (10⁻³)² / (0.1 − 10⁻³) = 10⁻⁶ / 0.099 = 1.01 × 10⁻⁵

Had you dropped the −x, you would have obtained 1.00 × 10⁻⁵—essentially the same result.

Calculating Percent Dissociation

The second part of this problem asks: what is the percent dissociation? In an ICE table, x represents the amount that dissociates. The percent dissociation formula is straightforward:

% dissociation = (x / [HA]_initial) × 100 = ([H⁺]_eq / [HA]_initial) × 100

Weak Base pH Calculations with K_b

Everything we established for weak acids has a direct analogue for weak bases. The shortcut formula for weak acids was [H⁺] = √(K_a × [HA]). For weak bases, the analogous formula is:

[OH⁻] = √(K_b × [B]_initial)

There are two key differences. First, instead of K_a you use K_b—the base dissociation constant for a weak base dissociating in water. Second, the formula gives you the hydroxide concentration rather than H⁺. That means you obtain the pOH first, then subtract from 14 to get the pH. One extra step compared to acids.

The Weak Base Dissociation Reaction

For weak bases, water must appear in the reaction. A Brønsted-Lowry base is a proton acceptor, and the base needs a source of H⁺—that source is water. (For weak acids, including water was optional. For weak bases, it is mandatory.)

A⁻ + H₂O → OH⁻ + HA

The base (A⁻, the conjugate base of acid HA) steals a proton from water. This produces hydroxide and converts A⁻ into its conjugate acid, HA.

The K_b Expression

The equilibrium constant for this reaction is K_b:

K_b = [OH⁻][HA] / [A⁻]

Water is a liquid and therefore excluded from the expression—just as with K_a.

K_a × K_b = K_w: The Conjugate Acid-Base Relationship

A powerful relationship emerges when you multiply the K_a of an acid by the K_b of its conjugate base. The A⁻ and HA terms cancel algebraically, leaving only [H⁺][OH⁻]—which equals K_w:

K_a × K_b = K_w = 1 × 10⁻¹⁴ (at 25°C)

This relationship also demonstrates mathematically what you already know conceptually: the stronger an acid, the weaker its conjugate base (and vice versa). A large K_a forces K_b to be small so that their product still equals 10⁻¹⁴.

Ammonia pH Calculation: Weak Base Example

Find the pH of a 0.1 M solution of ammonia (K_b = 1.75 × 10⁻⁵). This example was chosen deliberately: the K_b for ammonia (1.75 × 10⁻⁵) is nearly identical to the K_a for acetic acid (1.76 × 10⁻⁵), so the ICE table setup will look almost identical to the weak acid problem.

The critical difference: for a base, x equals [OH⁻], not [H⁺]. The K_b expression is:

K_b = x² / (0.1 − x)

Applying the 5% approximation and solving:

x² = 1.75 × 10⁻⁵ × 0.1 = 1.75 × 10⁻⁶

x = √(1.75 × 10⁻⁶) = 0.0013 M = [OH⁻]

Now comes the extra step that distinguishes base calculations from acid calculations:

pOH = −log(0.0013) = 2.9

pH = 14 − 2.9 = 11.1

How to Find K_b from pH for a Weak Base

What is the K_b of a weak base A⁻ if the pH of a 0.1 M solution is 10.2? This is arguably the trickiest problem type in this section, but it follows the same logic we used for finding K_a from pH—with extra conversions at the front end.

You are given pH = 10.2. What you need is x = [OH⁻] for the ICE table. The conversion path:

pOH = 14 − 10.2 = 3.8

[OH⁻] = 10⁻³·⁸ M

Now substitute into the K_b expression (keeping the −x for accuracy, since there is no quadratic to worry about):

K_b = (10⁻³·⁸)² / (0.1 − 10⁻³·⁸) = 2.5 × 10⁻⁷

pH Calculation Methods for 4 Solution Types

By this point in the course, you have encountered four distinct types of solutions that each require a different pH calculation approach. Correctly identifying which type you are dealing with is half the battle.

That covers four of the five major solution types. The fifth—buffers—is addressed in the next chapter, along with titration calculations (which often reduce to one of the four types above at different stages of the titration).