Weak acids only partially ionize in water, so their pH must be found by combining an equilibrium model with $K_a$ (or $pK_a$). This page outlines the essential setup and solution logic.

Weak acid equilibrium model

A weak acid reacts reversibly with water, establishing an equilibrium mixture of undissociated acid and ions.

The standard reaction model (for a monoprotic acid) is:

General monoprotic weak-acid equilibrium in water, showing $HA(aq)$ establishing equilibrium with $H_3O^+(aq)$ and $A^-(aq)$. This visual directly supports writing the equilibrium expression and setting up an ICE table for weak-acid pH calculations. Source

• $HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq)$

• Water is a pure liquid and is not included in the equilibrium expression.

The acid dissociation constant, $K_a$

A larger $K_a$ means more products at equilibrium and therefore a lower pH (more acidic).

Using $pK_a$ to simplify acid strength comparisons

Because $K_a$ values often span many powers of ten, chemists commonly use $pK_a$.

A lower $pK_a$ corresponds to a stronger acid (larger $K_a$).

Finding pH from initial concentration and $pK_a$

To determine pH, you connect the initial acid concentration (often called the formal concentration, $C_0$) to the equilibrium $[H_3O^+]$ produced by dissociation.

Core setup: ICE approach (no numbers required)

Use an ICE (Initial–Change–Equilibrium) framework:

• Initial: $[HA]=C_0$, $[H_3O^+] \approx 0$, $[A^-]\approx 0$

• Change: $HA$ decreases by $x$; $H_3O^+$ increases by $x$; $A^-$ increases by $x$

• Equilibrium: $[HA]=C_0-x$, $[H_3O^+]=x$, $[A^-]=x$

Substitute these equilibrium expressions into the $K_a$ expression:

• $K_a=\dfrac{x^2}{C_0-x}$

Common AP-level approximation

For many weak acids, ionization is small relative to the starting amount, so $x \ll C_0$. Then:

• $C_0-x \approx C_0$

• $K_a \approx \dfrac{x^2}{C_0}$

• $x \approx \sqrt{K_a C_0}$, and $x$ represents $[H_3O^+]$

This approximation must be checked conceptually: it is most reliable for small $K_a$ and not-too-dilute $C_0$.

Converting $[H_3O^+]$ to pH

Once equilibrium $[H_3O^+]$ is found (exactly or approximately), compute pH.

A labeled pH scale that pairs each integer pH value with the corresponding hydronium concentration, emphasizing the base-10 logarithmic nature of $pH=-\log[H_3O^+]$. It helps students interpret how a 1-unit change in pH corresponds to a tenfold change in $[H_3O^+]$. Source

Practical guidance for correct equilibrium reasoning

Choosing $K_a$ vs $pK_a$

• If $pK_a$ is given, convert to $K_a$ when you need to solve with algebra: $K_a=10^{-pK_a}$.

• If $K_a$ is given, you can proceed directly to the equilibrium setup.

What the equilibrium picture implies

• Because most $HA$ remains, $[H_3O^+]$ is much smaller than $C_0$ for a weak acid.

• The conjugate base forms in the same amount as hydronium for a monoprotic acid at equilibrium: $[A^-]=[H_3O^+]$ (within the model’s assumptions).

When the approximation is not sufficient

If $x$ is not negligible compared with $C_0$, you must treat $C_0-x$ exactly and solve the resulting equation rather than simplifying it. This is an equilibrium-math issue, not a different chemical model.