Mastering Weak Acid and Weak Base pH Calculations: The $K_a$ and $K_b$ Approach

Calculating the pH of a strong acid is straightforward: because it dissociates completely, the concentration of protons $[H^+]$ is equal to the initial acid concentration. Weak acids and bases, however, present a unique challenge. They exist in a state of chemical equilibrium, meaning only a fraction of molecules ionize in water.

To solve these problems, you cannot simply take the negative log of the concentration. Instead, you must utilize Equilibrium Constants ($K_a$ and $K_b$) and ICE Tables (Initial, Change, Equilibrium). This guide provides a rigorous, step-by-step framework to master these calculations, ensuring you can tackle any exam question on weak acid/base equilibria.

Defining $K_a$ and $K_b$ and the $K_a-K_b$ Relationship

The strength of a weak acid or base is quantified by its dissociation constant. These constants are simply special versions of the equilibrium constant ($K_{eq}$) applied to ionization reactions.

• Acid Dissociation Constant ($K_a$): Measures the extent to which an acid ($HA$) donates protons in water. $$K_a = \frac{[H_3O^+][A^-]}{[HA]}$$
• Base Dissociation Constant ($K_b$): Measures the extent to which a base ($B$) accepts protons from water to form hydroxide ions. $$K_b = \frac{[BH^+][OH^-]}{[B]}$$

The Conjugate See-Saw

There is a fundamental mathematical relationship between a weak acid and its conjugate base. Water has its own auto-ionization constant, $K_w$, which is $1.0 \times 10^{-14}$ at 25°C.

For any conjugate acid-base pair:

This equation is vital. Often, a problem will ask for the pH of a solution containing a weak base (like the nitrite ion, $NO_2^-$) but will only provide the $K_a$ of its conjugate acid (nitrous acid, $HNO_2$). You must use the relationship above to convert $K_a$ to $K_b$ before starting your calculations.

The ICE Table Method for Weak Acid pH

The standard tool for solving equilibrium problems is the ICE table. This method tracks the concentrations of species at the Initial stage, the Change that occurs, and the Equilibrium state.

Illustrative Example: Acetic Acid

Let's calculate the pH of a 0.100 M solution of acetic acid ($CH_3COOH$). The $K_a$ for acetic acid is $1.8 \times 10^{-5}$.

1. Write the Balanced Equation and $K_a$ Expression

$$CH_3COOH(aq) + H_2O(l) \rightleftharpoons CH_3COO^-(aq) + H_3O^+(aq)$$

$$K_a = \frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]} = 1.8 \times 10^{-5}$$

2. Construct the ICE Table

Water is a pure liquid, so we ignore it in the table. We assume the initial $[H_3O^+]$ is effectively zero.

3. Substitute and Solve

Plug the equilibrium values into the $K_a$ expression:

Because $K_a$ is very small ($10^{-5}$), the amount of acid that dissociates ($x$) is negligible compared to the initial concentration ($0.100$). We can apply the approximation $0.100 - x \approx 0.100$.

4. Calculate pH

Since $x = [H_3O^+]$:

Calculating $K_a$ from pH and Concentration

Sometimes you are given the experimental result (the pH) and asked to determine the acid's strength ($K_a$). This is common in laboratory scenarios where an unknown acid is analyzed.

Practice Problem: Hypochlorous Acid

Question: What is the $K_a$ if a 0.250 M solution of hypochlorous acid ($HClO$) has a pH of 4.07?

Step 1: Convert pH to Equilibrium Concentration

The pH tells us the equilibrium concentration of protons ($x$).

Step 2: Determine Equilibrium Concentrations

From the stoichiometry ($HClO \rightleftharpoons H^+ + ClO^-$), we know that produced ions are equal:

• $[H^+] = [ClO^-] = 8.51 \times 10^{-5} \text{ M}$
• $[HClO]_{eq} = [HClO]_{initial} - [H^+] = 0.250 - 8.51 \times 10^{-5} \approx 0.2499 \text{ M}$

Step 3: Solve for $K_a$

Calculating pH for Weak Base Solutions (Hydrolysis)

Weak bases (like ammonia or conjugate bases of weak acids) increase pH by generating hydroxide ions ($OH^-$). The process is nearly identical to weak acids, but with two critical differences:

1. You must use $K_b$, not $K_a$.
2. Solving for $x$ gives $[OH^-]$, from which you calculate pOH first, then pH.

Case Study: The Nitrite Ion

Problem: Calculate the pH of a 0.18 M solution of the weak base $NO_2^-$. (Given: $K_a$ for $HNO_2$ is $4.5 \times 10^{-4}$).

1. Convert $K_a$ to $K_b$

2. Set up the Equilibrium

Using the approximation method ($0.18 - x \approx 0.18$):

3. Solve for $x$ ($[OH^-]$)

4. Convert to pH

Percent Ionization and the 5% Rule

Percent ionization tells us what percentage of the initial acid or base molecules actually dissociated.

The 5% Rule

In our previous calculations, we assumed $C_{initial} - x \approx C_{initial}$. This approximation is scientifically valid only if the percent ionization is less than 5%. If it is greater than 5%, the approximation introduces too much error, and you must use the quadratic formula to solve for $x$.

Example: When the Rule Fails

Consider a $2.2 \times 10^{-3}$ M solution of Dimethylamine with a $K_b = 5.4 \times 10^{-4}$.

If we try the approximation:

Checking the percent ionization:

The rule fails (~50% > 5%). We cannot ignore $x$. We must solve the full quadratic equation:

Solving this accurately yields $x \approx 8.54 \times 10^{-4}$ M, which corresponds to a true percent ionization of 39%.

Key Takeaway: Always check your assumption ($x / C_{initial}$) at the end of the calculation. If it exceeds 5%, recalculate using the quadratic formula.