Problem 92 - Chemistry: The Central Science in SI Units, Global Edition

Question

Mathematically prove that the \(\mathrm{pH}\) at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to \(\mathrm{p} K_{a}\) for the acid.

Step-by-Step Solution

Answer

At the halfway point of titration of a weak acid with a strong base, the remaining concentration of the weak acid HA is \(\frac{1}{2}C_{A}\), and the concentration of the conjugate base MA is also \(\frac{1}{2}C_{A}\), since the stoichiometry of the reaction is 1:1. The equilibrium constant, \(K_{a}\), can be expressed as \(K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}\). At the halfway point of titration, the pH is calculated as \(\mathrm{pH} = -\log(K_{a})\), and since p\(K_{a} = -\log(K_{a})\), the pH at the halfway point is equal to the p\(K_{a}\) for the weak acid.

Step 1: Write the chemical reaction

For the titration of a weak acid HA with a strong base MOH, the chemical reaction can be written as: \[ HA + MOH \rightarrow MA + H_{2}O \] At the halfway point of the titration, exactly half of the weak acid has been neutralized by the strong base.

Step 2: Determine the relevant concentrations at the halfway point

Let the initial concentration of the weak acid HA be denoted by \(C_{A}\), and let the volume of the MOH solution used to reach the halfway point be \(V_{1}\). Since half of the acid has been neutralized, the remaining concentration of HA is \(\frac{1}{2}C_{A}\), and the concentration of the conjugate base MA is also \(\frac{1}{2}C_{A}\), since the stoichiometry of the reaction is 1:1.

Step 3: Write an expression for the equilibrium concentrations

Since the weak acid is incompletely ionized in water, we can write the ionization equilibrium as: \[ HA \rightleftharpoons H^{+} + A^{-} \] Then, the equilibrium constant, \(K_{a}\), can be expressed as: \[ K_{a} = \frac{[H^{+}][A^{-}]}{[HA]} \] At the halfway point, \([A^{-}] = \frac{1}{2}C_{A}\) and \([HA] = \frac{1}{2}C_{A}\).

Step 4: Solve for [H+] (hydrogen ion concentration)

Substitute the concentrations of \(A^{-}\) and \(HA\) at the halfway point into the equilibrium constant expression: \[ K_{a} = \frac{[H^{+}]\left(\frac{1}{2}C_{A}\right)}{\left(\frac{1}{2}C_{A}\right)} \] Since we have equal concentrations of weak acid HA and its conjugate base A-, we can simplify this expression as follows: \[ K_{a} = [H^{+}] \]

Step 5: Calculate the pH at the halfway point

Now that we have an expression for the hydrogen ion concentration, we can calculate the \(\mathrm{pH}\) at the halfway point of the titration using the definition of \(\mathrm{pH}\): \[ \mathrm{pH} = -\log ([H^{+}]) \] At the halfway point, \([H^{+}] = K_{a}\), so we can substitute this into the \(\mathrm{pH}\) equation: \[ \mathrm{pH} = -\log(K_{a}) \] By definition, p\(K_{a} = -\log(K_{a})\), so we can rewrite the equation as: \[ \mathrm{pH} = \mathrm{p} K_{a} \] Thus, at the halfway point of titration of a weak acid with a strong base, the \(\mathrm{pH}\) is equal to the p\(K_{a}\) for the weak acid.

Key Concepts

Weak Acid Ionization

In an acid-base titration, weak acids are characterized by their partial ionization in water. Unlike strong acids, which ionize completely, weak acids exist in equilibrium between their non-ionized and ionized forms. For a weak acid, which can be represented as HA, the ionization can be described by the chemical equilibrium:

• The non-ionized form is HA.
• It ionizes into the proton \(H^+\) and the ion \(A^-\).

This dynamic equilibrium is essential to understanding the behavior of weak acids in titrations because as the acid partially ionizes, some of it remains in the form HA, while another part is converted to its ions. This impacts how the pH changes during titration. The amount of ionization can be affected by factors like temperature and, importantly in titration, the addition of the strong base MOH. As the strong base is added, it reacts with the weak acid, which establishes another equilibrium in the solution, albeit a shifted one, because achieving full equivalence will tend to suppress ionization under the common ion effect.

pH Calculation

The concept of \( ext{pH}\) is central to acid-base chemistry. It describes the hydrogen ion concentration in a solution, quantified by the expression:\[\text{pH} = -\log([H^+])\]During a titration, knowing the pH helps determine the endpoint or equivalence point. At the halfway point in the titration of a weak acid with a strong base, an interesting phenomenon occurs: the concentrations of the weak acid and its conjugate base are equal. This equality can be used to simplify the equilibrium constant expression to solve for the pH:

• At the halfway point: \([HA] = [A^-]\)
• Therefore, \(K_a = [H^+]\)

Because at this stage the hydrogen ion concentration equals the acid dissociation constant (\(K_a\)), the pH becomes equal to the \( ext{p}K_a\) (which is \(-\log (K_a)\)). So, the pH can be directly calculated without needing complex math or equipment. This relationship is a cornerstone for performing titration calculations swiftly and accurately.

Equilibrium Constant

In a weak acid titration, the equilibrium constant \(K_a\) is crucial because it defines the extent of ionization of the weak acid in solution. The \(K_a\) is expressed with the formula:\[K_a = \frac{[H^+][A^-]}{[HA]}\]This value helps explain just how much the acid dissociates in water. Typically, weak acids have a \(K_a\) much less than 1, pointing to incomplete ionization. To determine the pH, knowing the \(K_a\) value is vital. It provides a mathematical way to calculate the hydrogen ion concentration, particularly at points like the halfway mark in titration.

• At this crucial stage, we say:
• \([H^+] = K_a\)

From equilibrium, the concentrations of HA and \(A^-\) are equal, simplifying the equation to focus on the amount of \(H^+\) ions in solution. Therefore, knowing the equilibrium constant helps predict pH shifts during the titration of a weak acid with a strong base.