Question

The ionization constant for acetic acid is \(1.8 \times 10^{-5}\). a) Calculate the concentration of \(\mathrm{H}^{+}\) ions in a \(0.10\) molar solution of acetic acid. b) Calculate the concentration of \(\mathrm{H}^{+}\) ions in a \(0.10\) molar solution of acetic acid in which the concentration of acetate ions has been increased to \(1.0\) molar by addition of sodium acetate.

Short answer

The concentration of \(\mathrm{H}^{+}\) ions in a \(0.10\) M solution of acetic acid is approximately \(1.34 \times 10^{-3}\) M. In a \(0.10\) M solution of acetic acid with increased acetate ions concentration to 1.0 M, the concentration of \(\mathrm{H}^{+}\) ions is approximately \(1.8 \times 10^{-6}\) M.

Step-by-step solution

a) Calculate the concentration of \(\mathrm{H}^{+}\) ions in a \(0.10\) molar solution of acetic acid.

To calculate the concentration of \(\mathrm{H}^{+}\) ions in a 0.10 M solution of acetic acid, we start by writing the dissociation reaction and expression for the ionization constant \(\mathrm{K_a}\): Dissociation reaction: \(\mathrm{CH_3COOH \leftrightharpoons H^+ + CH_3COO^-}\) Ionization constant expression: \(\mathrm{K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}}\) Given, \(\mathrm{K_a = 1.8 \times 10^{-5}}\), initial concentration of acetic acid is 0.10 M and we assume it dissociates into \(\mathrm{x\, mol/L}\) of \(\mathrm{H}^{+}\) and \(\mathrm{CH_3COO^-}\). Thus, the concentration of acetic acid left is \((0.10 - x)\). We can now substitute these values in the ionization constant expression: \[\mathrm{1.8 \times 10^{-5} = \frac{x \cdot x}{(0.10 - x)}}\] Since \(\mathrm{K_a}\) is very small, \(x\) will also be small, so we can approximate \((0.10 - x) \approx 0.10\): \[\mathrm{1.8 \times 10^{-5} = \frac{x^2}{0.10}\] Now, solve for \(x\) which represents the \(\mathrm{([H^+]})\): \[\mathrm{x^2 = 1.8 \times 10^{-5} \times 0.10}\] \[\mathrm{x = \sqrt{1.8 \times 10^{-6}}}\] \[\mathrm{[H^+] = x \approx 1.34 \times 10^{-3} \, M}\] So, the concentration of \(\mathrm{H}^{+}\) ions in a \(0.10\) M solution of acetic acid is approximately \(1.34 \times 10^{-3}\) M.

b) Calculate the concentration of \(\mathrm{H}^{+}\) ions in a \(0.10\) molar solution of acetic acid with increased acetate ions concentration of \(1.0\) molar.

We use the same ionization constant expression to determine the concentration of \(\mathrm{H}^{+}\) ions in the second scenario with increased concentration of acetate ions to 1.0 M: \[\mathrm{K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}}\] Initial concentrations are given as: \(\mathrm{[CH_3COOH] = 0.10 \, M}\) and \(\mathrm{[CH_3COO^-] = 1.0 \, M}\). Let \(\mathrm{[H^+]} = x \, M\). We substitute these values into the ionization constant expression: \[\mathrm{1.8 \times 10^{-5} = \frac{x \cdot (1.0)}{(0.10)}}\] Now, solve for \(x\) to find the concentration of \(\mathrm{H^+}\) ions: \[\mathrm{x = 1.8 \times 10^{-5} \times 0.10}\] \[\mathrm{[H^+] = x \approx 1.8 \times 10^{-6} \, M}\] So, the concentration of \(\mathrm{H}^{+}\) ions in a \(0.10\) molar solution of acetic acid with increased concentration of acetate ions to 1.0 M is approximately \(1.8 \times 10^{-6}\) M.

Key concepts

Acid Dissociation

Acid dissociation refers to the process where an acid releases a proton (H⁺ ion) into a solution, causing the acid to split into its ionic components. In acetic acid ( CH₃COOH) dissociation, the molecule releases a proton becoming acetate ion ( CH₃COO⁻). This process can be depicted in a chemical reaction as follows:
CH₃COOH ⇌ H⁺ + CH₃COO⁻
The equilibrium position of this dissociation will determine how much of the acid remains undissociated and how much has dissociated into its ions. The amount that dissociates is quantified by the acid dissociation constant ( K_a), a fundamental parameter indicating the strength of an acid in solution.

• Higher K_a value: stronger acid, more ionization.
• Lower K_a value: weaker acid, less ionization.

This constant provides insight into the degree of dissociation for weak acids, like acetic acid.

Equilibrium Expression

The equilibrium expression is a mathematical representation of the equilibrium state of a chemical reaction. It is formulated based on the concentrations of the reactants and products at equilibrium. For the dissociation of acetic acid, the equilibrium expression is given by the formula:
\(K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}\)
Where:

• \([H^+]\) is the concentration of hydrogen ions.
• \([CH_3COO^-]\) is the concentration of acetate ions.
• \([CH_3COOH]\) is the concentration of acetic acid.

In this expression, the equilibrium constant (K_a) helps us to calculate unknown concentrations when the system is at equilibrium. It is crucial for predicting how the concentrations will change when the initial mixture is disturbed, allowing deeper understanding of reaction dynamics.

Buffer Solution

A buffer solution is an aqueous solution that resists significant changes in pH upon the addition of small amounts of an acid or a base. It typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. Acetic acid and sodium acetate form a buffer system, where acetic acid ( CH₃COOH) is the weak acid and acetate ion ( CH₃COO⁻), provided by sodium acetate, acts as the conjugate base.
This buffer maintains its pH by absorbing excess H⁺ or OH⁻ ions without a significant change in the overall pH.

• When H⁺ ions are added: they are neutralized by the acetate ions ( CH₃COO⁻).
• When OH⁻ ions are added: they react with acetic acid to form more acetate ions.

The presence of a buffer in a solution stabilizes the pH, making it crucial in numerous chemical applications and biological systems.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a vital tool for calculating the pH of a buffer solution. It provides a straightforward way to determine pH when the concentrations of the acid and its conjugate base are known. The equation is given by:
\(pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)\)
Where:

• \(pK_a\) is the negative logarithm of the acid dissociation constant, indicating the strength of the acid.
• \([A^-]\) is the concentration of the conjugate base (acetate ions in this context).
• \([HA]\) is the concentration of the acid (acetic acid here).

This equation shows how the pH is influenced by the ratio of the concentrations of the conjugate base and the acid. It is immensely useful in preparing buffer solutions with desired pH levels in laboratories and industrial processes.