Question

The ionization constant of a very weak acid, HA, is \(4.0 \times 10^{-9} .\) Calculate the equilibrium concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{A}^{-},\) and \(\mathrm{HA}\) in a \(0.040 \mathrm{M}\) solution of the acid.

Short answer

[H3O+] = [A-] = 1.26 x 10^-5 M; [HA] ≈ 0.040 M.

Step-by-step solution

Write the Ionization Equation

The ionization equation for the weak acid HA in water is given by: \[ \mathrm{HA} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{A^-}. \] This equation indicates that the acid HA partially ionizes to form hydronium ions \(\mathrm{H_3O^+}\) and conjugate base \(\mathrm{A^-}\) in solution.

Write the Expression for the Ionization Constant

The ionization constant \(K_a\) for the reaction is given by the expression: \[ K_a = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}. \] We know from the problem that \(K_a = 4.0 \times 10^{-9}.\)

Set Initial Concentration and Changes at Equilibrium

Initially, the concentration of HA is \(0.040 \text{ M}\), and the concentrations of \(\mathrm{H_3O^+}\) and \(\mathrm{A^-}\) are \(0 \text{ M}\). Let \(x\) be the amount of HA that ionizes. Then, at equilibrium, the concentrations will be:- \([\mathrm{HA}] = 0.040 - x\),- \([\mathrm{H_3O^+}] = x\),- \([\mathrm{A^-}] = x\).

Substitute in the Ionization Constant Expression

Substitute the equilibrium concentrations back into the ionization constant expression:\[ K_a = \frac{x \cdot x}{0.040 - x} = \frac{x^2}{0.040 - x}. \] Since the acid is very weak, assume \(x\) is much smaller than 0.040, making \(0.040 - x \approx 0.040.\)

Solve for x

Substitute the approximated expression into the ionization constant equation:\[ 4.0 \times 10^{-9} = \frac{x^2}{0.040}. \]Solving for \(x\):\[ x^2 = 4.0 \times 10^{-9} \times 0.040, \]\[ x^2 = 1.6 \times 10^{-10}, \]\[ x = \sqrt{1.6 \times 10^{-10}} \approx 1.26 \times 10^{-5}. \]

Calculate Equilibrium Concentrations

Now use the value of \(x\) to find the equilibrium concentrations:- \([\mathrm{H_3O^+}] = x = 1.26 \times 10^{-5} \text{ M},\)- \([\mathrm{A^-}] = x = 1.26 \times 10^{-5} \text{ M},\)- \([\mathrm{HA}] = 0.040 - x \approx 0.040 \text{ M} \). Given that \(x\) is much smaller than 0.040, this approximation is very close to the exact value.

Key concepts

Understanding Equilibrium Concentrations

In the context of chemistry, equilibrium concentrations refer to the amounts of each species present in a reaction once the system has reached equilibrium. At this point, the forward and reverse reactions occur at the same rate, resulting in stable concentrations of reactants and products. When dealing with weak acids like HA, the equilibrium concentrations of the acid ([\mathrm{HA}]) and its ionized forms ([\mathrm{H_3O^+}] and [\mathrm{A^-}]) are of great interest.

• The equilibrium concentration of hydronium ions ([\mathrm{H_3O^+}]) and the conjugate base ([\mathrm{A^-}]) are found by solving for the variable that represents the amount of acid ionized, typically symbolized as \(x\).
• The initial concentration of the acid is given, and from this, using the ionization constant, the equilibrium concentrations are calculated, taking into account that weak acids ionize only slightly.

In our example with a 0.040 M solution of HA, the equilibrium concentration calculations show that both [\mathrm{H_3O^+}] and [\mathrm{A^-}] are approximately 1.26 x 10⁻⁵ M, while [\mathrm{HA}] remains nearly the same as its initial concentration due to minimal ionization.

Characteristics of Weak Acids

Weak acids are characterized by their limited ability to donate protons to water molecules. Unlike strong acids, which ionize completely, weak acids only partially ionize in solution. This leads to a dynamic equilibrium where only a fraction of the acid molecules dissociate into ions at any given time.

• Due to their partial ionization, weak acids have smaller ionization constants \(K_a\), often much less than 1.
• For very weak acids, like the one in our exercise with a \(K_a = 4.0 \times 10^{-9}\), the equilibrium concentration of ionized species is quite low compared to the initial concentration of the acid.
• This results in most of the acid remaining in its original, non-ionized form.

Understanding that weak acids do not fully dissociate explains why modifications and approximations, like assuming \(0.040 - x \approx 0.040\), are valid and useful in simplifying calculations for equilibrium concentrations.

Writing and Using an Ionization Equation

The ionization equation is fundamental when working with weak acids. It represents the chemical equilibrium established between the acid, its conjugate base, and hydronium ions in water:\[\mathrm{HA} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{A^-}.\]This equation shows that the acid (HA) reacts with water to produce hydronium ions \(\mathrm{H_3O^+}\) and the conjugate base \(\mathrm{A^-}\). Here’s why this equation is important:

• It helps visualize the weak acid's tendency to donate a proton, albeit to a limited extent, thus remaining largely as HA in solution.
• It serves as a foundation to derive the expression for the ionization constant \(K_a\):\[K_a = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}\]
• This expression helps us calculate the extent of ionization using the concentrations of the species at equilibrium.

Using the ionization equation and applying mathematical techniques makes it possible to predict and explain the behavior of weak acids in solution, confirming their small but significant ionization in aqueous environments.