Question

A \(0.20 \mathrm{M}\) aqueous solution of monoprotic acid HX has a pH of \(2.14\). (a) Is HX a strong acid or a weak acid? (b) Calculate \(K_{\text {eq }}\) for the reaction of HX with water. (Hint: What are the equilibrium concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{X}^{-}\), and HX?)

Short answer

(a) HX is a weak acid. (b) The equilibrium constant for the reaction of HX with water is approximately \(K_{\text{eq}} \approx 2.6 \times 10^{-4}\).

Step-by-step solution

Determine the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions from the pH

We're given the pH of the solution, which is a measure of acidity or the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions. We can use the pH equation to find the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions: \[ \text{pH} = -\log_{10} [\mathrm{H}_{3} \mathrm{O}^{+}] \] The pH is given as 2.14. Plugging the value into the equation, we can solve for the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions: \[-\log_{10} [\mathrm{H}_{3} \mathrm{O}^{+}] = 2.14\]

Calculate the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions

Rearranging the equation from Step 1, we get: \[ [\mathrm{H}_{3} \mathrm{O}^{+}] = 10^{-\text{pH}} \] Substituting the pH value, we solve for the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions: \[ [\mathrm{H}_{3} \mathrm{O}^{+}] = 10^{-2.14} \] \[ [\mathrm{H}_{3} \mathrm{O}^{+}] \approx 7.2 \times 10^{-3} \, \text{M} \]

Determine if HX is a strong acid or a weak acid

A strong acid, like HCl, would donate all its protons to water, resulting in a concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions equal to the initial concentration of the acid. In our case, the initial concentration of HX is given as 0.20 M, but the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions is much lower at approximately 7.2 x 10^{-3} M. Therefore, we can conclude that HX is a weak acid.

Write the dissociation reaction for HX and its equilibrium expression

The dissociation reaction of HX with water can be represented as: \[ \text{HX} + \text{H}_{2} \text{O} \rightleftharpoons \text{H}_{3} \text{O}^{+} + \text{X}^{-} \] The equilibrium constant expression for this reaction is: \[ K_{\text{eq}} = \frac{[\mathrm{H}_{3} \mathrm{O}^{+}] [\mathrm{X}^{-}]}{[\text{HX}]} \]

Determine the equilibrium concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{X}^{-}\), and HX

At equilibrium, the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{X}^{-}\) ions are equal to each other because one HX molecule dissociates into one \(\mathrm{H}_{3} \mathrm{O}^{+}\) ion and one \(\mathrm{X}^{-}\) ion. We already calculated the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions in Step 2, so we can use this value for both: \[ [\mathrm{H}_{3} \mathrm{O}^{+}] = [\mathrm{X}^{-}] \approx 7.2 \times 10^{-3} \, \text{M} \] As HX is a weak acid, we can assume that only a small portion of it will dissociate. Therefore, the concentration of HX at equilibrium can be approximated as its initial concentration: \[ [\text{HX}] \approx 0.20 \, \text{M} \]

Calculate the equilibrium constant \(K_{\text {eq}}\)

Now that we have the equilibrium concentrations of all species involved in the reaction, we can plug them into the equilibrium constant expression we derived in Step 4: \[ K_{\text{eq}} = \frac{[\mathrm{H}_{3} \mathrm{O}^{+}] [\mathrm{X}^{-}]}{[\text{HX}]} \] \[ K_{\text{eq}} = \frac{(7.2 \times 10^{-3} \, \text{M}) (7.2 \times 10^{-3} \, \text{M})}{0.20 \, \text{M}} \] Calculate the value of K_eq: \[ K_{\text{eq}} \approx 2.6 \times 10^{-4} \] Answers: (a) HX is a weak acid. (b) The equilibrium constant for the reaction of HX with water is approximately \(2.6 \times 10^{-4}\).

Key concepts

Weak Acids

Weak acids are fascinating because they don't fully dissociate in water. Unlike strong acids, which donate nearly all their protons to water, weak acids only partially ionize. This means that in a solution, both the undissociated acid and the ions it breaks into exist simultaneously.
For example, acetic acid (vinegar) is a common weak acid. When you dissolve it in water, only a small portion of it donates a proton to form hydronium ions ([\(\text{H}_3\text{O}^+\)]) and acetate ions ([\(\text{CH}_3\text{COO}^-\)]).
Some key characteristics of weak acids include:

• They have a higher pH compared to strong acids of the same concentration.
• Only a small fraction of their molecules dissociate into ions in solution.
• The strength of a weak acid is measured by its dissociation constant \(K_a\).

In the exercise, HX is identified as a weak acid because the concentration of \([\text{H}_3\text{O}^+]\) ions is much lower than its initial concentration.

Dissociation Reaction

A dissociation reaction is where compounds split into smaller molecules or ions when added to a solvent. For weak acids, this reaction is not complete, meaning the acid is only partially ionized.
When HX dissociates in water, it undergoes the following reaction:
\[ \text{HX} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{X}^- \]
The double arrows indicate that the reaction is reversible and reaches equilibrium. Not all HX molecules transfer their protons to water; some remain intact.
Important aspects of dissociation reactions for weak acids include:

• They create dynamic equilibrium, where the forward and reverse reactions occur at the same rate.
• Both reactants, such as HX, and products like \(\text{H}_3\text{O}^+\) and \(\text{X}^-\), are present simultaneously.
• The extent of dissociation can be influenced by the acid's concentration and the solvent's nature.

Understanding dissociation helps in calculating equilibrium constants, giving insight into acid strength and behavior in solution.

Equilibrium Constant (K_eq)

The equilibrium constant, \(K_{eq}\), is fundamental in understanding the balance between reactants and products in a chemical reaction at equilibrium. For weak acids, the equilibrium constant tells us how much the acid dissociates in solution.
In the exercise, the reaction
\[ \text{HX} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{X}^- \] has its equilibrium constant given by:
\[ K_{eq} = \frac{[\text{H}_3\text{O}^+][\text{X}^-]}{[\text{HX}]} \]
In this expression:

• \([\text{H}_3\text{O}^+]\) is the concentration of hydronium ions at equilibrium.
• \([\text{X}^-]\) is the concentration of the conjugate base ions at equilibrium.
• \([\text{HX}]\) is the remaining concentration of the weak acid at equilibrium.

A low \( K_{eq} \) value, like \(2.6 \times 10^{-4}\), indicates minimal dissociation, confirming our acid's weak nature. This provides insights into the strength and reactivity of HX in aqueous solutions.