Question

Without doing any calculations, determine if \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) \(\mathrm{pH}>\mathrm{p} K_{\mathrm{a}},\) or \(\mathrm{pH}<\mathrm{p} K_{\mathrm{a}}\). Assume that \(\mathrm{HA}\) is a weak monoprotic acid. a. \(0.10 \mathrm{~mol}\) HA and \(0.050 \mathrm{~mol}\) of \(\mathrm{A}^{-}\) in \(1.0 \mathrm{~L}\) of solution b. \(0.10 \mathrm{~mol}\) HA and \(0.150 \mathrm{~mol}\) of \(\mathrm{A}^{-}\) in \(1.0 \mathrm{~L}\) of solution c. \(0.10 \mathrm{~mol}\) HA and \(0.050 \mathrm{~mol}\) of \(\mathrm{OH}^{-}\) in \(1.0 \mathrm{~L}\) of solution d. \(0.10 \mathrm{~mol}\) HA and \(0.075 \mathrm{~mol}\) of \(\mathrm{OH}^{-}\) in \(1.0 \mathrm{~L}\) of solution

Short answer

a) pH < pKa, b) pH > pKa, c) pH > pKa, d) pH ≈ pKa (slightly less)

Step-by-step solution

Analyze Case a

With 0.10 mol of HA and 0.050 mol of A- in 1.0 L of solution, the ratio of A- to HA is 1:2. Since there is less A- (the conjugate base) than HA (the weak acid), the solution is more acidic, and therefore, the pH will be less than the pKa.

Analyze Case b

With 0.10 mol of HA and 0.150 mol of A- in 1.0 L of solution, the ratio of A- to HA is 3:2. Since there is more A- than HA, the solution is more basic, and therefore, the pH will be greater than the pKa.

Analyze Case c

With 0.10 mol of HA and 0.050 mol of OH- in 1.0 L of solution, the OH- will react with HA to form A- and water. Since OH- is a strong base, it will increase the amount of A-, shifting the balance towards a more basic solution, and hence the pH will be greater than the pKa.

Analyze Case d

With 0.10 mol of HA and 0.075 mol of OH- in 1.0 L of solution, the OH- will partially neutralize the HA to form more A-. Despite some neutralization, there will still be more HA than A- left. However, due to the reaction with OH-, this case is less acidic than case a, and it's the closest to neutrality among all cases. The pH could be close to, but slightly less than, the pKa.

Key concepts

pH and pKa Relationship

Understanding the relationship between pH and pKa is crucial when dealing with weak acids and their dissociation in solution. The pH of a solution indicates its acidity or basicity, with lower values being more acidic and higher values more basic.

pH is a measure of the hydrogen ion concentration \[\mathrm{[H^+]}\], whereas pKa is the negative logarithm of the acid dissociation constant (Ka) for a particular acid, representing the strength of the acid. A low pKa value indicates a strong acid, which dissociates completely in water, and higher pKa values indicate weaker acids.

The important point to remember is that when the pH of a solution equals the pKa of the weak acid present, the concentrations of the acid and its conjugate base are equal. This is due to the Henderson-Hasselbalch equation, which connects pH, pKa, and the ratio of the conjugate base \[\mathrm{[A^-]}\] to the acid \[\mathrm{[HA]}\]:
\[\text{pH} = \text{pKa} + \log\left(\frac{\mathrm{[A^-]}}{\mathrm{[HA]}}\right)\]
Understanding this relationship helps to predict the behavior of the acid-base balance in solution and is essential for solving problems like the exercise given.

Weak Monoprotic Acids

Weak monoprotic acids, such as the hypothetical HA in the given exercise, only partially dissociate in solution. This incomplete dissociation creates a dynamic equilibrium between the undissociated acid \(HA\) and the products of its dissociation, which are the hydrogen \(H^+\) and conjugate base ions \(A^-\).

Because these acids don’t fully dissociate, the solution’s pH is influenced by both the acid and its conjugate base, unlike strong acids which dissociate completely. The degree of dissociation is determined by the acid's strength, represented by its pKa value.

In practical terms, for a weak monoprotic acid solution, when \(\mathrm{[HA]} > \mathrm{[A^-]}\), the pH will generally be less than the pKa, indicating a more acidic solution. Conversely, when \(\mathrm{[A^-]} > \mathrm{[HA]}\), the pH will typically be greater than the pKa, indicating a more basic solution. As such, the molar ratios of HA to A- are informative indicators of the solution's pH in relation to its pKa value.

Acid-base Neutralization

Acid-base neutralization involves the reaction between an acid and a base to produce a salt and often water. In terms of the given exercise, we see instances where the strong base hydroxide ion \(OH^-\) is introduced to the weak acid HA.

When \(OH^-\) ions are added to the solution, they react with \(HA\) to produce the conjugate base \(A^-\) and water. This reaction diminishes the \(HA\) concentration while increasing the \(A^-\) concentration, thereby affecting the acid-base balance and shifting the pH.

This is what happens in scenarios (c) and (d) from the exercise. Enough \(OH^-\) will shift the pH above the pKa, reflecting a more basic solution. Lesser \(OH^-\) still triggers a shift towards basicity, but the final pH will depend on the extent of that shift and may end up being close to the pKa if the reaction leads to near-equivalent concentrations of \(HA\) and \(A^-\).

Remembering these principles of acid-base neutralization can help students predict the outcome of such reactions and understand the resulting pH changes in the context of the pKa of the acid involved.