Problem 5 - Chemical Principles: The Quest for Insight

Question

Determine the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of (a) a solution that is \(0.40 \mathrm{M} \mathrm{NaHSO}_{4}(\mathrm{aq})\) and \(0.080 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\); (b) a solution that is \(0.40 \mathrm{M} \mathrm{NaHSO}_{4}(\mathrm{aq})\) and \(0.20 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}) ;\) (c) a solution that is \(0.40 \mathrm{M} \mathrm{NaHSO}_{4}(\mathrm{aq})\) and \(0.40 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\).

Step-by-Step Solution

Answer

For (a) calculate the pH with excess NaHSO4 using the dissociation constant and ignore the contribution of Na2SO4 to pH. For (b) and (c), the pH would be higher than in (a) due to the common ion effect from SO4^2-.

Step 1: Understand the Nature of the Compounds

Recognize that NaHSO4 is sodium bisulfate, a weak acid, and Na2SO4 is sodium sulfate, a salt derived from the neutralization of sulfuric acid H2SO4.

Step 2: Determine the Main Species in Solution

In a NaHSO4 solution, the main species are HSO4- (bisulfate anion) and SO4^2- (sulfate anion). Both come from the dissociation of H2SO4. Na2SO4 is a neutral salt that fully dissociates into 2Na+ and SO4^2-.

Step 3: Calculate the pH of the Solution for (a)

Since NaHSO4 is in excess compared to Na2SO4, HSO4- will partially dissociate into H+ and SO4^2-. The Ka value for HSO4- is needed. Assuming Ka = 1.2 x 10^-2, write the dissociation expression and set up an equation considering the initial concentration of HSO4- as 0.40 M - x, where x is the concentration of H+ produced.

Step 4: Solve for x, the Concentration of H+

Ka = 1.2 x 10^-2 = x^2 / (0.40 - x). Assuming x is small and 0.40 - x approximately equals 0.40, solve for x to find the concentration of H+. x = sqrt(Ka * [HSO4-]) = sqrt(1.2 x 10^-2 * 0.40).

Step 5: Calculate the pH and pOH for Solution (a)

pH = -log([H+]). Calculate the pOH by using the relation pOH = 14 - pH.

Step 6: Repeat the Process for (b) and (c)

For (b), Na2SO4 is half the concentration of NaHSO4. The common ion effect from SO4^2- will shift equilibrium, reducing the concentration of H+. For (c), the concentrations are the same, further reducing the concentration of H+.

Step 7: Consider the Common Ion Effect

Adjust the equation from Step 4 to include the common ion effect. The initial concentration of SO4^2- from the dissociation of Na2SO4 needs to be considered in the equilibrium expression.

Key Concepts

Acid-Base Equilibrium

Acid-base equilibrium is a fundamental concept in chemistry that involves the balance between an acid and its conjugate base in a solution. In such a dynamic equilibrium, acids donate protons (H⁺) while bases accept them, creating a delicate balance defined by the acid dissociation constant (K_a) for weak acids or the base dissociation constant (K_b) for weak bases.

Understanding this equilibrium is crucial for calculating pH, which measures the acidity or basicity of a solution. For a weak acid like HSO₄^-, which partially dissociates into H⁺ and SO₄^2-, the K_a provides insight into the degree of dissociation and thus the concentration of H⁺ in solution. The formula for calculating pH, which is the negative logarithm of the H⁺ concentration, depends directly on the position of the acid-base equilibrium.

Common Ion Effect

The common ion effect occurs when a solution contains two compounds that share a common ion. This effect influences the dissociation of weak acids or bases through Le Chatelier’s principle, which states that a system at equilibrium will respond to dampen an applied change.

For instance, in a solution with NaHSO₄ and Na₂SO₄, the SO₄^2- ion is common to both. The presence of additional SO₄^2- from Na₂SO₄ shifts the equilibrium of the dissociation of HSO₄^- towards the left, resulting in less dissociation, fewer free protons (H⁺), and a higher pH than expected if NaHSO₄ were alone in solution. This is crucial to consider when calculating the pH of solution (b), where the common ion effect is introduced.

Dissociation of Weak Acids

The dissociation of weak acids is an incomplete reversible reaction where the acid donates protons to water, forming its conjugate base and hydronium ions (H₃O⁺). This can be represented by the generic equilibrium expression HA <--> H⁺ + A^-, where HA is a weak acid.

The extent of this dissociation is described by the acid’s K_a, with a smaller K_a indicating a weaker acid that dissociates less. To solve for the concentration of H⁺, the equilibrium expression is often rearranged into K_a = [H⁺][A^-]/[HA]. In our NaHSO₄ scenario, assuming 'x' is small simplifies the mathematics and allows us to solve for K_a without complex quadratic equations, a trick commonly used when x is small relative to the initial concentration.

Neutralization Reactions

Neutralization reactions are chemical reactions where an acid and a base react to form a salt and water. Such reactions are often exothermic, releasing heat as a product along with the salt. In the case of sulfuric acid (H₂SO₄), it can neutralize with sodium hydroxide (NaOH) to produce sodium sulfate (Na₂SO₄) and water.

In exercise scenarios like these, we typically do not deal with a neutralization reaction directly, but we do deal with the products of such a reaction. Sodium sulfate is a product of the neutralization between sulfuric acid and sodium hydroxide. Its presence in solution can affect the pH, as it alters the concentration of ions in the solution, which ties back to the common ion effect. Understanding neutralization is vital when interpreting the ionic components in a solution and predicting the resulting pH.