Question

Is an aqueous solution of \(\mathrm{NaHSO}_{4}\) acidic, basic, or neutral? What reaction occurs with water? Calculate the \(\mathrm{pH}\) of a \(0.10-M\) solution of \(\mathrm{NaHSO}_{4}\).

Short answer

An aqueous solution of NaHSO4 is acidic, as the bisulfate ion (HSO4-) acts as a weak acid when dissolved in water. The reaction that occurs with water is: \[ \mathrm{HSO^{-}_{4(aq)}} + \mathrm{H}_{2}\mathrm{O_{(l)}} \rightleftharpoons \mathrm{H}_{3}\mathrm{O^{+}_{(aq)}} + \mathrm{SO^{2-}_{4(aq)}} \] The pH of a 0.10 M solution of NaHSO4 can be calculated by first finding the concentration of H3O+ ions, which is approximately \(3.5\times10^{-2}\,\mathrm{M}\). Using the pH formula, we find that the pH of this solution is approximately 1.5.

Step-by-step solution

Identify ions from NaHSO4 in water

When sodium bisulfate (NaHSO4) dissolves in water, it forms the following ions: \[ \mathrm{NaHSO}_{4(s)} \rightarrow \mathrm{Na^+_{(aq)}} + \mathrm{HSO^{-}_{4(aq)}} \] The sodium ion (Na+) is a neutral ion, as it does not have the ability to act as either an acid or a base. However, the bisulfate ion (HSO4-) can act as either an acid (by donating a proton, H+) or a base (by accepting a proton, H+).

Determine acid-base nature

Based on the acidic dissociation constant of sulfuric acid (H2SO4), which is around \(K_{a1}=10^3\), it can lose a proton easily, resulting in HSO4- which has a lower Ka around \(K_{a2}=1.2 \times 10^{-2}\). Since HSO4-has a significant Ka value, it will act as a weak acid in water, as shown in the following reaction: \[ \mathrm{HSO^{-}_{4(aq)}} + \mathrm{H}_{2}\mathrm{O_{(l)}} \rightleftharpoons \mathrm{H}_{3}\mathrm{O^{+}_{(aq)}} + \mathrm{SO^{2-}_{4(aq)}} \] As HSO4- acts as an acid, the reaction with water will result in an acidic solution.

Set up the Ka expression

For the equilibrium constant\(K_{a2}\) for the reaction above, we can write: \[ K_{a2} = \frac{ [\mathrm{H}_{3}\mathrm{O^+}][\mathrm{SO^{2-}_{4} }] }{[\mathrm{HSO^{-}_{4}]} \]

Calculate the H3O+ concentration

At equilibrium, we assume the initial concentration of H3O+ ions and sulfate ions (SO42-) to be zero. Since the NaHSO4 solution has an initial concentration of 0.10 M, we can write the following changes in concentration at equilibrium: - \([\mathrm{HSO^{-}_{4}}] = 0.10 - x\) - \([\mathrm{H}_{3}\mathrm{O^+}] = x\) - \([\mathrm{SO^{2-}_{4}}] = x\) Plugging these values in the Ka expression, we get: \(1.2 \times 10^{-2} = \frac{x^2}{0.10 - x}\) Assuming that the value of x is much smaller than 0.10, we can simplify the equation to: \(1.2\times10^{-2} \approx \frac{x^2}{0.10}\) Now, solve for x: \(x^2 = 1.2\times10^{-2} \times 0.10\) \(x \approx \sqrt{1.2\times10^{-3}}\) \(x \approx 3.5\times10^{-2}\) Therefore, the concentration of H3O+ ions at equilibrium is \(\approx 3.5\times10^{-2} \,\mathrm{M}\).

Calculate the pH

Now that we have the concentration of H3O+ ions, we can calculate the pH using the formula: \[ \mathrm{pH} = -\log{[\mathrm{H_{3}O^{+}}]} \] \[ \mathrm{pH} = -\log{(3.5\times10^{-2})} \] \[ \mathrm{pH} \approx 1.5 \] The pH of a 0.10 M solution of NaHSO4 is approximately 1.5, which confirms that the aqueous solution of NaHSO4 is acidic.

Key concepts

Acid-Base Reaction

In chemistry, acid-base reactions are fundamental. An acid-base reaction involves the transfer of a proton (H⁺) from one substance to another.

Let's break this down with sodium bisulfate (NaHSO₄) as an example. In water, NaHSO₄ dissociates into sodium ions (Na⁺) and bisulfate ions (HSO₄⁻). Here, Na⁺ is neutral because it doesn't participate in the acid-base balance. Meanwhile, HSO₄⁻ can either donate a proton to water, acting as an acid, or accept a proton, functioning as a base.

Given its ability to donate a proton, HSO₄⁻ acts as an acid by transferring a proton to water, forming hydronium ions (H₃O⁺). This results in an acidic solution:

• HSO₄⁻ + H₂O → H₃O⁺ + SO₄²⁻

This nature of HSO₄⁻ makes it crucial in determining the pH of solutions involving sodium bisulfate.

Sodium Bisulfate

Sodium bisulfate (NaHSO₄) is a salt derived from sulfuric acid. It's commonly used in various applications, such as pH adjustment in pools and household cleaning.

When NaHSO₄ dissolves in water, it separates into sodium ions (Na⁺) and bisulfate ions (HSO₄⁻). Na⁺, as a spectator ion, does not change the pH. However, HSO₄⁻ can interact actively with water due to its acidic nature.

• Sodium ions do not contribute to the acidity or basicity of the solution.
• Bisulfate ions act as weak acids, slightly increasing the concentration of H₃O⁺.

The role of NaHSO₄ in solutions is mainly dictated by the behavior of the bisulfate ion, making it important in understanding the acidic characteristic of the solution.

Equilibrium Constant

In chemical reactions, the equilibrium constant ( K_a) is a measure of the strength of an acid in solution, reflecting the extent of its dissociation into ions.

For sodium bisulfate, the dissociation of the bisulfate ion in water is characterized by its equilibrium constant, denoted as K⁽a₂⁾.

• K⁽a₂⁾ = 1.2 × 10⁻² for HSO₄⁻.

This suggests it is a weak acid since its K⁽a₂⁾ value is much smaller compared to strong acids.

The equilibrium constant expression for HSO₄⁻ with water is:
K⁽a₂⁾ = [H₃O⁺][SO₄²⁻] / [HSO₄⁻]

This equation shows the relationship between the concentration of reactants and products at equilibrium. Understanding this concept helps to predict the behavior of HSO₄⁻ in water and its impact on the solution's pH.

Hydronium Ion Concentration

The concentration of hydronium ions (H₃O⁺) is directly related to the acidity of a solution. It is a key factor in calculating pH.

When NaHSO₄ dissolves in water, the HSO₄⁻ ions increase the concentration of H₃O⁺ through their reaction with water:
HSO₄⁻ + H₂O ⇌ H₃O⁺ + SO₄²⁻

Calculating the concentration of H₃O⁺ involves using the equilibrium expression for K⁽a₂⁾, where we simplify the equation to find x, representing the concentration of H₃O⁺ at equilibrium.

• Initially, HSO₄⁻ = 0.10 M, H₃O⁺ = 0 M.
• At equilibrium, H₃O⁺ concentration becomes x.

From the calculations, we find x ≈ 3.5 × 10⁻² M.

This concentration directly impacts pH calculation, as pH = -log[H₃O⁺].
The higher the H₃O⁺ concentration, the lower (more acidic) the pH of the solution.