Question

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

Short answer

The aqueous solution of NaHSO₄ is acidic due to the HSO₄⁻ ions donating a proton to water molecules, forming H₃O⁺ ions. The pH of a 0.10 M NaHSO₄ solution is approximately 6.5.

Step-by-step solution

Write the dissociation reaction of NaHSO₄ in water

NaHSO₄ is a salt formed from the reaction of a strong base (NaOH) and a weak acid (H₂SO₄). Let's write the dissociation reaction of NaHSO₄ in water: NaHSO₄ (aq) ⇌ Na⁺ (aq) + HSO₄⁻ (aq)

Determine the reaction of HSO₄⁻ ions with water

As NaHSO₄ dissociates in water, we should consider the possible reaction of HSO₄⁻ ions with water molecules. If HSO₄⁻ donates a proton (H⁺) to water, the solution will be acidic. If HSO₄⁻ accepts a proton from water, the solution will be basic. Since HSO₄⁻ is the conjugate base of a weak acid (H₂SO₄), it will tend to donate a proton to water: HSO₄⁻ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + SO₄²⁻ (aq)

Write the equilibrium expression and calculate the pH

Now, we need to write the expression for the equilibrium constant, Kₐ, for the reaction of HSO₄⁻ with water and use it to calculate the pH of the 0.10 M NaHSO₄ solution. First, we need to find the value of Kₐ for the reaction. The Kₐ of H₂SO₄ is 1.0 x 10², and since KₐKₐ₁ is approximately equal to Kₐ(Kw), we can find Kₐ₁ for HSO₄⁻: Kₐ₁ = Kw / Kₐ = 1.0 x 10⁻¹₄ / 1 x 10² = 1.0 x 10⁻₁₂ Now, we can write the equilibrium expression for the reaction: Kₐ₁ = [H₃O⁺][SO₄²⁻] / [HSO₄⁻] Since the concentration of NaHSO₄ is 0.10 M and it dissociates completely in water, the initial concentration of HSO₄⁻ ions can be assumed to be 0.10 M. Let x represent the equilibrium concentration of H₃O⁺: 1.0 x 10⁻₁₂ = x² / (0.10 - x) Assuming x << 0.10, we can simplify the expression: x² ≈ 1.0 x 10⁻₁³ x ≈ √(1.0 x 10⁻₁³) = 1.0 x 10⁻⁶.⁵ Since x represents the concentration of H₃O⁺, the pH can be calculated as: pH = -log([H₃O⁺]) = -log(1.0 x 10⁻⁶.⁵) ≈ 6.5

Conclusion

The aqueous solution of NaHSO₄ is acidic, as the HSO₄⁻ ions donate a proton to water molecules, forming H₃O⁺ ions. The pH of a 0.10 M NaHSO₄ solution is approximately 6.5.

Key concepts

Dissociation of NaHSO4

When sodium hydrogen sulfate (NaHSO4) dissolves in water, it undergoes a process called dissociation. This is where NaHSO4 breaks down into its constituent ions. The equation for this dissociation is:

NaHSO4 (aq) NaH ⇌ Na^+ (aq) + HSO4^- (aq).

The dissociation of NaHSO4 is essential in determining the nature of its solution because it releases hydrogen sulfate ions (HSO4^-), which in turn can affect the solution's pH. As NaHSO4 comes from the neutralization of a strong base and a weak acid, the resulting solution can be unambiguously understood as having a pH that is not neutral.

Understanding the behavior of the HSO4^- ion in water is crucial, as its ability to donate a proton is what ultimately leads to the acidic nature of the NaHSO4 solution.

Acidic and Basic Solutions

Determining whether a solution is acidic, basic, or neutral is based upon the concentration of hydronium ions (H3O^+) relative to hydroxide ions (OH^-). Acidic solutions have more hydronium ions than hydroxide ions, while basic solutions have more hydroxide ions.

In the case of NaHSO4, the dissociated hydrogen sulfate ions (HSO4^-) can donate protons to water, forming hydronium ions (H3O^+). This proton-donating reaction can be represented as:

HSO4^- (aq) + H2O (l) ⇌ H3O^+ (aq) + SO4^2- (aq).

This shows that the NaHSO4 solution is indeed acidic due to the formation of H3O^+ ions. So, when a solution of NaHSO4 is mixed with water, inspecting the behavior of HSO4^- reveals a clear tendency for the solution to be more acidic.

pH Calculation

The pH of a solution is a numeric scale used to specify the acidity or basicity of an aqueous solution. It is defined as the negative logarithm of the concentration of hydronium ions (H3O^+).

For the NaHSO4 solution, the pH can be calculated using the expression:
pH = -log([H3O^+]).

By using the equilibrium constant (Ka1), we can estimate the concentration of those hydronium ions in the solution and subsequently determine the pH. The provided steps in the textbook solution show how to derive the concentration of H3O^+ from the equilibrium expression, which greatly simplifies to X When x is smaller than the initial concentration of HSO4^- (0.10 M).

Following the simplification, we can calculate the pH, as shown:

pH [ x H3O^+nly consider the significant figures and rationale behind the approximation of X in the final answer. This detailed look into the pH calculation provides students with a deeper understanding of the concept and its application in the exercise.