Question

Consider the sparingly soluble salt Zinc Sulfide, ZnS, dissolving in an aqueous buffer solution containing 0.1 M acetic acid (CH₃COOH) and 0.2 M sodium acetate (CH₃COONa).

Given:

• Ksp(ZnS) = 2.5 × 10⁻²²
• For H₂S: Ka₁ = 1.0 × 10⁻⁷, Ka₂ = 1.0 × 10⁻¹⁴
• For CH₃COOH: Ka = 1.8 × 10⁻⁵
• Kw = 1.0 × 10⁻¹⁴

Assume ideal solution behavior and that the buffer components do not significantly interact with Zn²⁺ ions other than controlling pH. Neglect the autoionization of water in pH calculation but consider it for relevant equilibria if necessary (though in this case, buffer controls pH strongly).

Which of the following statements are CORRECT?

• A. The pH of the buffer solution is approximately 5.04.
• B. The molar solubility of ZnS in this buffer is approximately 4.6 × 10⁻⁶ M.
• C. In the saturated solution, the concentration of S²⁻ ions is negligible compared to HS⁻ and H₂S.
• D. If the buffer concentration were doubled (0.2 M CH₃COOH and 0.4 M CH₃COONa), the solubility of ZnS would approximately double.

Answer 1

Answer: (A), (B), (C)

The correct options are (A), (B), and (C).

Answer 2

Let S be the molar solubility of ZnS in the buffer.

ZnS(s) <=> Zn²⁺(aq) + S²⁻(aq) ; Ksp = [Zn²⁺][S²⁻]

Since ZnS is the only source of Zn²⁺ and the total sulfide species (S²⁻, HS⁻, H₂S), we have:

[Zn²⁺] = S

Total sulfide concentration = [S²⁻] + [HS⁻] + [H₂S] = S

The sulfide species are involved in the following equilibria:

H₂S <=> H⁺ + HS⁻ ; Ka₁ = [H⁺][HS⁻]/[H₂S]

HS⁻ <=> H⁺ + S²⁻ ; Ka₂ = [H⁺][S²⁻]/[HS⁻]

From these equilibria, we can express [HS⁻] and [H₂S] in terms of [S²⁻] and [H⁺]:

[HS⁻] = [H⁺][S²⁻] / Ka₂

[H₂S] = [H⁺][HS⁻] / Ka₁ = [H⁺] ([H⁺][S²⁻]/Ka₂) / Ka₁ = [H⁺]²[S²⁻] / (Ka₁Ka₂)

Now, substitute these into the total sulfide expression:

S = [S²⁻] + [H⁺][S²⁻]/Ka₂ + [H⁺]²[S²⁻]/(Ka₁Ka₂)

S = [S²⁻] (1 + [H⁺]/Ka₂ + [H⁺]²/(Ka₁Ka₂))

From the Ksp expression, [S²⁻] = Ksp / [Zn²⁺] = Ksp / S.

Substitute this into the equation for S:

S = (Ksp / S) (1 + [H⁺]/Ka₂ + [H⁺]²/(Ka₁Ka₂))

S² = Ksp (1 + [H⁺]/Ka₂ + [H⁺]²/(Ka₁Ka₂))

First, calculate the pH of the buffer solution using the Henderson-Hasselbalch equation:

pH = pKa(CH₃COOH) + log([CH₃COO⁻]/[CH₃COOH])

pKa(CH₃COOH) = -log(1.8 × 10⁻⁵) ≈ 4.74

pH = 4.74 + log(0.2 M / 0.1 M) = 4.74 + log(2) = 4.74 + 0.301 ≈ 5.04

So, [H⁺] = 10⁻⁵.⁰⁴ ≈ 9.12 × 10⁻⁶ M.

Now, calculate the term in the parenthesis:

1 + [H⁺]/Ka₂ + [H⁺]²/(Ka₁Ka₂) = 1 + (9.12 × 10⁻⁶) / (1.0 × 10⁻¹⁴) + (9.12 × 10⁻⁶)² / (1.0 × 10⁻⁷ × 1.0 × 10⁻¹⁴) = 1 + 9.12 × 10⁸ + (8.32 × 10⁻¹¹) / (1.0 × 10⁻²¹) = 1 + 9.12 × 10⁸ + 8.32 × 10¹⁰

The terms 9.12 × 10⁸ and 8.32 × 10¹⁰ are much larger than 1. The dominant term is 8.32 × 10¹⁰.

So, 1 + [H⁺]/Ka₂ + [H⁺]²/(Ka₁Ka₂) ≈ 8.32 × 10¹⁰

Now calculate S²:

S² = Ksp (1 + [H⁺]/Ka₂ + [H⁺]²/(Ka₁Ka₂))

S² ≈ (2.5 × 10⁻²²) × (8.32 × 10¹⁰)

S² ≈ 20.8 × 10⁻¹² = 2.08 × 10⁻¹¹

Calculate S:

S = sqrt(2.08 × 10⁻¹¹) = sqrt(20.8 × 10⁻¹²) ≈ 4.56 × 10⁻⁶ M

Let's evaluate the options:

(A) The pH of the buffer solution is approximately 5.04.

Calculated pH ≈ 5.04. This statement is CORRECT.

(B) The molar solubility of ZnS in this buffer is approximately 4.6 × 10⁻⁶ M.

Calculated S ≈ 4.56 × 10⁻⁶ M, which is approximately 4.6 × 10⁻⁶ M. This statement is CORRECT.

(C) In the saturated solution, the concentration of S²⁻ ions is negligible compared to HS⁻ and H₂S.

We found that S = [S²⁻] (1 + [H⁺]/Ka₂ + [H⁺]²/(Ka₁Ka₂)).

The term in parenthesis is approximately 8.32 × 10¹⁰.

S = [S²⁻] × 8.32 × 10¹⁰

[S²⁻] = S / (8.32 × 10¹⁰)

The total sulfide S is distributed among [S²⁻], [HS⁻], and [H₂S].

The ratios are:

[HS⁻] / [S²⁻] = [H⁺] / Ka₂ ≈ 9.12 × 10⁸

[H₂S] / [S²⁻] = [H⁺]² / (Ka₁Ka₂) ≈ 8.32 × 10¹⁰

Since both ratios are very large, it means [S²⁻] is much smaller than [HS⁻] and [H₂S]. The dominant species is H₂S, followed by HS⁻, and S²⁻ is the least abundant. This statement is CORRECT.

(D) If the buffer concentration were doubled (0.2 M CH₃COOH and 0.4 M CH₃COONa), the solubility of ZnS would approximately double.

If concentrations are doubled, [CH₃COOH] = 0.2 M and [CH₃COONa] = 0.4 M.

The ratio [CH₃COO⁻]/[CH₃COOH] is still 0.4/0.2 = 2.

The pH of the buffer would remain approximately 5.04 (assuming ideal behavior and no significant change in activity coefficients at higher concentrations, which is a reasonable approximation for this level).

Since the pH ([H⁺]) remains essentially unchanged, the value of (1 + [H⁺]/Ka₂ + [H⁺]²/(Ka₁Ka₂)) remains unchanged.

The solubility S is proportional to the square root of this term (multiplied by Ksp).

S = sqrt( Ksp * (1 + [H⁺]/Ka₂ + [H⁺]²/(Ka₁Ka₂)) )

As Ksp and the term in the parenthesis do not change, the solubility S will remain approximately the same. It will not double. This statement is INCORRECT.

Therefore, the correct statements are (A), (B), and (C).

The solubility of ZnS is enhanced in an acidic buffer because the S²⁻ ions released from the dissolution react with H⁺ ions from the buffer, forming HS⁻ and H₂S. This removal of S²⁻ shifts the solubility equilibrium ZnS(s) <=> Zn²⁺(aq) + S²⁻(aq) to the right, increasing the solubility. The extent of this enhancement depends on the pH (specifically, the [H⁺] concentration), which is controlled by the buffer. The multiple Ka values of H₂S are crucial for determining the distribution of total sulfide among S²⁻, HS⁻, and H₂S at the given pH, which is necessary to relate the total solubility S to the equilibrium concentration of S²⁻ required by the Ksp expression. The Henderson-Hasselbalch equation is used to find the buffer pH, and then this pH is used to calculate the solubility using the derived relationship between S, Ksp, and [H⁺] involving Ka₁ and Ka₂. Option (D) tests the understanding that buffer pH primarily depends on the ratio of conjugate base to acid, not the absolute concentrations (within limits).