Selective Sulfide Precipitation

Introduction

Metal sulfides include some of the least soluble compounds known, yet their Ksp values range over more than twenty powers of ten. That enormous spread makes sulfide an ideal separating reagent — provided its concentration can be dialled up or down with great precision. Hydrogen sulfide is a weak diprotic acid, so the free sulfide ion concentration depends on the square of [H⁺]. Changing the pH therefore acts as a very sensitive control knob for which metal sulfides precipitate.

Core explanation

Sulfide availability from H₂S. Hydrogen sulfide loses protons in two steps:

H₂S ⇌ H⁺ + HS⁻, Ka₁ ≈ 1.0 × 10⁻⁷ HS⁻ ⇌ H⁺ + S²⁻, Ka₂ (a traditionally tabulated value is about 1.3 × 10⁻¹³)

Multiplying the two expressions gives

Ka₁Ka₂ = [H⁺]²[S²⁻] ÷ [H₂S] ≈ 1.3 × 10⁻²⁰

A solution saturated with the gas at 25 °C has [H₂S] ≈ 0.10 mol dm⁻³, held constant by the gas phase. Then

[H⁺]²[S²⁻] ≈ 1.3 × 10⁻²¹, so [S²⁻] = 1.3 × 10⁻²¹ ÷ [H⁺]²

Every tenfold increase in [H⁺] lowers [S²⁻] a hundredfold. Between pH 0.5 and pH 8, free sulfide changes by about fourteen orders of magnitude (slope 2 in log[S²⁻] up to about pH 7, then roughly slope 1 as HS⁻ becomes significant).

Onset pH for a metal sulfide. For a divalent metal forming MS, precipitation begins when [M²⁺][S²⁻] = Ksp. Combining with the expression above:

[H⁺]² at onset = 1.3 × 10⁻²¹ × [M²⁺] ÷ Ksp

Metals with very small Ksp values precipitate even in strong acid; those with larger Ksp values require a less acidic solution.

Typical Ksp values (25 °C, traditional tabulations):

Sulfide Approximate Ksp --- --- CuS 6 × 10⁻³⁶ PbS 3 × 10⁻²⁸ FeS 6 × 10⁻¹⁸ MnS 3 × 10⁻¹³

The separation. In about 0.30 mol dm⁻³ acid, [S²⁻] ≈ 1.3 × 10⁻²¹ ÷ 0.090 = 1.4 × 10⁻²⁰ mol dm⁻³. For 0.010 mol dm⁻³ Cu²⁺, Q = 1.4 × 10⁻²² which vastly exceeds Ksp, so CuS precipitates. For 0.010 mol dm⁻³ Fe²⁺, Q = 1.4 × 10⁻²² is far below 6 × 10⁻¹⁸, so iron stays dissolved. Only after the acid is removed and the solution is made weakly basic do FeS and MnS precipitate. This is the logic of the historical "acid-insoluble" and "base-insoluble" sulfide groups in qualitative analysis.

Safety at a conceptual level. Hydrogen sulfide is a highly toxic gas that dulls the sense of smell at dangerous concentrations. Modern teaching laboratories avoid it and use safer sulfide sources only under strict fume-hood control, or study the chemistry through simulation and data analysis.

Formulae

[H⁺]²[S²⁻] = Ka₁Ka₂[H₂S] ≈ 1.3 × 10⁻²¹ (saturated H₂S, 25 °C). Onset: [H⁺] = √(1.3 × 10⁻²¹ × [M²⁺] ÷ Ksp). Kspa = Ksp ÷ (Ka₁Ka₂) = [M²⁺][H₂S] ÷ [H⁺]².

Step-by-step reasoning

1. Write [S²⁻] in terms of [H⁺] for saturated H₂S. 2. Write Ksp for the metal sulfide and insert the metal concentration. 3. Solve for the sulfide needed at onset. 4. Convert that to [H⁺] and then pH. 5. Compare onset pH values: metals with lower onset pH precipitate first as the pH is raised.

Visual explanation

In the solubility simulation, plot log[S²⁻] against pH for saturated H₂S: a straight line with slope +2 below about pH 7. Draw a horizontal line for each metal at log(Ksp ÷ [M²⁺]). Where the sulfide line crosses a metal's horizontal line is that metal's onset pH. CuS crosses far to the left of pH 0; FeS crosses near pH 3; MnS near pH 5.

Real-world analogy

Think of a dimmer switch controlling a room full of light-sensitive sensors, each with its own trigger brightness. Turning the dimmer slightly changes the light enormously, because the dimmer works on a square law. Only the most sensitive sensors fire at low light; raising it step by step triggers the others in turn.

Real-world example

In the mining industry, copper, zinc and nickel are recovered from acidic leach solutions and mine drainage by controlled sulfide precipitation. Adjusting pH lets operators precipitate copper sulfide first, then other metals in separate stages, producing concentrates that are easier to refine and reducing metals released to the environment.

Why?

Why does [S²⁻] depend on [H⁺]² rather than [H⁺]? Two protons must be removed from H₂S to make each sulfide ion. The overall equilibrium includes two H⁺ on the product side, so the equilibrium expression contains [H⁺] squared.

Common misconception

"Adding acid dissolves all metal sulfides." Acid lowers [S²⁻], but for extremely insoluble sulfides such as CuS the remaining sulfide is still far above the threshold. These sulfides dissolve only when an oxidising acid destroys sulfide itself, which is a different process.

Worked example

Question: Estimate the pH at which FeS begins to precipitate from 0.010 mol dm⁻³ Fe²⁺ saturated with H₂S. Use Ksp(FeS) = 6 × 10⁻¹⁸.

Reasoning: Onset [S²⁻] = 6 × 10⁻¹⁸ ÷ 0.010 = 6 × 10⁻¹⁶ mol dm⁻³. [H⁺]² = 1.3 × 10⁻²¹ ÷ 6 × 10⁻¹⁶ = 2.2 × 10⁻⁶, so [H⁺] = 1.5 × 10⁻³ mol dm⁻³.

Answer: pH ≈ 2.8; FeS forms only above about pH 2.8.

Quick check

1. If the pH of a saturated H₂S solution is raised by one unit, by what factor does [S²⁻] change? Answer: [H⁺] falls tenfold, so [S²⁻] rises by a factor of 10² = 100.

Exam focus

Learn the combined expression [H⁺]²[S²⁻] = Ka₁Ka₂[H₂S] and use the value given in the question — data books differ widely. Show the square root when finding [H⁺]. A complete answer names which metal precipitates at the stated pH and which remains, with Q versus Ksp comparisons.

Advanced insight

Modern measurements suggest Ka₂ for HS⁻ is much smaller (perhaps 10⁻¹⁷ to 10⁻¹⁹), meaning free S²⁻ is almost nonexistent in water. For this reason many modern data tables list the acid solubility product Kspa, based on MS + 2H⁺ ⇌ M²⁺ + H₂S, which avoids [S²⁻] altogether. Ratios between metals, and hence the order of precipitation, are unaffected. Kinetic factors also matter: some sulfides, such as NiS, dissolve far more slowly once aged than their Ksp suggests.

Summary

In saturated H₂S, free sulfide depends on 1 ÷ [H⁺]², so pH controls [S²⁻] over many orders of magnitude. Each metal sulfide has an onset pH from Ksp and its metal concentration. Very insoluble sulfides such as CuS precipitate even from strong acid; more soluble ones such as FeS and MnS need weakly acidic or basic conditions, enabling stepwise separation.

Practice questions

1. Calculate [S²⁻] in saturated H₂S at pH 1.00, using [H⁺]²[S²⁻] = 1.3 × 10⁻²¹. Answer: [S²⁻] = 1.3 × 10⁻²¹ ÷ (0.10)² = 1.3 × 10⁻¹⁹ mol dm⁻³. 2. At pH 1.00, does PbS precipitate from 1.0 × 10⁻³ mol dm⁻³ Pb²⁺ (Ksp 3 × 10⁻²⁸)? Answer: Q = 1.0 × 10⁻³ × 1.3 × 10⁻¹⁹ = 1.3 × 10⁻²², which exceeds 3 × 10⁻²⁸, so PbS precipitates. 3. Estimate the onset pH for MnS from 0.010 mol dm⁻³ Mn²⁺ (Ksp 3 × 10⁻¹³). Answer: [S²⁻] = 3 × 10⁻¹¹; [H⁺]² = 1.3 × 10⁻²¹ ÷ 3 × 10⁻¹¹ = 4.3 × 10⁻¹¹; [H⁺] = 6.6 × 10⁻⁶; pH ≈ 5.2. 4. Calculate the residual Cu²⁺ in 0.30 mol dm⁻³ acid saturated with H₂S, using Ksp(CuS) = 6 × 10⁻³⁶. Answer: [S²⁻] = 1.4 × 10⁻²⁰, so [Cu²⁺] = 6 × 10⁻³⁶ ÷ 1.4 × 10⁻²⁰ ≈ 4 × 10⁻¹⁶ mol dm⁻³ — removal is essentially complete.