Selective Hydroxide Precipitation

Introduction

Raising pH raises hydroxide concentration, allowing metal hydroxides to precipitate. Different metals can reach their precipitation thresholds at different pH values, making selective removal possible. The threshold depends on both Ksp and the current free metal concentration; comparing Ksp numbers alone may be misleading when concentrations or hydroxide stoichiometries differ.

Core explanation

For a divalent metal M²⁺, the dissolution equilibrium is M(OH)₂(s) ⇌ M²⁺ + 2 OH⁻ with Ksp = [M²⁺][OH⁻]² in a dilute concentration model. Precipitation begins when Qsp reaches Ksp from below. At a specified pre-precipitation metal concentration cM, the onset hydroxide concentration is [OH⁻]onset = √(Ksp/cM). Convert to pOH = −log[OH⁻], then pH = pKw−pOH at the stated temperature. At 25 °C, pKw is approximately 14; at other temperatures use the supplied value.

For M(OH)₃(s) from a trivalent metal, Ksp = [M³⁺][OH⁻]³ and onset [OH⁻] = (Ksp/cM)^(1/3). The exponent changes the comparison. A smaller Ksp does not by itself guarantee lower onset pH if another metal has a different cM or hydroxide stoichiometry. Always derive each threshold from its own balanced equation.

Suppose two divalent ions M²⁺ and N²⁺ each start at 0.010 M, with hypothetical Ksp values 10⁻¹² for M(OH)₂ and 10⁻⁸ for N(OH)₂. M begins precipitating at [OH⁻] = √(10⁻¹²/10⁻²) = 10⁻⁵ M, pH about 9 at 25 °C. N begins at [OH⁻] = √(10⁻⁸/10⁻²) = 10⁻³ M, pH about 11. Thus a pH region between these onsets can precipitate much of M while N remains below its own threshold.

At the moment N begins to precipitate, the residual free M²⁺ permitted by M's Ksp is 10⁻¹²/(10⁻³)² = 10⁻⁶ M. Relative to initial 0.010 M, this is 0.01% remaining, or approximately 99.99% removed if other equilibria and volume changes are negligible. This residual calculation is stronger than merely saying “M comes first,” because it measures whether the separation is substantial.

Actual hydroxide precipitation may be complicated by hydrolysis, complex formation and amphoteric redissolution. Free metal may differ from total metal if ligands bind it. Some metal hydroxides form soluble hydroxo complexes in excess base, so increasing pH indefinitely can redissolve a precipitate. The next page examines that behaviour. A controlled separation therefore chooses a pH window, not simply “as basic as possible.”

Mixing and pH control also matter. Adding a strong base locally can create high-OH⁻ regions before the bulk solution equilibrates, potentially causing temporary coprecipitation. In calculation problems, assume well-mixed equilibrium only when stated. In a real experiment, gradual controlled addition and monitoring would be needed, but the note's task is to understand the underlying thresholds rather than prescribe a procedure.

Step-by-step reasoning

1. Write each hydroxide's dissolution equation and Ksp expression. 2. Use each free-metal concentration before precipitation begins. 3. Solve Qsp = Ksp for onset [OH⁻]. 4. Convert to pH using the temperature-specific pKw. 5. At the second onset, calculate residual concentration of the first metal and check side reactions.

Visual explanation

Draw a pH axis with an M(OH)₂ onset marker at pH 9 and N(OH)₂ marker at pH 11 for the hypothetical example. Shade the interval where M can precipitate while N remains dissolved.

Real-world analogy

Two alarms trigger at different dial settings. Raising the dial into the interval between triggers activates only the first alarm; turning it all the way up defeats selective control.

Real-world example

Selective metal removal from water can exploit pH-dependent hydroxide formation, provided other ligands and metal concentrations are considered. The measured pH window matters more than a simple solubility label.

Why?

Why must initial metal concentration enter the onset calculation? Qsp is a product of free metal and hydroxide powers, so a more concentrated metal reaches Ksp at a lower hydroxide level.

Common misconception

“The metal hydroxide with the smallest Ksp always precipitates first.” Compare actual onset thresholds using each ion's concentration and stoichiometry; Ksp alone is insufficient in general.

Worked example

For hypothetical M(OH)₂ with Ksp = 10⁻¹² and [M²⁺] = 0.010 M, set 10⁻¹² = 0.010[OH⁻]². Then [OH⁻] = 10⁻⁵ M, pOH = 5 and pH ≈ 9 at 25 °C. If [M²⁺] were tenfold lower, the onset hydroxide level would rise by √10, showing concentration dependence.

Quick check

1. What power of [OH⁻] appears in the Ksp expression for M(OH)₃? Answer: The third power, because each formula unit releases three hydroxide ions.

Exam focus

Compute thresholds from Qsp = Ksp, then residual first-metal level at the second threshold. Use pKw appropriate to temperature.

Advanced insight

OpenStax discusses selective precipitation at https://openstax.org/books/chemistry-2e/pages/15-1-precipitation-and-dissolution. Activity and complexation effects can shift practical onset pH away from an ideal concentration estimate.

Summary

Metal-hydroxide precipitation begins when free-metal and hydroxide activities reach Ksp. Distinct thresholds can create a pH range for selective separation. A useful separation requires not only earlier onset for one metal but a low residual level before the second begins precipitating.

Practice questions

1. Derive onset [OH⁻] for M(OH)₂ at free metal concentration c. Answer: [OH⁻]onset = √(Ksp/c). 2. What is pH at 25 °C if onset [OH⁻] = 10⁻⁴ M? Answer: pOH = 4, so pH ≈ 10. 3. What quantity measures how much first metal remains when a second starts precipitating? Answer: Its residual free concentration from Ksp divided by the current hydroxide concentration raised to the proper power. 4. Why might very high pH undermine hydroxide separation? Answer: Some amphoteric hydroxides can form soluble hydroxo complexes in excess base.