Lesson 4 · 30 min

Selective precipitation and separating ions

The more their solubility products differ, the more cleanly ions can be precipitated one at a time. pH control separates Fe³⁺ from Cu²⁺, and metered Ag⁺ makes a chloride titration its own indicator.

In this lesson

Use Ksp to calculate the precipitant concentration or pH at which each ion starts to precipitate and is “completely” precipitated
Decide whether two ions can be separated by selective precipitation, and find the workable pH window
Explain why chromate can serve as the indicator in the Mohr titration of chloride
Understand how sulfides are grouped by acidity

In hydrometallurgical copper production, the leach solution holds $\ce{Cu^2+}$ together with plenty of $\ce{Fe^3+}$ . No elaborate trick is needed: bring the pH to between 3 and 4 and almost all the iron drops out as red-brown iron(III) hydroxide while none of the copper does. The secret is the huge difference between solubility products — and turning it into conditions you can actually set.

Predict first

A solution is 0.010 M in both $\ce{Fe^3+}$ and $\ce{Cu^2+}$ . You raise the pH slowly with NaOH. Which precipitates first, and can you remove the iron (to below 10⁻⁵ M) before any copper precipitates?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Cu(OH)₂ first, because Cu²⁺ has the lower charge Fe(OH)₃ first, and the iron can be removed completely between about pH 3 and 5 Fe(OH)₃ first, but copper starts before the iron is complete Both at once

Who precipitates first? Compare the precipitant needed

As precipitant is added, the ion that needs the least precipitant precipitates first — not necessarily the one with the smallest Ksp (the two differ when formula types differ). For hydroxides, solving the onset condition $Q = K_\mathrm{sp}$ for pH gives the formula below. By analytical convention an ion is completely precipitated once $10^{-5}$ M remains; if one ion is complete before the other starts, the two can be separated.

[\ce{OH-}] = \left(\frac{K_\mathrm{sp}}{c}\right)^{1/n} \qquad \mathrm{pH} = 14 + \frac{1}{n}\lg\frac{K_\mathrm{sp}}{c}

The pH at which M(OH)ₙ starts to precipitate (c = initial concentration) or is complete (c = 10⁻⁵ M), at 25 °C.

$n$: charge of the metal ion (number of hydroxides)
$c$: metal-ion concentration in solution, mol/L

Worked example

The iron–copper separation window

A solution holds 0.010 M $\ce{Fe^3+}$ and 0.010 M $\ce{Cu^2+}$ . Find the pH at which the iron is complete and the pH at which copper starts ( $K_\mathrm{sp}$ : Fe(OH)₃ $1.6\times10^{-39}$ , Cu(OH)₂ $4.8\times10^{-20}$ ).

Fractional precipitation by titration: the Mohr method

To determine chloride in food or water, analysts titrate with $\ce{AgNO3}$ and add a little $\ce{K2CrO4}$ as indicator. In 0.10 M $\ce{Cl-}$ , AgCl starts to form at $[\ce{Ag+}] = 1.8\times10^{-9}$ M, whereas $5\times10^{-3}$ M chromate gives brick-red $\ce{Ag2CrO4}$ only at $[\ce{Ag+}] = \sqrt{1.2\times10^{-12}/0.005} = 1.5\times10^{-5}$ M. So white AgCl precipitates first; only when the chloride is nearly used up and $[\ce{Ag+}]$ shoots up does the brick red appear — the endpoint.

pAg curve · the Mohr indicator

Fractional precipitation: titrating halides with Ag⁺

0.100 M AgNO₃ added · 49.50 mL Chromate indicator concentration · 0.005 M

\ce{AgCl}

pAg: 6.44
Precipitates so far: AgCl 4.950 mmol
Brick-red Ag₂CrO₄ appears: 50.014 mL
Endpoint error: +0.014 mL (0.03 %)
Cl⁻ still unprecipitated when Ag₂CrO₄ appears: 0.017 %

Drag the volume past the 50 mL equivalence point: pAg drops sharply and about 0.014 mL later brick-red Ag₂CrO₄ appears. Raise the chromate concentration and the endpoint comes earlier; then switch to the I⁻/Br⁻/Cl⁻ mixture and watch three steps appear in the order AgI, AgBr, AgCl.

Sulfide grouping uses the same idea. In saturated $\ce{H2S}$ (about 0.10 M), the most metal ion that can stay dissolved is $K_\mathrm{spa}[\ce{H+}]^2/[\ce{H2S}]$ , where $K_\mathrm{spa}$ is the acid solubility product ( $\ce{MS(s)} + \ce{2H+} \mathrel{\ce{<=>}} \ce{M^2+} + \ce{H2S}$ ). In 0.3 M HCl, $\ce{CuS}$ ( $K_\mathrm{spa} = 6\times10^{-16}$ ), $\ce{PbS}$ ( $3\times10^{-7}$ ) and $\ce{CdS}$ precipitate while $\ce{FeS}$ ( $6\times10^2$ ), $\ce{NiS}$ and $\ce{MnS}$ stay dissolved — which is how groups II and III of classical qualitative analysis are separated. Switch the bench to “Sulfides” to see it.

Concept check

$\ce{AgNO3}$ is added dropwise to a solution 0.010 M in each of $\ce{I-}$ , $\ce{Br-}$ and $\ce{Cl-}$ . In what order do the precipitates appear?

AgCl → AgBr → AgI AgI → AgBr → AgCl All three at once AgBr → AgI → AgCl

Try the calculation

At what pH does 0.010 M $\ce{Mg^2+}$ begin to precipitate as $\ce{Mg(OH)2}$ ? ( $K_\mathrm{sp} = 7.1\times10^{-12}$ )

Key takeaways

The ion that needs the least precipitant precipitates first; salts of different types cannot be ranked by Ksp alone.
A hydroxide starts to precipitate at pH = 14 + (1/n)·lg(Ksp/c); if one ion is complete (10⁻⁵ M) before another starts, they can be separated.
In the Mohr method AgCl precipitates first; once chloride is used up, [Ag⁺] jumps and brick-red Ag₂CrO₄ marks the endpoint.
Controlling acidity separates very insoluble sulfides (CuS, PbS) from more soluble ones (FeS, MnS).

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Data and references

A. E. Martell & R. M. Smith, Critical Stability Constants, Vol. 4 (1976) and Vol. 5 (1982), via D. Harvey, Analytical Chemistry 2.1, Appendices 10 and 12.
D. C. Harris, Quantitative Chemical Analysis, 9th ed. (2016): precipitation titration curves and the Mohr method.
R. H. Petrucci et al., General Chemistry, 11th ed. (2017), Ch. 18 (Kspa of metal sulfides), after R. J. Myers, J. Chem. Educ. 63, 687 (1986).