Chemistry

Problem

A solution initially contains

0.010\,\mathrm{M}\,\mathrm{Fe^{3+}}

and

0.010\,\mathrm{M}\,\mathrm{Cu^{2+}}

. As the pH is raised slowly from 1 to 9, hydroxide salts may precipitate. Given

K_{sp}(\mathrm{Fe(OH)_3}) = 4 \times 10^{-38}

and

K_{sp}(\mathrm{Cu(OH)_2}) = 2 \times 10^{-20}

, which statement best predicts the order of precipitation? a)

\mathrm{Cu(OH)_2}

precipitates before

\mathrm{Fe(OH)_3}

because its larger

K_{sp}

indicates lower solubility. b)

\mathrm{Cu(OH)_2}

precipitates before

\mathrm{Fe(OH)_3}

because Cu

^{2+}

needs only two

\mathrm{OH^-}

ions, whereas Fe

^{3+}

needs three. c)

\mathrm{Fe(OH)_3}

precipitates before

\mathrm{Cu(OH)_2}

because it reaches its

K_{sp}

at a lower

\mathrm{[OH^-]}

. d) Both hydroxides begin to precipitate at the same pH since the metal-ion concentrations are equal.

Intuitive explanation

This problem is about selective precipitation — using $K_{sp}$ to predict which metal hydroxide forms first as pH rises. The key insight is that each metal ion will precipitate when $[\mathrm{OH^-}]$ reaches the threshold where $K_{sp}$ is exceeded; whichever ion needs the lower $[\mathrm{OH^-}]$ to trigger precipitation will come out of solution first. Calculate the critical $[\mathrm{OH^-}]$ for each using $K_{sp}$ and the given metal-ion concentration, then convert to pH to see which precipitates at lower pH.

Intuitive explanation

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