What are the equilibrium concentrations of the dissolved ions in a saturated solution of $F e {(O H)}_{2}$ $Fe(OH)_2$ at 25°C?

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Answer 1 · 2017-08-03T21:32:30

We need $K_{sp}$ $K_"sp"$ for $F e {(O H)}_{2}$ $Fe(OH)_2$ .........

And we assess the equilibrium......

$F e {(O H)}_{2} (s) ⇌ F e^{2 +} + 2 H O^{-}$ $Fe(OH)_2(s)rightleftharpoonsFe^(2+) + 2HO^-$

For which this site quotes $K_{sp} = 8 \times 10^{- 16}$ $K_"sp"=8xx10^-16$ .

So we set the solubility of $F e {(O H)}_{2} \equiv S$ $Fe(OH)_2-=S$ .

And thus $K_{sp} = [F e^{2 +}] {[H O^{-}]}^{2} = S \times {(2 S)}^{2} = 4 S^{3}$ $K_"sp"=[Fe^(2+)][HO^-]^2=Sxx(2S)^2=4S^3$

And thus...... $S =^{3} \sqrt{\frac{8.0 \times 10^{- 16}}{4}}$ $S=""^(3)sqrt((8.0xx10^-16)/4)$

$= 5.85 \times 10^{- 6} \cdot m o l \cdot L^{- 1}$ $=5.85xx10^-6*mol*L^-1$

And so $F e^{2 +} = 5.85 \times 10^{- 6} \cdot m o l \cdot L^{- 1}$ $Fe^(2+)=5.85xx10^-6*mol*L^-1$

And $[H O^{-}] = 1.17 \times 10^{- 5} \cdot m o l \cdot L^{- 1}$ $[HO^-]=1.17xx10^-5*mol*L^-1$

Can you work out the solubilities in $g \cdot L^{- 1}$ $g*L^-1$ ?

What are the equilibrium concentrations of the dissolved ions in a saturated solution of Fe(OH)_2Fe(OH)2Fe(OH)_2 at 25°C?