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Gay Lussac's Law, Avagadro's Law

The conc. of ...

Question

The conc. of ${F e}^{3 +}$ ions in a sample of water is found to be $50 \times 10^{- 5} M$ . Calculate the $p H$ (nearest integer) at which $99$ % of ${F e}^{3 +}$ will be precipitated. $K_{{s p}_{F e {(O H)}_{3}}} = 10^{- 36}$ .

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Solution

99 % of $F e^{3 +}$ ions are precipitated. This means 1 % is remaining in the solution.
Hence, $[F e^{3 +}] = \frac{50 \times 10^{- 5}}{100} M$
The expression for the solubility product is $K_{s p} = [F e^{3 +}] [O H^{-}]^{3}$
Substitute values in the above equation.
$\frac{50 \times 10^{- 5}}{100} \times [O H^{-}]^{3} = 1 \times 10^{- 36}$
$[O H^{-}] = 5.85 \times 10^{- 11} M$ .
Hence, the pOH of the solution is $p O H = - l o g [O H^{-}] = - l o g (5.85 \times 10^{- 11}) = 10.23$ .
The pH of the solution is $p H = 14 - p O H = 14 - 10.23 = 3.767 \approx 4$

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