The conc- of Fe3- ions in a sample of water is found to be 50-10-5M- Calculate the pH at which 99

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Solubility Product

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 $p H$ 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

The correct option is B 3.7
99 % of $F e^{3 +}$ ions are precipitated. This mans 1 % are remaining in the solution.
Hence, $F e^{3 +} = \frac{50 \times 10^{- 5}}{100}$
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 $p O H$ 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 $p H = 14 - p O H = 14 - 10.23 = 3.767$ .
Hence, the pH of the solution is $3.767$

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The conc. of Fe3+ ions in a sample of water is found to be 50×10−5M. Calculate the pH at which 99% of Fe3+ will be precipitated. KspFe(OH)3=10−36.

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