A solution is 0.10 M in Fe^3+ and 0.10 M in Mg^2+. K_sp(Fe(OH)3) = 1.0x10^-38 and K_sp(Mg(OH)2) = 4.0x10^-12; log 2 = 0.301. Alkali is added without volume change. Select the correct statement(s).

Mg(OH)2 starts to precipitate at pH approx 8.80

At pH = 3.00, 99.99% of the Fe^3+ is precipitated while no Mg(OH)2 forms

Fe(OH)3 starts to precipitate at pH approx 1.67

Mg(OH)2 starts to precipitate at pH approx 8.80

At pH = 3.00, 99.99% of the Fe^3+ is precipitated while no Mg(OH)2 forms

Fe(OH)3 starts to precipitate at pH approx 1.67

Raising the pH to 9.0 separates Fe^3+ from Mg^2+ cleanly

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ia212

A solution is $0.10 M$ in $Fe X^{3 +}$ and $0.10 M$ in $Mg X^{2 +}$ . $K_{s p} (Fe (OH) X_{3}) = 1.0 \times 1 0^{- 38}$ and $K_{s p} (Mg (OH) X_{2}) = 4.0 \times 1 0^{- 12}$ ; $lo g 2 = 0.301$ . Alkali is added without volume change. Select the correct statement(s).

A $Mg (OH) X_{2}$ starts to precipitate at $pH \approx 8.80$
BAt $pH = 3.00$ , $99.99%$ of the $Fe X^{3 +}$ is precipitated while no $Mg (OH) X_{2}$ forms
C $Fe (OH) X_{3}$ starts to precipitate at $pH \approx 1.67$
DRaising the pH to $9.0$ separates $Fe X^{3 +}$ from $Mg X^{2 +}$ cleanly

Fe (OH) X_{3}

[OH X^{-}] = (1 0^{- 38} /0.10)^{1/3} = 4.6 \times 1 0^{- 13}

, pH

1.67

Mg (OH) X_{2}

[OH X^{-}] = 4 \times 1 0^{- 12} /0.10 = 6.3 \times 1 0^{- 6}

, pH

8.80

. For

1 0^{- 5} M

Fe X^{3 +}

left (

99.99%

gone),

[OH X^{-}] = (1 0^{- 33})^{1/3} = 1 0^{- 11}

, pH

3.00

, far below

8.80

. At pH

9.0

the

Mg (OH) X_{2}

also precipitates, so separation fails.