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Ways to Define Concentration

If the concen...

Question

If the concentration of $O H^{-}$ ions in the reaction $F e (O H)_{3} (s) ⇌ F e^{3 +} (a q) + 3 O H^{-} (a q)$ is decreased by $1 / 4$ times, then equilibrium concentration of $F e^{3 +}$ will increased by:

64

times

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4

times

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8

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16

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Solution

The correct option is A $64$ times
$F e (O H)_{3} (s) ⇌ F e^{3 +} (a q) + 3 O H^{-} (a q)$
$K = \frac{[F e^{3 +}] [O H^{-}]^{3}}{[F e (O H)_{3}]}$
$K = [F e^{3 +}] [O H^{-}]^{3}$ ( $∵$ activity of solid is taken unity)
Concentration of $O H^{-}$ ion in the reaction is decreased by $1 / 4$ times, then equilibrium concentration of $F e^{3 +}$ will increased by $64$ times in order to keep the value of $K$ constant.

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Similar questions

{O H}^{-}

F e {(O H)}_{3} (s) ⇌ {F e}^{3 +} (a q) + 3 {O H}^{-} (a q .)

1 / 4

{F e}^{3 +}

O H^{-}

F e (O H)_{3} (s) ⇋ F e^{3 +} + (a q) + 3 O H^{-} (a q)

1 / 4

F e^{3 +}

O H^{-}

F e (O H)_{3} (s) ⇋ F e^{3 +} + (a q) + 3 O H^{-} (a q)

1 / 4

F e^{3 +}

O H^{-}

F e (O H)_{3} (s) ⇌ F e^{+ 3} (a q .) + 3 O H^{-} (a q .)

\frac{1}{4}

F e^{+ 3}

F e {(O H)_{3}}_{(s)}

⇌ F e_{a q}^{3 +} + 3 O H_{a q}^{-}

O H^{-}

\frac{1}{4}

o f F e^{3 +}

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