You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4

times

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8

times

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16

times

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

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 +}

Join BYJU'S Learning Program