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Standard XII

Chemistry

Derivation of Kp and Kc

If concentrat...

Question

If 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 increase by

8 times

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16 times

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64 times

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4 times.

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Solution

The correct option is C 64 times
$K = [F e^{3 +}] [O H^{-}]^{3}$ If concentration of $O H^{-}$ ion is decreased to $\frac{1}{4}$ th to keep $K$ constant
Concentration of $F e^{3 +}$ ions has to be increased by $64$ times.

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Similar questions

Q. 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 increase by:

Q. 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:

Q. If 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 increase by

Q. In the reaction

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

⇌ F e_{a q}^{3 +} + 3 O H_{a q}^{-}

, if the concentration of

O H^{-}

is decreased by

\frac{1}{4}

times then the equilibrium concentration

o f F e^{3 +}

increased by how many times?

Q. 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 upto

\frac{1}{4}

times, then equilibrium concentration of

F e^{+ 3}

will increase upto:

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If concentration of OH−ions in the reaction Fe(OH)3(s) ⇋ Fe3++(aq)+ 3 OH− (aq) is decreased by 1/4 times, then equilibrium concentration of Fe3+will increase by

The correct option is C 64 times K = [Fe3+] [OH−]3 If concentration of OH− ion is decreased to 14 th to keep K constant Concentration of Fe3+ions has to be increased by 64 times.

If 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 increase by

The correct option is C 64 times
$K = [F e^{3 +}] [O H^{-}]^{3}$ If concentration of $O H^{-}$ ion is decreased to $\frac{1}{4}$ th to keep $K$ constant
Concentration of $F e^{3 +}$ ions has to be increased by $64$ times.