Applying Le Chateliers principle, what can you conclude about the change $2 C r O_{4}^{- 2} \to C r_{2} O_{7}^{- 2}$ $2CrO_4^(-2) -> Cr_2O_7^(-2)$ and its dependence on hydrogen ions, $H^{+}$ $H^+$ ?

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Applying Le Chateliers principle, what can you conclude about the change $2 C r O_{4}^{- 2} \to C r_{2} O_{7}^{- 2}$ $2CrO_4^(-2) -> Cr_2O_7^(-2)$ and its dependence on hydrogen ions, $H^{+}$ $H^+$ ?

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Answer 1 · 2018-01-03T13:22:45

Adding more hydrogen ions would favour the formation of $C r_{2} O_{7}^{- 2}$ $Cr_2O_7^(-2)$

Removing hydrogen ions would favour the formation of $C r O_{4}^{- 2}$ $CrO_4^(-2)$

Explanation:

The net ionic equation is:

$2 C r O_{4}^{- 2} + {2H}^{+} ⇌ C r_{2} O_{7}^{- 2} + H_{2} O$ $2CrO_4^(-2) + "2H"^"+"⇌ Cr_2O_7^(-2) + "H"_2"O"$

There are $2 H^{+}$ $2 H^+$ ions on the LHS and none on the RHS.

Based on Le Chateliers Principle;

Adding more hydrogen ions (making conditions more acidic) would favour the formation of $C r_{2} O_{7}^{- 2}$ $Cr_2O_7^(-2)$ because the equilibrium would shift to the side with fewer hydrogen ions to balance the system.

Removing hydrogen ions (making conditions more basic) would favour the formation of $C r O_{4}^{- 2}$ $CrO_4^(-2)$ because the equilibrium would shift to the side with more hydrogen ions to balance the system.

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Applying Le Chateliers principle, what can you conclude about the change $2 C r O_{4}^{- 2} \to C r_{2} O_{7}^{- 2}$ $2CrO_4^(-2) -> Cr_2O_7^(-2)$ and its dependence on hydrogen ions, $H^{+}$ $H^+$ ?

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Explanation:

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Applying Le Chateliers principle, what can you conclude about the change 2CrO−24→Cr2O−272CrO_4^(-2) -> Cr_2O_7^(-2) and its dependence on hydrogen ions, H+H^+?

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Applying Le Chateliers principle, what can you conclude about the change $2 C r O_{4}^{- 2} \to C r_{2} O_{7}^{- 2}$ $2CrO_4^(-2) -> Cr_2O_7^(-2)$ and its dependence on hydrogen ions, $H^{+}$ $H^+$ ?