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

Jan 3, 2018

Adding more hydrogen ions would favour the formation of $C {r}_{2} {O}_{7}^{- 2}$

Removing hydrogen ions would favour the formation of $C r {O}_{4}^{- 2}$

#### Explanation:

The net ionic equation is:

$2 C r {O}_{4}^{- 2} + \text{2H"^"+"⇌ Cr_2O_7^(-2) + "H"_2"O}$

There are $2 {H}^{+}$ 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 $C {r}_{2} {O}_{7}^{- 2}$ 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 $C r {O}_{4}^{- 2}$ because the equilibrium would shift to the side with more hydrogen ions to balance the system.