10 When vanadium(II) compounds are dissolved in water, the following equilibrium is established.
V
2+
+ H2O V
3+
+ H2
2
1
+ OH
–
What would alter the composition of the equilibrium mixture in favour of the V
2+
ions?
A adding an acid
B adding a reagent that selectively precipitates V
3+
ions
C allowing the hydrogen to escape as it forms
D making the solution more alkaline

Question

10 When vanadium(II) compounds are dissolved in water, the following equilibrium is established.
V
2+
 + H2O        V
3+
 + H2
2
1
+ OH
–
What would alter the composition of the equilibrium mixture in favour of the V
2+
 ions?
A adding an acid
B adding a reagent that selectively precipitates V
3+
 ions
C allowing the hydrogen to escape as it forms
D making the solution more alkaline

vincent-lyon.fr · Answer

Hi Saad! Is your equation:
$\large V^{2+}+H_2O\rightarrow V^{3+}+\frac {1}{2}H_2+OH^-$       ?

anonymous · Answer

yes :)
i understand why the option is D but i dont know why is option B wrong
pls help me with that :)

jfraser · Answer

forcing the precipitation of the V+3 ions would lower the concentration at equilibrium. The reaction will then shift to make MORE V+3 ions to relieve the stress.

anonymous · Answer

thanks alot :) 
i didnt realise the compund was being dissolved in water so obviously formaiton of precipitate was moving it out of the reaction mixture