Question

(b) Saturated Barium Chromate. Watch Video-3b where to 3 mL of 0.1 M BaCl2, 5-6 drops of 1 MK2CrO4 are added, and then 10 to 12 drops of 6 M HCI. Write down the chemical reaction(s) involved (you might look back at Part la), being sure to include the state of matter of all reactants and products so it is clear if there is a solid/precipitate expected. (1 point) Specify the color after K Cro, is added and whether the reaction shifts toward the products or reactants for your reaction(s) above. Specify the color / change in state after HCI is added and explain how the HCI perturbs the equilibrium. (1 point)

Answer 1

1. When mixed, solutions of barium chloride, BaCl₂, and potassium chromate, K₂CrO₄, form a yellow precipitate of barium chromate, BaCrO₄. The balanced equation is:

BaCl₂(aq) + K₂CrO₄(aq) → BaCrO₄(s) + 2KCl(aq)

and after adding HCl the equation would be

2K₂CrO₄ + BaCl₂ + 2HCl → BaCr₂O₇ + 4KCl + H₂O

2. After addition of K₂CrO₄, a yellow precipitate of barium chromate, BaCrO₄ is formed . The reaction will proceed in forward direction towards the product, due to formation of yellow precipitate of barium chromate, BaCrO_4, the backward reaction will not take place.

After adding HCl the ppt of barium chromate, BaCrO₄ dissolves and reaction is shifted to backward direction .

2K₂CrO₄ + BaCl₂ + 2HCl → BaCr₂O₇ + 4KCl + H₂O

However, because HCl was added, there is also an increase in [H⁺], 2CrO₄^2- + 2H⁺ <----> Cr₂O₇^2- + H₂O . Eqm for this reaction would shift to the right to produce Cr₂O₇^2- . that is why a orangey yellow solution was formed due to Cr₂O₇^2- .