I'm a little stumped of this one... IV) 2CrO4 –2(aq) + 2H+ (aq)  Cr2O7 –2(aq)...

I'm a little stumped of this one...

IV) 2CrO4 –2(aq) + 2H+ (aq)  Cr2O7 –2(aq) + H2O(l)

(7) *Add 2 drops of 6M HNO3 to about 3mL of 0.1M K2CrO4.

(8) Add 10% NaOH to (7) dropwise until the original color of K2CrO4 is restored.

(9) *Add 2 drops of 6M H2SO4 to about 3mL of 0.1M K2CrO4.

(10) Add 10% NaOH to (9) dropwise until the original color of K2CrO4 is restored.

Question:

In step (10), which direction DID the equilibrium shift? Using balanced equations, explain the direction of this shift.

Expert Solution

2CrO₄^2-(aq) + 2H⁺ (aq) ⇌ Cr₂O₇^2-(aq) + H₂ O(l)

In step (10), the equilibrium shift towards the left (reactants).

Explanation- In step (9) 2 drops of 6M H₂SO₄ is added to 3mL of 0.1M K2CrO4. Originally the solution is yellow (more CrO₄^2-) as H⁺ are added to the solution, the equilibrium shifts towards the products, resulting in the solution turning orange (more Cr₂O₇^2-). When NaOH is added to the solution in step (10), it neutralizes the acid; removing H⁺ ions and equilibrium sifted towards left (the reactants) and the solutions turns yellow (more CrO₄^2-) again.

Balanced equations for step (10)

Cr₂O₇^2- + 2 OH^- <====> 2 CrO₄^2- + H₂O

queen_honey_blossom answered 2 years ago

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