A solution of HNO2 is at equilibrium:
HNO2(aq) + H2O(l) ⇔ H3O+(aq) + NO2-(aq)
Water is added. After the water is added, is there any shift
in the equilibrium?
Your question, due to the wording, doesn't make any sense to me. The problem states that the solution is at equilibrium so how can there be a shift? From the wording, the water has been added already.
Yea so you see what kind of instructor I am working with...at least I know I'm not the only one confused by her questions...
I may have been a little harsh. After re-reading the problem, I see the "more water is added" statement.
Here what I'm going to do. This is the way I interpret the problem. We have a solution at equilibrium as shown by the following equation.
HNO2(aq) + H2O(l) <--> H3O^+(aq) + NO2^-(aq)
In a separate step, I will add more water. Is there any shift in the equilibrium?
Answer: Yes, adding more water causes the reaction to shift to the right. Generally, adding more water to a weak acid causes it to ionize more; i.e., a greater percentage of ionization occurs.
Additional comment: Many texts and some teachers go to great lengths to caution students that "pure" substances, such as H2O(l) do not appear in Ka or Kb or Keq expressions. The pure material, by definition, is defined with an activity of 1(no unit) and it should not even appear in the Ka, Kb, or Keq expression. And they also caution, that adding PURE substances to either side will NOT shift the equilibrium. But those groups of people fail to add to the student that, in cases like this, where the water actually enters into the reaction, it will shift the equilibrium.
To determine whether there will be a shift in the equilibrium after water is added, you need to consider Le Chatelier's principle. According to this principle, a system at equilibrium will respond to an external stress by shifting the equilibrium position to counteract the effect of the stress.
When water is added to the system, the concentration of H2O will increase. Since H2O is a pure liquid, increasing its concentration does not have an effect on the equilibrium. This is because the concentration of a pure liquid does not appear in the equilibrium expression.
Therefore, adding water to the system will not cause a shift in the equilibrium position. The equilibrium will remain unaffected.