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?
3 answers
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.
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.
0 answers