3. Consider the following equilibrium for nitrous acid, HNO2, a weak acid:?

Question

3. Consider the following equilibrium for nitrous acid, HNO2, a weak acid:
HNO2(aq) + H2O(l) ß à H3O+(aq) + NO2-(aq)

In which direction will the equilibrium shift if
a) NaOH is added?
b) NaNO2 is added
c) HCl is added
d) The acid solution is made more dilute?

Answer 1

(a) right as a competing equilibrium uses up much of the H3O+(aq) so forward reaction exceeds reverse reaction until equilibrium re-established (consult Le Chatelier's Principle)
(b) left as NO2-(aq); a product of the equilibrium, is added....
(c) left as H3O+(aq) added as a constituent of the HCl solution; and if HCl gas is used instead, it dissolves in the water to form H3O+(aq) as well. The amount of H2O consumed is such that it still stays in huge excess anyway
(d) one assumes this is done by adding water, so the concentrations of all ions is decreased, and reaction will proceed to the right to increase the number of particles in solution back towards a level that gets more particles/ions....

The last one is best described mathematically. If we double the volume of water, [H2O] is basically unchanged - still over 55M whereas the concentration of the remaining three partners in the equilibrium is instantly halved.

Using the equilibrium law:

K...=..[H3O+]..x..[NO2-]
.........._____________

..........[HNO2]..x..[H2O]

becomes 0.5 x 0.5
.............._______
.............0.5 x constant

which is a fraction that is a half of the equilibrium constant (0.25/0.5 = 0.5)

the top line values have to increase and the bottom line ones decrease to get back to unity (the equilibrium constant)

Product concentrations have to increase or reactant concentrations have to decrease, or both. Both it is, so reaction shifts to the right to re-establish equilibrium.

Answer 2

a. right
b. left
c. left
d. right