What is the pH of a 0.42 M NH4NO3 aqueous solution?
What is the percent ionization?
Hydrolyze NH4^+.
NH4^+ + H2O ==> H3O^+ + NH3
Ka = (Kw/Kb) = (H3O^+)(NH3)/(NH4^+)
Set up an ICE chart and solve for (H3O^+), then convert to pH.
To determine the pH of the NH4NO3 solution, we need to consider the dissociation of the compound in water and the subsequent formation of ions.
First, let's write the balanced equation for the dissociation of NH4NO3 in water:
NH4NO3 (aq) -> NH4+ (aq) + NO3- (aq)
In this equation, NH4NO3 breaks apart into NH4+ ions and NO3- ions.
To find the pH of the NH4NO3 solution, we need to examine the behavior of NH4+ in water. NH4+ is the conjugate acid of the weak base NH3 (ammonia), which can undergo the following equilibrium reaction with water:
NH4+ (aq) + H2O (l) <-> NH3 (aq) + H3O+ (aq)
Here, NH4+ donates a proton (H+) to water, forming NH3 (ammonia) and H3O+ ions. As NH4+ is a weak acid, this equilibrium lies more towards the left side.
To find the pH, we need to calculate the concentration of H3O+ ions. In this case, we can assume that all NH4+ ions dissociate into NH3 and H3O+ ions. Therefore, the concentration of H3O+ ions is equal to the initial concentration of NH4+ ions.
Next, let's calculate the concentration of NH4+ ions:
0.42 M of NH4NO3 means we have 0.42 moles of NH4NO3 in 1 liter of solution.
Since NH4NO3 dissociates into NH4+ ions and NO3- ions in a 1:1 ratio, the concentration of NH4+ ions is also 0.42 M.
Therefore, the concentration of H3O+ ions is 0.42 M.
Now, we can calculate the pH using the formula:
pH = -log[H3O+]
pH = -log(0.42) = 0.38
So, the pH of the 0.42 M NH4NO3 solution is approximately 0.38.
To determine the percent ionization, we can compare the concentration of the ionized form (NH4+) to the initial concentration of NH4NO3.
The percent ionization is calculated using the formula:
Percent Ionization = (concentration of ionized form / initial concentration of compound) x 100
In this case, since all the NH4+ ions come from the dissociation of NH4NO3, the concentration of ionized form (NH4+) is 0.42 M.
Percent Ionization = (0.42 / 0.42) x 100 = 100%
Therefore, the percent ionization of the 0.42 M NH4NO3 solution is 100%.