What is the pH of 0.144M aqueous solution of ammonium nitrate?
I am not sure what to do. Would I do 10^-0.144^ = 0.71?
No. You go through the hydrolysis equation.
.........NH4^+ + H2O ==> H3O^+ + NH3
initial..0.144............0.......0
change....-x...............x......x
equil...0.144-x............x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (NH3)(H3O^+)/(NH4^+)
Substitute for NH3 and H3O^+ in the above. Substitute 0.144-x for NH4 and solve for x = (H3O^+), then convert to pH.
so the values for both NH3 and H3O are 0?
No. They are x. And you're solving for x. And you convert x to pH.
For the Ka of NH4^+^ I did 1.0e-14/1.83-5 = 5.55e-10. What do I do with this value?
So for (NH3)(H3O) / NH4 it would be
x^2/0.144-x
Kb = 5.55E-10 so
5.55E-10 = x^2/0.144-x
oh ok I get it, thank you!!
To determine the pH of a solution, you need to use the concept of pOH, which is the negative logarithm of the hydroxide ion concentration (OH-) in the solution, and then convert it to pH using the relationship pH = 14 - pOH.
Ammonium nitrate (NH4NO3) is a salt that dissociates in water to produce ammonium ions (NH4+) and nitrate ions (NO3-). However, only the ammonium ions will contribute to the pH of the solution because they can react with water to form hydronium ions (H3O+).
The dissociation reaction is as follows:
NH4+ (aq) + H2O (l) -> H3O+ (aq) + NH3 (aq)
To find the concentration of the ammonium ions, you need to consider that one mole of ammonium nitrate will produce one mole of ammonium ions.
Given that the concentration of the ammonium nitrate solution is 0.144 M, the concentration of the ammonium ions will also be 0.144 M.
Since ammonium ions are weakly acidic, you need to consider the ionization of water. Water can auto-ionize to produce hydronium and hydroxide ions. At 25°C, the concentration of hydronium ions is approximately 1.0 x 10^-7 M.
Now, you can calculate the pOH:
pOH = -log10 (OH-)
OH- concentration is given by the dissociation of water, so OH- concentration = 1.0 x 10^-7 M
pOH = -log10 (1.0 x 10^-7)
pOH ≈ 7
Finally, you can find the pH:
pH = 14 - pOH
pH = 14 - 7
pH = 7
Therefore, the pH of the 0.144 M aqueous solution of ammonium nitrate is approximately 7.