The ph of the given solution of NH4Cl is -0.504.
Explanation:
Ka [tex]\times[/tex] Kb = Kw
Ka = Kw / Kb,
Kw = 1 [tex]\times[/tex] 10^-14
Kb of NH3 = 1.8 [tex]\times[/tex] 10^-5
We eventually want [H3O+] to calculate pH, so determine the Ka of NH3 from Kb
The Kb for NH3 is 1.8 [tex]\times[/tex] 10^-5
Ka = (1 [tex]\times[/tex] 10^-14) / Kb
Ka = (1 [tex]\times[/tex] 10^-14) / ( 1.8 [tex]\times[/tex] 10^-5)
Ka = 5.56 [tex]\times[/tex] 10^-10
NH4+ + H2O ↔ NH3 + H3O+
Calculate [H3O+] using the Ka equation:
Ka = [NH3] [H3O+) / [NH4+]
We know that [NH3] = [H3O+] , and that [NH4+] = 2.0 M
Ka = [H3O+]² / 2.0
5.56 [tex]\times[/tex] 10^-10 = [H3O+]²/ 2.0
[H3O+]² = (5.56 [tex]\times[/tex] 10^-10) [tex]\times[/tex] 2.0
[H3O+]² = 10.227
[H3O+] = 3.198
pH = -log [H3O+]
pH = -log 3.198
pH = -0.504
