Question
calculate the pH of 0.1M NH4NO3
Answers
Anonymous
You need the Kb value for Ammonium Hydroxide to complete the problem. Kb(NH4OH) = 1.8 x 10^-5
Calculate the Ka for NH4^+ Hydrolysis reaction = Kw/Kb
Set equal to the hydrolysis Ka expression for ...
NH4^+ + HOH <=> NH4OH + H^+
C(eq) 0.10 - x x
Ka =[NH^4OH][H^+]/[NH4^+] = Kw/Kb
=> x^2/0.1 = 1x10^-14/1.8x10^-5
=> solve for 'x' = [H^+]
=> pH = -log[H^+]
Calculate the Ka for NH4^+ Hydrolysis reaction = Kw/Kb
Set equal to the hydrolysis Ka expression for ...
NH4^+ + HOH <=> NH4OH + H^+
C(eq) 0.10 - x x
Ka =[NH^4OH][H^+]/[NH4^+] = Kw/Kb
=> x^2/0.1 = 1x10^-14/1.8x10^-5
=> solve for 'x' = [H^+]
=> pH = -log[H^+]