Question

determine the pH of a 0.62 M NH4NO3 solution at 25 degrees Celcius. The Kb for NH3 is 1.76*10^-5

Answer 1

1) Determine the pH of a 0.62 M NH4NO3 solution at 25C. The Kb for NH3 is 1.76 x 10^-5.
NH4+ is the conjugate acid oc the base NH3
Ka for NH4+ = K water / K NH3 conjugate base
Ka = 1 e-14 / 1.76 e-5 = 5.68 e-10

which is he Ka for
NH4)+ in water --> NH4OH &H+

Ka = [NH4OH] [H+] / [NH4)+]

5.68 e-10 = [X] [X] / [0.62]

X2 = 3.52 e-10

X = [H+] = 1.88 e=5

p of that H+ = 4.73

Answer 2

Determine the pH of a 0.62 M NH4NO3 solution at 25C. The Kb for NH3 is 1.76 x 10^-5.
NH4+ is the conjugate acid oc the base NH3
Ka for NH4+ = K water / K NH3 conjugate base
Ka = 1 e-14 / 1.76 e-5 = 5.68 e-10

which is he Ka for
NH4)+ in water --> NH4OH &H+

Ka = [NH4OH] [H+] / [NH4)+]

5.68 e-10 = [X] [X] / [0.62]

X2 = 3.52 e-10

X = [H+] = 1.88 e=5

p of that H+ = 4.73
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