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Consider the titration of 27.8 mL of 0.447 M NH3 (Kb = 1.79 ×10-5) with 0.267 M HNO3. What is the pH of 100% titration?

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HNO3 + NH3 ==> NH4NO3 ==> NH4+ + NO3- ... balanced equation

moles NH3 = 0.0278 L x 0.447 mol/L = 0.01243 moles

moles HNO3 = 0.01243 moles

volume HNO3 needed = (x L)(0.267 mol/L) = 0.01243 and x = 0.0466 L = 46.6 mls

Total volume at neutralization = 27.8 ml + 46.6 mls = 74.4 ml = 0.0744 L

[NH4+] at neutralization = 0.01243 moles/0.0744 L = 0.167 M

NH4+ + H2O ==> H3O+ + NH3

Ka for NH4 acting as an acid = 1x10-14 / Kb NH3 = 1x10-14/1.79x10-5

Ka = 5.6x10-10

Ka = 5.6x10-10 = [H3O+][NH3]/[NH4+] = (x)(x)/0.167-x and if we assume x is small vs. 0.167 we can ignore

x2 = 9.35x10-11

x = 9.7x10-6 = [H3O+] (note: this is small vs. 0.167 so our assumption was correct)

pH = -log [H3O+] = -log 9.7x10-6

pH = 5.01

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