Consider the titration of 30.0mL of 0.050 M NH_3 with 0.025 M HCl. Calculate the

Question

Consider the titration of 30.0mL of 0.050 M NH_3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added.

a) 0mL b)20.0 c)59.3 d)60 e)72.6 f)73.1

Explanation / Answer

Kb of NH3 = 1.8 x 10^-5 a) NH3 + H2O NH4+ + OH- Kb = 1.8 x 10^-5 = x^2/ 0.030-x x = [OH-]= 0.00073 M pOH = 3.1 pH = 10.9 b) Moles NH3 = 0.0300 L x 0.030 M=0.00090 moles HCl = 0.0100 L x 0.025 M= 0.00025 NH3 + H+ = NH4+ moles NH3 = 0.00090 - 0.00025=0.00065 moles NH4+ = 0.00025 total volume = 30.0 + 10.0 = 40.0 mL => 0.0400 L [NH3]= 0.00065 / 0.0400=0.016 M [NH4+]= 0.00025/ 0.0400=0.0063 M pKb = - log Kb = 4.7 pOH = 4.7 + log 0.0063/ 0.016 = 4.3 pH = 14 - 4.3 = 9.7 c) moles HCl= 0.0200 L x 0.025 M=0.00050 moles NH3 = 0.00090 - 0.00050=0.00040 moles NH4+ = 0.00050 total volume = 0.050 L [NH3]= 0.00040/ 0.050=0.0080 M [NH4+]= 0.00050/ 0.050 =0.010 M pOH = 4.7 + log 0.010/ 0.0080=4.8 pH = 9.2

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