In a titration of a 100.0mL 1.00M NH3 solution with 1.00M HCl, what is the pH...

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In a titration of a 100.0mL 1.00M NH3 solution with1.00M HCl, what is the pH of the solution after the addition of 127mL of HCl? Kb = 1.8 x 10-5 forNH3

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millimoles of NH3 = 100 x 1 = 100

millimoles of HCl = 127 x 1= 127

NH3 + HCl ---------------------> NH4Cl

100       127                          0 ----------------------------> initial

0            27                             100 ------------------------> after reaction

here strong acid remained in the solution. so pH can be decided by strong acid

HCl millimoles remained = 27

HCl molarity = millimoles / total volume

                     = 27 / (100+127)

                     = 0.119 M

[H+] = 0.119M

pH = -log[H+]

pH = -log(0.119)

pH = 0.925


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