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

In a titration of a 100.0mL 1.00M NH₃ solution with 1.00M HCl, what is the pH of the solution after the addition of 127 mL of HCl? K_b = 1.8 x 10^-5 for NH₃

Expert Solution

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

queen_honey_blossom answered 1 year ago

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