Question

URGENT QUESTION!!! Please answer this question with detailed explanation. Thank you.

A solution is made by mixing 100mL of 0.1 M HCl with 100 mL of 0.1 M NH₃ at 25C (degrees celcius). What is the pH of the final solution? (K_b for NH₃ = 1.8 x 10^-5)

Answer 1

moles of NH3= MxV/1000= 0.1x100/1000=0.01

NH3 + H2O -------------> NH4⁺ + OH^-

0.01                               0           0     initial

0.01-x                              x         x    equlibrium

Kb= [NH4⁺][OH^-]/[ NH3]

1.8 x 10^-5 = x²/0.01-x

x= 4.2 x10^-4

OH^- moles = 4.2 x10^-4

H⁺ moles = 0.1 x 100/1000= 0.01

now [H⁺ ] >[OH^-]

[H⁺ ] remainig = [H⁺ ] -[OH^-]/total volume

                       = 0.01-0.00042/100+100

                         =0.01-0.00042/200

                         = 47x10^-6

      pH = -log [H+]

     = - log[47 x 10^-6 ]

     = 4.32