Calculate the pH of a 0.0677 M anilinium chloride solution. A list of ionization

Question

Calculate the pH of a 0.0677 M anilinium chloride solution. A list of ionization constants can be found here. (https://sites.google.com/site/chempendix/ionization)

Calculate the concentrations of the conjugate acid and base at equilibrium. The conjugate acid and conjugate base are represented as HA and A–, respectively.

Map aplingearning Calculate the pH of a 0.0677 M anilinium chloride solution. A list of ionization constants can be found here. Number Calculate the concentrations of the conjugate acid and base at equilibrium. The conjugate acid and conjugate base are represented as HA and A, respectively. anilinium chloride Number Number

Explanation / Answer

Kb for C6H5NH2 = 4*10-10

The conjugate base is anilinium chloride, (C6H5NH3Cl) is a strong acid

Ka= 10-14/(4*10-10)=0.000025

C6H5NH2

HA----> H+ +A-

Ka= 0.000025

x2/(0.0677-x)= 0.000025 x = [H+] =[A-]

x= 0.003991

pH= -log (0.003991)= 2.4

[A-] =0.003991 M

[HA] =0.0677-0.003991=0.63709M

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