Question

The pH in a 0.13 M solution of aniline (Kb=4.3×10−10). pH=8.87

Calculate the concentration of C6H5NH+3 in a 0.13 M solution of aniline.

Calculate the concentration of OH− in a 0.13 M solution of aniline.

Calculate the concentration of H3O+ in a 0.13 M solution of aniline.

Answer 1

Aniline is C₆H₅NH₂

Kb=4.3x10^-10

Dissociation of aniline in solution is as shown

Initial concentration (M) 0.13 0 0

Change -y +y +y

Final concentration 0.13-y y y

Kb=4.3x10^-10==(y x y)/(0.13-y)

4.3x10^-10=(y x y)/(0.13-y)

As the value of Kb is very small, we can ignore y as compared to 0.13 as y<<<<0.13 M

So 4.3x10^-10=(y x y)/(0.13)

4.3x10^-10 x 0.13=y²=0.559 x 10^-10

y=0.75 x 10^-5 M

So =0.75 x 10^-5 M

We know that ionic product of water Kw=1.0x10^-14=[H₃O⁺][OH^-]

Substituting the value of [OH^-]

1.0x10^-14=[H₃O⁺] x 0.75 x 10^-5 M

[H₃O⁺]=1.0x10^-14/0.75 x 10^-5 M=1.33 x 10^-9

pH=-log [H₃O⁺]=-log (1.33 x 10^-9)=8.87