Question

Calculate the concentration of all species in a 0.240 M C₆H₅NH₃Cl solution. For C₆H₅NH₂, Kb=7.5×10⁻¹⁰.

[C₆H₅NH₃⁺],[Cl^-],[C₆H₅NH₂],[H₃O⁺],[OH^-]

Express your answer using two significant figures.

Answer 1

use:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/7.56*10^-10

Ka = 1.323*10^-5

C6H5NH3+ + H2O -----> C6H5NH2 + H+

0.24 0 0

0.24-x x x

Ka = [H+][C6H5NH2]/[C6H5NH3+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.323*10^-5)*0.24) = 1.782*10^-3

since c is much greater than x, our assumption is correct

so, x = 1.782*10^-3 M

So, [H3O+] = x = 1.782*10^-3 M

[C6H5NH2] = x= 1.782*10^-3 M

[C6H5NH3+] = 0.240-x = 0.240 - 1.782*10^-3 = 0.238 M

use:

[OH-] = Kw/[H3O+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(1.782*10^-3)

[OH-] = 5.61*10^-12 M

C6H5NH3Cl dissociates into C6H5NH3+ and Cl-. So, [Cl-] = 0.240 M

[Cl-] = 0.240 M

[H3O+] = 1.8*10^-3 M

[C6H5NH2] = 1.8*10^-3 M

[C6H5NH3+] = 0.24 M

[OH-] = 5.6*10^-12 M