Question

Calculate the pH of a .215 M aqueous solution of aniline C6H5NH2, Kb= 7.4x10^-10 and the equilibrium concentrations of the weak base and it’s conjugate acid

Calculate the pH of a 0.0454 M aqueous solution of triethelyene C2H5(3)N, Kb= 5.2x10^-4 and the equilibrium concentrations of the weak base and its conjugate acid

Answer 1

1)

a)

C6H5NH2 dissociates as:

C6H5NH2 +H2O -----> C6H5NH3+ + OH-

0.215 0 0

0.215-x x x

Kb = [C6H5NH3+][OH-]/[C6H5NH2]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((7.4*10^-10)*0.215) = 1.261*10^-5

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

so, x = 1.261*10^-5 M

So, [OH-] = x = 1.261*10^-5 M

use:

pOH = -log [OH-]

= -log (1.261*10^-5)

= 4.8992

use:

PH = 14 - pOH

= 14 - 4.8992

= 9.1008

Answer: 9.10

b)

[C6H5NH2] = 0.215-x

= 0.215 - 1.261*10^-5 M

= 0.215 M

[C6H5NH3+] = x

= 1.261*10^-5 M

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