Question

Consider a solution that contains both C6H5NH2 and C6H5NH3+. Calculate the ratio [C6H5NH2]/[C6H5NH3+] if the solution has the following pH values. (Assume that the solution is at 25°C.) (a) pH = 4.69 (b) pH = 5.19 (c) pH = 5.03 (d) pH = 5.44

Answer 1

Given, a solution of C₆H₅NH₂ and C₆H₅NH₃⁺.

We know, the Henderson-Hesselbalch equation,

pH = pK_a + log[ Conjugate Base / Acid]

Also, We know,

K_b for C₆H₅NH₂ at 25^oC = 4.2 x 10^-10

Thus,

K_a = (1.0 x 10^-14) / 4.2 x 10^-10

K_a= 2.3809 x 10^-5

pK_a = -log K_a

pK_a = -log (2.3809 x 10^-5)

pK_a = 4.62

Now,

a) pH = 4.69

Thus,

pH = pK_a + log[ Conjugate Base / Acid]

4.69 = 4.62 + log [C₆H₅NH₂/C₆H₅NH₃⁺.]

[C₆H₅NH₂/C₆H₅NH₃⁺] = 1.17 Or 1.16

b) pH = 5.19

Thus,

pH = pK_a + log[ Conjugate Base / Acid]

5.19 = 4.62 + log [C₆H₅NH₂/C₆H₅NH₃⁺.]

[C₆H₅NH₂/C₆H₅NH₃⁺] = 3.69

c) pH = 5.03

Thus,

pH = pK_a + log[ Conjugate Base / Acid]

5.03 = 4.62 + log [C₆H₅NH₂/C₆H₅NH₃⁺.]

[C₆H₅NH₂/C₆H₅NH₃⁺] = 2.55

d) pH = 5.44

Thus,

pH = pK_a + log[ Conjugate Base / Acid]

5.44 = 4.62 + log [C₆H₅NH₂/C₆H₅NH₃⁺.]

[C₆H₅NH₂/C₆H₅NH₃⁺] = 6.56