Question

3. A 30.0-mL sample of 0.10 M C2H3NH2 (ethylamine) is titrated with 0.15 M HCl. What is the pH of the solution after 20.00 mL of acid have been added to the amine? Kb = 6.5 × 10–4

Answer 1

Given:

M(HCl) = 0.15 M

V(HCl) = 20 mL

M(C2H3NH2) = 0.1 M

V(C2H3NH2) = 30 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.15 M * 20 mL = 3 mmol

mol(C2H3NH2) = M(C2H3NH2) * V(C2H3NH2)

mol(C2H3NH2) = 0.1 M * 30 mL = 3 mmol

We have:

mol(HCl) = 3 mmol

mol(C2H3NH2) = 3 mmol

3 mmol of both will react to form C2H3NH3+ and H2O

C2H3NH3+ here is strong acid

C2H3NH3+ formed = 3 mmol

Volume of Solution = 20 + 30 = 50 mL

Ka of C2H3NH3+ = Kw/Kb = 1.0E-14/6.5E-4 = 1.538*10^-11

concentration ofC2H3NH3+,c = 3 mmol/50 mL = 0.06 M

for simplicity lets write acid ion as BH+

C2H3NH3+ + H2O -----> C2H3NH2 + H+

6*10^-2 0 0

6*10^-2-x x x

Ka = [H+][C2H3NH2]/[C2H3NH3+]

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 ((1.538*10^-11)*6*10^-2) = 9.608*10^-7

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

so, x = 9.608*10^-7 M

[H+] = x = 9.608*10^-7 M

use:

pH = -log [H+]

= -log (9.608*10^-7)

= 6.02

Answer: 6.02