Question

64.31 mL of 0.224 M hydrochloric acid is added to 154.24 mL of 0.096 M ethylamine solution. What is the approximate pH of the resulting solution?

Answer 1

Kb for ethylamine = 4.3*10^-4

Given:

M(HCl) = 0.224 M

V(HCl) = 64.31 mL

M(C2H5NH2) = 0.096 M

V(C2H5NH2) = 154.24 mL

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

mol(HCl) = 0.224 M * 64.31 mL = 14.4054 mmol

mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)

mol(C2H5NH2) = 0.096 M * 154.24 mL = 14.807 mmol

We have:

mol(HCl) = 14.4054 mmol

mol(C2H5NH2) = 14.807 mmol

14.4054 mmol of both will react

excess C2H5NH2 remaining = 0.4016 mmol

Volume of Solution = 64.31 + 154.24 = 218.55 mL

[C2H5NH2] = 0.4016 mmol/218.55 mL = 0.0018 M

[C2H5NH3+] = 14.4054 mmol/218.55 mL = 0.0659 M

They form basic buffer

base is C2H5NH2

conjugate acid is C2H5NH3+

Kb = 4.3*10^-4

pKb = - log (Kb)

= - log(4.3*10^-4)

= 3.367

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 3.367+ log {6.591*10^-2/1.838*10^-3}

= 4.921

use:

PH = 14 - pOH

= 14 - 4.9213

= 9.08

Answer: 9.08