Question

Determine the pH of the solution when 20.00 mL of 0.1563 M aniline hydrochloride (C6H5NH3+Cl-) is titrated with 15.00 mL of 0.1249 M NaOH. (Ka(aniline)=2.40 x 10-5)

Answer 1

Given:

M(C6H5NH3+) = 0.1563 M

V(C6H5NH3+) = 20 mL

M(NaOH) = 0.1249 M

V(NaOH) = 15 mL

mol(C6H5NH3+) = M(C6H5NH3+) * V(C6H5NH3+)

mol(C6H5NH3+) = 0.1563 M * 20 mL = 3.126 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1249 M * 15 mL = 1.8735 mmol

We have:

mol(C6H5NH3+) = 3.126 mmol

mol(NaOH) = 1.8735 mmol

1.8735 mmol of both will react

excess C6H5NH3+ remaining = 1.2525 mmol

Volume of Solution = 20 + 15 = 35 mL

[C6H5NH3+] = 1.2525 mmol/35 mL = 0.0358M

[C6H5NH2] = 1.8735/35 = 0.0535M

They form acidic buffer

acid is C6H5NH3+

conjugate base is C6H5NH2

Ka = 2.4*10^-5

pKa = - log (Ka)

= - log(2.4*10^-5)

= 4.62

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.62+ log {5.353*10^-2/3.579*10^-2}

= 4.795

Answer: 4.80