Question

An analytical chemist is titrating 81.1 mL of a 0.5900 M solution of trimethylamine ((CH3)3N) with a 0.1300 M solution of HNO3. The pKb of trimethylamine is 4.19. Calculate the pH of the base solution after the chemist has added 415.2 mL of HNO3 solution to it. Round your answer to 2 decimal places.

Answer 1

Number of moles = molarity* volume

millimoles of (C2H5)3N (base) = 0.5900*81.1= 47.849 mmoles

millimoles of HNO3 (acid) = 415.2 x 0.1300 = 53.976 m moles

total volume = 81.1 ml +415.2 ml

= 496.3 ml

HNO3 moles present more as compare to trimethyl amine .

After neutralize, number of moles of acid = 53.976 m moles - 47.849 mmoles

=6.127 m moles

Or 6.127 *10^-3 moles

New moalrity of acid = number of moles / volume in L

=6.127 *10^-3 moles/ 0.4963 L

= 0.012345 M

pH = - log [H+]

= - log 0.012345

=- (-1.91)

= 1.91