Aniline hydrochloride, (C_6H_5NH_3)CI, is a weak acid. (Its conjugate base is th
ID: 976123 • Letter: A
Question
Aniline hydrochloride, (C_6H_5NH_3)CI, is a weak acid. (Its conjugate base is the weak base aniline, (C_6H_5NH_2.) The acid can be titrated with a strong base such as NaOH. Assume 40.0 mL of 0.1000 M aniline hydrochloride is titrated with 0.185 M NaOH. What pH of the solution before the titration begins? (three decimal places) What is the pH at the equivalence point? (three decimal places) What is the pH at the half-way point of the titration? (three decimal places) The K_a for anilinium ion is 2.400 times 10^-5. pk_9 = 4.66Explanation / Answer
Qa) before adding base the solution is of anilinium hydrochloride , which is a salt of weak base and strong acid.
Its pH is given by pH = 1/2 pKw - 1/2 pKb -1/2 log c
Given ka of anilinium ion = 2.4 x 10-5 that is pka = 4.62
Thus pkb of aniline = 14-4.62 = 9.38
Hence pH of solution before titration = 7 -1/2 (9.38) - 1/2 log 0.1
= 2.810
b) At the equivalence point
BH+ + NaOH -------> B + H2O + NaCl
40x0.1 vx0.185 0 initial mmoles
=4
0 0 4 at equivalence mmoles
0 0 4/(40+21.62) =0.065 eq. concentration
At equivalence we have only free weak base in solution.The pH of weak base is given by
pH = 14 -( 1/2 pKb -1/2 logC)
= 14 -( 4.69 -1/2 log 0.065) =8.716
Qc)
At half equivalence point
BH+ + NaOH -------> B + H2O + NaCl
40x0.1 vx0.185 0 initial mmoles
=4
2 0 2 at half equivalence mmoles
At this stage the solution has the acid anilinium ion and its conjugate base aniline in equal quantities. thus it acts as a buffer.
The pH of this buffer is given by Hendersen equation
pH = pKa + log [conjugate base]/[acid]
= 4.619 + log 2/2
= 4.619