X. XX = 8.09 Aniline, C_6H_5NH_2, is a weak base which is used extensively in th

Question

X. XX = 8.09

Aniline, C_6H_5NH_2, is a weak base which is used extensively in the dye industry. Its conjugate acid, Aniline hydrochloride, [C_6H_5NH_3]C1, can be titrated by a strong base such as NaOH. Assume you titrate 25.0 mL of 0.100 M aniline hydrochloride with 0.115 M NaOH. Assume the K_b for aniline is X.XX times 10^-10. What is the pH of the aniline hydrochloride solution before the titration begins? What is the pH at the equivalence point? What is the pH halfway to the equivalence point?

Explanation / Answer

Before titration begins, the reaction is as follow and ICE chart is as follow:

r: C6H5NH3+ + H2O --------> C6H5NH2 + H3O+   Ka = 1x10-14 / 8.09x10-10 = 1.24x10-5
i: 0.100 0 0
e: 0.1-x x x

1.24x10-5 = x2 / 0.1-x --> Ka is small, so we can neglect the substract of 0.1-x to 0.1 only:
1.24x10-5 * 0.1 = x2
x = 1.114x10-3 M = [H3O+]
pH = -log(1.114x10-3)
pH = 2.95

The pH at halfway equivalence point would be the value of pKa, this is because at this point, half of the aniline is consumed, and the concentration of the product formed and the base is the same. So, pH = pKa

Calculating pKa:
pKa = -log(1.24x10-5)
pKa = pH = 4.91

Finally, the pH at equivalence point, the aniline is consumed at all, and the acid begins to be in excess so the reaction taking place and the moles involved are:

Vb = 0.100 * 25 / 0.115 = 21.74 mL of base required to reach the equivalence point.
moles OH- = 0.02174 * 0.115 = 0.0025 moles

These moles are produced in the equivalence point, the concentration is:
[OH-] = [C6H5NH2] = 0.0025 / 0.04674 = 0.0535 M

and the reaction taking place is:
r: C6H5NH2 ----------> C6H5NH3+ + OH-
i: 0.0535 0 0
e: 0.0535-x x x

8.09x10-10 = x2 / 0.0535-x
8.09x10-10* 0.0535 = x2
x = 6.58x10-6 M = [OH-] = [C6H5NH2]
pOH = -log(6.58x10-6) = 5.18
pH = 14-5.18
pH = 8.82

Hope this helps

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