Calculate the pH at the equivalence points for titrating 0.200 M solutions of each of the following bases with 0.200 M HBr.

Question

a. hydroxylamine (NH₂OH)

b. aniline (C₆H₅NH₂)

Answer 1

In a titration, at the equivalence point all of the analyte will be converted to the product as you will have equal moles of HBr and the analyte.

NH₂OH + HBr ==> ⁺NH₃OH + Br^-

The ⁺NH₃OH formed is from a weak base and a strong acid, so it should have an acidic pH @ equivalence

⁺NH₃OH + H2O ==> H3O⁺ + NH₂OH

Ka = [H₃O⁺][NH₂OH] / [⁺NH₃OH] so we need the Ka for ⁺NH₃OH. Looked up and found Kb for NH₂OH

Kb NH₂OH = 1.1x10^-8

Ka = 1x10^-14 / 1.1 x10^-8 = 9.09x10^-7

9.09x10^-7 = (x)(x) / 0.1

x = [H₃O⁺] = 3.01x10^-4 M

pH = -log [H₃O⁺]

pH = 3.52

C₆H₅NH₂ + HBr ==> C₆H₅NH₃⁺ + Br^-

Same process as above. At equivalence all of the C₆H₅NH₂ is converted to C₆H₅NH₃⁺

Look up the Kb for aniline, convert to Ka for C₆H₅NH₃⁺ and then hydrolyze the C₆H₅NH₃⁺ as follows:

C₆H₅NH₃⁺ + H2O ==> H3O+ + C₆H₅NH₂

Ka = [H3O+][C₆H₅NH₂] / [C₆H₅NH₃⁺]

Follow above calculations. pH @ equivalence should be acidic (< 7).

Answer 2

The previous answer covers the titration process and the pH calculations well but is missing that during the titration the solution is diluted as the titer is added to the aliquot. Since both solutions were the same concentration, then the two volumes will be the same and the resultant concentration of the conjugate acid will be 0.1 M, leading to a more dilute solution and a higher pH

[H₃O⁺]= 3.15x10^-4

pH = 3.50