A chemist titrates of a aniline solution with solution at . Calculate the pH at equivalence. The of aniline is . Round your answer to decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of solution added.

Explanation:

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In this case, since the volume of the aniline (Kb = 7.41x10⁻¹⁰) is 160.0mL, its concentration 0.3403 M and the HNO3 solution is 0.0501 M, we can first compute the employed volume of the acid via:

Because at the equivalence point, the moles of acid equals the moles of base, which causes that at equilibrium the mayor species are NO₃⁻, C₆H₅NH₃⁺ and H₂O which means that the following ionization occurs:

And the pH at the equivalence point is lower than 7 due to the presence of the hydronium ion, that is why we use Ka rather than Kb. Next, we write the equilibrium expression:

Which based on the ICE chart, knowing that Kw=Ka*Kb and Kb is so small, we write:

Whereas the initial concentration of C₆H₅NH₃⁺ is:

Therefore, the concentration of hydronium un solution which equals is:

Thus the pH is:

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explanation:

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