Answer:
[tex]pH=5.25[/tex]
Explanation:
Hello.
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:
[tex]V_{acid}=\frac{V_{base}M_{base}}{M_{acid}} =\frac{160mL*0.3403M}{0.0501M}=1087mL[/tex]
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:
[tex]H_2O+C_6H_5NH_3^+\rightarrow C_6H_5NH_2+H_3O^+[/tex]
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:
[tex]K_a=\frac{[C_6H_5NH_2][H_3O^+]}{[C_6H_5NH_3^+]}[/tex]
Which based on the ICE chart, knowing that Kw=Ka*Kb and Kb is so small, we write:
[tex]Ka=\frac{K_w}{K_b}=\frac{1x10^-14}{1.35x10^{-5}}=7.41x10^{-10}\\\\7.41x10^{-10}=\frac{x*x}{[C_6H_5NH_3^+]_0}[/tex]
Whereas the initial concentration of C₆H₅NH₃⁺ is:
[tex][C_6H_5NH_3^+]_0=\frac{0.16L*0.3403mol/L}{(1.087+0.16)L} =0.0437M[/tex]
Therefore, the concentration of hydronium un solution which equals [tex]x[/tex] is:
[tex][H_3O^+]=x=\sqrt{7.41x10^{-10}*0.0437M}=5.69x10^{-6}M[/tex]
Thus the pH is:
[tex]pH=-log([H_3O^+])=-log(5.67x10^{-6})\\\\pH=5.25[/tex]
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