Question

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.

Answer 1

`pH=5.25`

Step-by-step 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:

`V_(acid)=(V_(base)M_(base))/(M_(acid)) =(160mL*0.3403M)/(0.0501M)=1087mL`

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:

`H_2O+C_6H_5NH_3^+\rightarrow C_6H_5NH_2+H_3O^+`

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:

`K_a=([C_6H_5NH_2][H_3O^+])/([C_6H_5NH_3^+])`

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

`Ka=(K_w)/(K_b)=(1x10^-14)/(1.35x10^(-5))=7.41x10^(-10)\\\\7.41x10^(-10)=(x*x)/([C_6H_5NH_3^+]_0)`

Whereas the initial concentration of C₆H₅NH₃⁺ is:

`[C_6H_5NH_3^+]_0=(0.16L*0.3403mol/L)/((1.087+0.16)L) =0.0437M`

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

`[H_3O^+]=x=\sqrt{7.41x10^(-10)*0.0437M}=5.69x10^(-6)M`

Thus the pH is:

`pH=-log([H_3O^+])=-log(5.67x10^(-6))\\\\pH=5.25`

Best regards!