The pH of 0.726 M anilinium hydrochloride solution in water is 5.36.
C₆H₅NH₃Cl dissociates completely in water according to the following equation.
C₆H₅NH₃Cl(aq) ⇒ C₆H₅NH₃⁺(aq) + Cl⁻(aq)
Cl⁻ is a very weak base so it can't react with water, whereas C₆H₅NH₃⁺ reacts with water according to the following acid equation.
C₆H₅NH₃⁺(aq) + H₂O(l) ⇒ C₆H₅NH₂(aq) + H₃O⁺
Given Kb for C₆H₅NH₂ is 3.83 × 10⁻⁴, we can calculate Ka for C₆H₅NH₃⁺ using the following expression.
[tex]Ka \times Kb = Kw\\\\Ka = \frac{Kw}{Kb} = \frac{1.00 \times 10^{-14} }{3.83 \times 10^{-4} } = 2.61 \times 10^{-11}[/tex]
Given the concentration of C₆H₅NH₃⁺ (Ca) is 0.726 M and the acid dissociation constant (Ka) is 2.61 × 10⁻¹¹, we can calculate [H⁺] using the following expression.
[tex][H^{+} ] = \sqrt{Ka \times Ca } = \sqrt{(2.61 \times 10^{-11} ) \times 0.726 } = 4.35 \times 10^{-6} M[/tex]
The pH of the solution is:
[tex]pH = -log[H^{+} ] = -log (4.35 \times 10^{-6}) = 5.36[/tex]
The pH of 0.726 M anilinium hydrochloride solution in water is 5.36.
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Calculate the pH of 0.726 M anilinium hydrochloride (C₆H₅NH₃Cl) solution in water, given that Kb for aniline is 3.83 × 10⁻⁴.
