What is the ph of a 0.45 m solution of aniline (c6h5nh2)? (pkb  9.40)?

Answer is: pH of aniline is 9.13.
Chemical reaction: C₆H₅NH₂(aq)+ H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq).

pKb(C₆H₅NH₂) = 9.40.

Kb(C₆H₅NH₂) = 10∧(-9.4) = 4·10⁻¹⁰.

c₀(C₆H₅NH₂) = 0.45 M.

c(C₆H₅NH₃⁺) = c(OH⁻) = x.

c(C₆H₅NH₂) = 0.45 M - x.

Kb = c(C₆H₅NH₃⁺) · c(OH⁻) / c(C₆H₅NH₂).

4·10⁻¹⁰ = x² /  (0.45 M - x). 

Solve quadratic equation: x = c(OH⁻) = 0.0000134 M.

pOH = -log(0.0000134 M.) = 4.87.

pH = 14 - 4.87 = 9.13.

i think it is true since you can test the answer and can use observations to complete it.

the answer is:  

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