Answer to Question #303664 in General Chemistry for Jacey

Question

Calculate the concentration of C6H5NH2, H3O+, and OH− in a 0.230 M C6H5NH3Cl solution. (Kb(C6H5NH2)=3.9×10^−10

Answer 1

K_a × k_b = 10^-14

k_a = 10^-14 ÷ 3.9×10^-10

=2.564×10^-5

Let [C₆H₅NH₂] be x

[H₃O⁺] is also x and

[C₆H₅NH₃⁺] is (0.230-x)

K_a = ( [C₆H₅NH₂] [H₃O⁺] ÷ [C6H5NH3+] )

2.564×10^-5 = x² ÷ (0.230-x)

Solving for x quadratically,

X= 2.4251×10^-3

Therefore, [H₃O⁺] = 2.4251×10^-3 M

[C₆H₅NH₂] = 2.4251×10^-3M

[C₆H₅NH₃⁺] = 0.230 - 2.4251×10^-3

= 0.2275M