Question

Calculate the concentration of all species in a 0.230 M M C6H5NH3Cl C 6 H 5 N H 3 C l solution. Enter your answers numerically separated by commas. Express your answer using two significant figures.

[C6H5NH3+], [Cl−], [C6H5NH2], [H3O+], [OH−]

Answer 1

C₆H₅NH₃Cl (aq) + H₂O (l) C₆H₅NH₂ (aq) + Cl^- (aq) + H₃O⁺ (aq)

So, [H₃O⁺] = [Cl^-] = [C₆H₅NH₂]

C₆H₅NH₃Cl is salt of weak base (C₆H₅NH₂) and strong acid (HCl).

Pkb of aniline = 9.4

pH of salt solution of weak base strong acid is

pH = 7 - ( pKb - logC_salt)

Given concentration of salt (C_salt) = 0.230 M.

Then,

pH = 7 - {9.4 - log(0.230)}

= 7 - ( 9.4 + 0.64)

= 7 - 5.0

= 2.0

Then, [H₃O⁺] = 10^-2.0 = 0.01 M

[OH^-] = Kw/[H₃O⁺]

[OH^-] = (10^-14/ 0.01) = 1.0 × 10^-12 M

[H₃O⁺] = [Cl^-] = [C₆H₅NH₂] = 0.01 M.

[C₆H₅NH₃⁺] = 0.23 M