Question

An aqueous mixture of phenol and ammonia has initial concentrations of 0.200 M C6H5OH(aq) and 0.120 M NH3(aq). At equilibrium, the C6H5O–(aq) concentration is 0.050 M. Calculate the equilibrium constant, K, for the reaction below.
C6H5OH(aq) + NH3(aq) C6H5O– + NH4+(aq)

Answer 1

Answer:

The equilibrium constant, K, for the reaction = 0.23

Explanation:

Given that

               initial concentrations of phenol(C₆H₅OH) = 0.2 Molar

                                  & concentration of ammonia = 0.12 Molar

At equilibrium the phenoxide ion concentration (C₆H₅O⁻) = 0.05 Molar

                        C₆H₅OH(aq) + NH₃(aq)    ⇄     C₆H₅O⁻(aq) + NH₄⁺(aq)

Initial conc.         0.2               0.12                        0                    0

Equilibrium  (0.2 - 0.05)      (0.12 - 0.05)           0.05              0.05

                     = 0.15               = 0.07  

                According to law of mass action

                           $K_{c} =\frac{[C_{6}H_{5}O^{-} ][NH_{4}^{+} ]}{[C_{6}H_{5}OH ][NH_{3} ]}$

                                 $=\frac{0.05X0.05}{0.15X0.07}$

                                 $= 0.23$