Question

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5.

A. Calculate the equilibrium concentration of H3O+ in the solution if the initial concentration of C6H5COOH is 5.1×10⁻²M

B. Calculate the equilibrium concentration of C6H5COO− in the solution if the initial concentration of C6H5COOH is 5.1×10⁻² M .

C. Calculate the equilibrium concentration of C6H5COOH in the solution if the initial concentration of C6H5COOH is 5.1×10⁻² M .

Answer 1

              C6H5COOH (aq) + H2O(l) ---------------> C6H5COO^- (aq) + H3O^+ (aq)

I              5.1*10^-2                                                    0                             0

C             -x                                                                +x                           +x

E            5.1*10^-2 -x                                                  +x                           +x

                   Ka      =    [C6H5COO^-][H3O^+]/[C6H5COOH]

                  6.3*10^-5   = x*x/(5.1*10^-2-x)

                 6.3*10^-5 *(5.1*10^-2 -x)   = x^2

                           x    = 0.00176

A.   [H3O^+]    = x    = 0.00176M

B.   [C6H5COO^-]    = x    = 0.00176M

C.   [C6H5COOH]    = 5.1*10^-2 - x     = 5.1*10^-2 -0.00176   = 0.04924M