Question

the acid-dissociation constant for benzoic acid is 6.3*10^-5. calculate the equilibrium constant concentration of h30+, C6H5COO-, and C6h5COOH in the solution if the initial concentration of C6H5COOH is 0.050 M

Answer 1

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

Initial                  0.05                       -                  0                0

change                  -x                         -                      +x                    +x

equilibrium        0.05 - x                      -                       x                     x

The acid dissociation constant, K_a is given by:

Assuming x to be very small compared to 0.05, we get:

6.3 x 10^-5 = x²/0.05

solving for x, we get:

x = 0.00177 M

Therefore equilibrium concentrations are:

[H₃O⁺] = x = 0.00177 M

[C₆H₅COO^-] = x = 0.00177 M

[C₆H₅COOH] = 0.05 - x = 0.05 - 0.00177 = 0.0482 M