Question

Determine the equilibrium concentrations for all species and the pH of 0.310 M CH3CH(OH)COOH (lactic acid).

Options:
Option 1: [H+]: 4.9 x 10^-3 M; [CH3CH(OH)COOH]: 0.305 M; [CH3CH(OH)COO-]: 0.005 M; pH: 2.31
Option 2: [H+]: 3.9 x 10^-3 M; [CH3CH(OH)COOH]: 0.210 M; [CH3CH(OH)COO-]: 0.100 M; pH: 2.41
Option 3: [H+]: 5.9 x 10^-3 M; [CH3CH(OH)COOH]: 0.105 M; [CH3CH(OH)COO-]: 0.205 M; pH: 2.21
Option 4: [H+]: 2.9 x 10^-3 M; [CH3CH(OH)COOH]: 0.010 M; [CH3CH(OH)COO-]: 0.300 M; pH: 2.51

Answer 1

The equilibrium concentrations for the species in the 0.310 M CH3CH(OH)COOH (lactic acid) solution are [H+]: 4.9 x 10^-3 M, [CH3CH(OH)COOH]: 0.305 M, and [CH3CH(OH)COO-]: 0.005 M. The pH of the solution is 2.31.

Step-by-step explanation:

The equilibrium concentrations for the species in the 0.310 M CH3CH(OH)COOH (lactic acid) solution can be determined using the ionization constant (Ka).

Using an ICE table, we can calculate the equilibrium concentrations as follows:

• [H+]: 4.9 x 10^-3 M
• [CH3CH(OH)COOH]: 0.305 M
• [CH3CH(OH)COO-]: 0.005 M

The pH of the solution can be calculated using the formula pH = -log[H+]. Therefore, the pH of the solution is 2.31.