Lactic acid, which builds up in muscle tissue upon strenuous exercise, is partia

Question

Lactic acid, which builds up in muscle tissue upon strenuous exercise, is partially dissociated in aqueous solution:

1. What is the equilibrium constant expression for Kc?

Kc = 1[C3H6O3]

2. What is the value of KC if the extent of dissociation in 0.100 M lactic acid is 3.68 % at 25 C?

Kc = [C3H6O3] Kc = 1[H+][C3H5O3] Kc = [H+][C3H5O3]       Kc = [C3H5O3][C3H6O3] Kc = [C3H6O3][H+][C3H5O3] Kc = [H+][C3H5O3][C3H6O3] Kc = [H+]2[C3H5O3][C3H6O3]

Kc = 1[C3H6O3]

2. What is the value of KC if the extent of dissociation in 0.100 M lactic acid is 3.68 % at 25 C?

Explanation / Answer

LActic acid:

C3H5O3H

In solution:

C3H5O3H <--> H+ and C3H5O3-

The K expression:

K = [H+][C3H5O3-]/[C3H5O3H ]

2)

Find value of Kc

% = 3.68% and M = 0.1

K = [H+][C3H5O3-]

% ion = [H+]/[C3H5O3H] * 100%

[H+] = (%ion)/100% * [C3H5O3H] = 3.68/100 * 0.1 = 0.00368

since 1 H+ is free when C3H5O3- is free

[C3H5O3-] = 0.00368

substitute in Ka

K = [H+][C3H5O3-]/[C3H5O3H ]

K = (0.00368)(0.00368) / (0.1-0.00368) =1.41*10^-4

K = 1.41*10^-4

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