Part A: Calculate the pH of 1.99 M CH 3 CO 2 H. Part B: Calculate the concentrat

Question

Part A: Calculate the pH of 1.99 M CH3CO2H.

Part B: Calculate the concentration of H3O+ in 1.99 M CH3CO2H.

Part C: Calculate the concentration of CH3CO2- in 1.99 M CH3CO2H.

Part D: Calculate the concentration of CH3CO2H in 1.99 M CH3CO2H.

Part E: Calculate the concentration of OH in 1.99 M CH3CO2H.

Part F: Calculate the pH of 0.0224 M CH3CO2H.

Part G: Calculate the concentration of H3O+ in 0.0224 M CH3CO2H.

Part H: Calculate the concentration of CH3CO2- in 0.0224 M CH3CO2H.

Part I: Calculate the concentration of CH3CO2H in 0.0224 M CH3CO2H.

Part J: Calculate the concentration of OH- in 0.0224 M CH3CO2H

Explanation / Answer

Part A:

concentration = 1.99 M

Ka = 1.8 x 10^-5

CH3COOH + H2O ---------------> CH3COO- +   H3O+

1.99                                                  0                  0

1.99 - x                                                x                  x

Ka = x^2 / 1.99 - x

1.8 x 10^-5 = x^2 / 1.99 - x

x = 5.98 x 10^-3

[H3O+] = 5.98 x 10^-3 M

pH = -log (5.98 x 10^-3 )

pH = 2.22

Part B:

[H3O+] = 5.98 x 10^-3 M

Part C:

[CH3COO-] = 5.98 x 10^-3 M

Part D:

[CH3COOH] = 1.98 M

Part E:

[OH-] = 1.67 x 10^-12 M

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