a solution of acetic acid on a labratory shelf was ofundetermined concentration. if the PH...

a solution of acetic acid on a labratory shelf was of undetermined concentration. if the PH of the solution was found to be 2.57 what was the concentration of the acetic acid. The Ka of the Acidic acid is 1.7x10^-5

Expert Solution

Given the pH of acetic acid=2.57

pH= -log[H3O⁺]=2.57

[H3O⁺]=10^-2.57=2.69 x 10^-3 M

Ka=1.7 x 10^-5 for acetic acid.

Since acetic acid is weak acid then the equilibrium reaction is

CH3COOH + H2O <------> H3O⁺ + CH3COO^-

Ka=[H3O⁺][CH3COO^-]/[CH3COOH]

In this equilibrium, [H3O⁺]=[CH3COO^-]

Therefore Ka=[H3O⁺]²/[CH3COOH]

1.7 x 10^-5=(2.69 x 10^-3 )²/[CH3COOH]

[CH3COOH]=0.426 M.

queen_honey_blossom answered 2 years ago

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