When acetic acid to dissolved in wafer, it reacts to form hydronium and acetate

Question

When acetic acid to dissolved in wafer, it reacts to form hydronium and acetate anions. This reaction will reach equilibrium, which favors heavily the loft side of this equation. a) What can be said about the ratio of acetic acid to acetate at equilibrium? b) If you were to add NaOH (sodium hydroxide) to this solution, how would this affect the equilibrium and the proportion of acetic acid and acetate? c) If the concentration of acetic acid at equilibrium is determined to be 1.0 M and the concentration of acetate and hydronium is 0.010 M, what is the equilibrium constant at 21 degreeC. (The concentration of water is ignored in this calculation!)

Explanation / Answer

a) ratio of acetate to acetic acid decreases due to decrease in concentration of acetate.

b) CH3COOH + H2O <---------------------> CH3COO- + H3O+

NaOH --------------------------> Na+ + OH-

on addition of NaOH to the equilibirum OH- react with H3O+ from the products side forms acetate ion more . that means equilibrium shifts to forward direction (right side)

c)

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

Ka = [CH3COO-][H3O+]/[CH3COOH]

Ka = 0.01 x 0.01 / 1.0

Ka = 10^-4

Ka = 1.0 x 10^-4

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