What is the equilibrium conversion for a reversible reaction in a batch reactor?

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The discussion revolves around calculating the equilibrium conversion for a reversible reaction involving acetic acid and water. The initial concentration of CH3COOH is 0.01 mol/L, with a constant H3O+ concentration of 0.001 mol/L and an equilibrium constant of 4.7 x 10^-4. The user ultimately determines that the equilibrium conversion is 32%. They also suggest that similar problems should be posted in the homework forum for better assistance. The conversation highlights the importance of understanding equilibrium constants in reversible reactions.
cruckshank
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I am struggling with the following problem:

A reversible reaction where: CH3COOH + H2O--><--CH3COO- + H3O+
The overall solution initially has a CH3COOH concentration of 0.01mol/L and no acetate. The solution has 0.001mol/L of H3O+. Assume that this is constant throughout the experiment. K(equilibrium constant)=4.7*10^-4.

Show that the equilibrium conversion is 32%.
 
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Never mind, finally figured it out.
 
Next time, I suggest you place it in the homework forum
 
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