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

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The equilibrium conversion for the reversible reaction CH3COOH + H2O <--> CH3COO- + H3O+ is calculated to be 32%. The initial concentration of acetic acid (CH3COOH) is 0.01 mol/L, and the concentration of hydronium ions (H3O+) is maintained at 0.001 mol/L throughout the experiment. The equilibrium constant (K) for this reaction is 4.7 x 10^-4, which is essential for determining the conversion at equilibrium.

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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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