myvow
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For this question,why Ka =(ch3coo-)(h+)/(ch3cooh) instead of Ka =(ch3coo +x)(x)/(ch3cooh-x) ?
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The discussion centers on the equilibrium expression for the dissociation constant (Ka) of acetic acid (CH3COOH) and its conjugate base acetate (CH3COO-). The correct formula for Ka is given as Ka = (CH3COO-)(H+)/(CH3COOH), which utilizes equilibrium concentrations. This contrasts with an incorrect approach that modifies initial concentrations with an extent of dissociation variable (x). The confusion arises from the representation of sodium acetate (CH3COONa) in solution and its interaction with hydrochloric acid (HCl).
PREREQUISITESChemistry students, educators, and professionals involved in acid-base chemistry, particularly those focusing on buffer solutions and equilibrium calculations.
myvow said:why Ka =(ch3coo-)(h+)/(ch3cooh) instead of Ka =(ch3coo +x)(x)/(ch3cooh-x)
myvow said:For this question,why Ka =(ch3coo-)(h+)/(ch3cooh) instead of Ka =(ch3coo +x)(x)/(ch3cooh-x) ?
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