## Chemistry: The Molecular Science (5th Edition)

This student made the assumption that "x" is much smaller than the acid concentration (which isn't in this case), and he didn't considered the $[H_3O^+]$ produced by pure water.
1. When the student was calculating the "x" in the $K_a$ expression, he made the assumption that it is much less than $[CH_3COOH]$, which isn't true in this case. -$[H_3O^+] = [CH_3COO^-] = x$ -$[CH_3COOH] = [CH_3COOH]_{initial} - x = 1 \times 10^{- 7} - x$ For approximation, the student considered: $[CH_3COOH] = 1 \times 10^{- 7}M$ $Ka = \frac{[H_3O^+][CH_3COO^-]}{ [CH_3COOH]}$ $Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 1\times 10^{- 7}}$ $Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 1\times 10^{- 7}}$ $1.8 \times 10^{- 12} = x^2$ $x = 1.342 \times 10^{- 6}M$ As we can see, "x" is greater than $[CH_3COOH]$, so, this is a incorrect approximation. 2. Pure water has $[H_3O^+] = 1.0 \times 10^{-7}M$, so, when we have a very small acid concentration, we should consider it in the expression: Equilibrium: $[CH_3COO^-] = x$ $[H_3O^+] \ne x$ Normally, we consider $[H_3O^+] \approx x$, but, in this case, its concentration is affected by the autoionization of water, so, this approximation is incorrect.