Answer
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.
Work Step by Step
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.
