Calculating molecular weight of an acid based on given values of mass, volume, pH and pKa
Ok here's what you're doing wrong.

When you write $$C = \frac{n}{V}$$

$C$ is meant to be the concentration of the monoprotic acid dissolved initially, not at equilibrium. Plugging in the equilibrium concentration is disasterous, as it is bound to be different from the actual ratio of amount of substance in moles of $\ce{HA}$ to the volume of solution in liters.

This approach will better suit the problem:

$$\ce{HA \ &lt;=&gt; \ H+ \ + \ A-}$$
$[\ce{HA}]_{\mathrm{initial}}=\frac{m}{M}.\frac{1}{V}$

Assuming $x$ moles/liter to be consumed at equilibrium,

$[\ce{HA}]_{\mathrm{eq}}=\frac{m}{M}.\frac{1}{V}-x$

Now $x$ moles/liter each of $\ce{H+}$ and $\ce{A-}$ would have appeared.

$[\ce{A-}]_{\mathrm{eq}}=[\ce{H+}]_{eq}=x$

$x$ happens to be $10^{-5}$, as $\mathrm{pH}$ is given to be 5.

$K_a$ would be $10^{-6}$, as you have already mentioned.

$$K_a = \ce{ \frac{[H^+]_{eq}[A^-]_{eq}}{[HA]_{eq}}=10^{-6}}$$

Plugging in, I get

$$10^{-6}=\frac{10^{-5} \times 10^{-5}}{\frac{0.011}{M \times 0.5}-10^{-5}}$$

Simplifying a little, I get

$$M=\pu{200 g/mol}$$