Calculating molecular weight of an acid based on given values of mass, volume, pH and pKa This is what I thought was an easy problem but according to a given answer I am doing it wrong:


  $\pu{11 mg}$ of a monoprotic acid was dissolved in $\pu{0.5 l}$ of water. The resulting $\mathrm{pH}$ of said solution was measured to be $5$. What is the molecular weight of said monoprotic acid? $\mathrm{p}K_\mathrm{a} = 6.$ $(t = \pu{25°C})$


My solution

Given values are:

$$m = \pu{0.011 g} \qquad V = \pu{0.5 l} \qquad \mathrm{pH} = 5 \qquad \mathrm{p}K_\mathrm{a} = 6$$

Formula used to calculate molecular weight:

$$M = \frac{m}{n}$$ 

Formula used to calculate concentration of acid in the solution, that is, the (hopefully correct) formula for calculating the acid equilibrium constant: 

$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$

Formula used to calculate molar concentration:

$$C = \frac{n}{V}$$

Putting them together:

$$M = \frac{m}{\frac{[\ce{H+}][\ce{A-}]}{K_\mathrm{a}}\cdot V} \label{eqn:alpha}\tag{α}$$

Concentration of $\ce{H+}$ and $\ce{A-}$ based on $\mathrm{pH}$ value:

$$[\ce{H+}] = [\ce{A-}] = 10^{-5}$$

Calculating the acid equilibrium constant from $\mathrm{p}K_\mathrm{a}$:

$$K_\mathrm{a} = 10^{-6}$$

Result:

Entering the above vaules for $m,$ $V,$ $K_\mathrm{a},$ $[\ce{H+}]$ and $[\ce{A-}]$ into equation \eqref{eqn:alpha} yields the molecular weight of $\pu{220 g mol-1}$. This is answer is wrong, supposedly. Did I make any mistakes?
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

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