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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?

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  • $\begingroup$ I find your lack of faith disturbing.. 200 it is. $\endgroup$ – William R. Ebenezer May 14 at 18:34
  • $\begingroup$ Sorry, it was meant to say 220. I re-edited the post, it now says 220 g/mole which is the result I get when I plug in the values. 200, however is the correct answer that I was given. Could my math simply be wrong? I am literally plugging in those number into ($\alpha$). $\endgroup$ – Pettersson May 14 at 18:39
  • $\begingroup$ You don't have enough significant figures to say 220. Half a liter is a very coarse measurement for volume. $\endgroup$ – Zhe May 14 at 18:58
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    $\begingroup$ @Zhe I think you have to be careful with significant figures in these toy problems. If I write $\frac{1}{2}$ in a formula, is that one significant figure? If the pKa is given as 6 and the pH as 5, does that mean we have less than a significant figure (maybe just answer in terms of order of magnitude, i.e. molar mass is on the order of hundreds to thousands of g/mol)? $\endgroup$ – Karsten Theis May 14 at 22:05
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    $\begingroup$ @KarstenTheis Yes, pH 6 and 5 have less than 1 significant figure. I don't take significant figures as gospel (since you should better handle error with error bars), but it is frequently a cause of confusion in such problems. This is simply something I noticed while looking superficially at the statement of the problem and the proposed answer. Hence, I added a comment instead of writing an answer. $\endgroup$ – Zhe May 14 at 22:10
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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:

$$\ce{HA \ <=> \ H+ \ + \ A-}$$ $[\ce{HA}]_{\mathrm{initial}}=\frac{m}{M}.\frac{1}{V}$

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

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

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