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Given is an aqueous solution of mandelic acid (M = 152.2 g/mol; $pK_a$ = 3.4) with a molar concentration of 0.3 mol/ liter. A $\mathrm{pH}$ value of $2$ is measured for the aqueous mandelic acid solution. What percentage of the dissolved mandelic acid is deprotonated?


I know deprotonation means the removal of a $\ce{H+}$ from the acid. After googling , I found the deprotonation percentage can be obtained by dividing the $\ce{H+}$ concentration by the initial acid concentration. Deprotonation percentage shows what percent of $\ce{H+}$ is removed from the initial acid

However, I am unsure about how to proceed because it seems to me there are different ways that give a similar results, is this normal ? For example, we could just do

$$\mathrm{pH}=-\log_{10}([\ce{H+}])\rightarrow10^{-\mathrm{pH}}=10^{-2}=[\ce{H+}] \rightarrow[\ce{H+}]/0.3= 0.0333 = 3.33 \% $$

or we could do

$$K_\mathrm a=10^{-\mathrm pK_\mathrm a}=10^{-3.4}$$

$$K_\mathrm a=[\ce{H+}][\ce{A-}]/[\ce{HA}]\rightarrow K_\mathrm a=x^2/0.3\rightarrow\sqrt{0.3\times10^{-3.4}}=x=[\ce{H+}]$$

$$[\ce{H+}]/0.3 = 0.0364 = 3.64 \%$$

Are both ways correct ? Or is there another way to proceed ? I googled quite a bit but was unable to find a clear explanation

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    $\begingroup$ @wengen . The following information is absurd :$\ce{10^{−2}=[H+] → [H+]/0.3=3.33 }$ $\endgroup$
    – Maurice
    Dec 17, 2022 at 9:55
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    $\begingroup$ Is there any difference between "deprotonation percentage" and "dissociation degree $\ce{\alpha}$" ? $\endgroup$
    – Maurice
    Dec 17, 2022 at 16:22
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    $\begingroup$ @Maurice IMHO, the difference is just the proportionality constant 100 and the wording. I am not sure if I have ever seen the former until now. // Generally, if one searches A by searching its rarely used/unused synonym B, one may not find A. $\endgroup$
    – Poutnik
    Dec 17, 2022 at 18:19
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    $\begingroup$ pH is an extra info there. You can get the result directly from the provided pH, or you can calculate pH from the other data. Note that both values may differ. $\endgroup$
    – Poutnik
    Dec 17, 2022 at 19:23
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    $\begingroup$ $$K_\mathrm a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{(c \alpha)(c \alpha)}{c(1-\alpha)} = c \frac{\alpha^2 }{ 1-\alpha} $$ $\endgroup$
    – Poutnik
    Dec 17, 2022 at 22:15

1 Answer 1

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As @Poutnik stated, one of the values you provided is extra. Let's try to make everything tidier.

Complex protonated entities aside, what chemical species are we really dealing with? $\ce{[HA], [A^-], [H_3O+]}$. The concentration of $\ce{H2O}$ shall remain constant. Right off the bat, we are dealing with 3 unknowns: the concentrations of the aforementioned species. Wait! - you protest - I know $\ce{[H3O+]}$, it's given in the problem as the $\mathrm{pH}$ value! Let's try to ignore that for the time being. Let's pretend we were given only the initial concentration (I shall denote it as $\ce{C^0_{HA}}$ where $\ce{HA}$ and $\ce{A-}$ are, of course, the acid and its conjugate base) and the $\mathrm{pK_a}$. Would one still be able to calculate this mystifying deprotonation percentage without being force-fed a $\mathrm{pH}$ value?

At any point, we could define the deprotonation percentage intuitively as: $$p = 100\cdot \frac{\ce{[A-]}}{\ce{[HA]}+\ce{[A-]}} = \frac{\ce{[A-]}}{\ce{C^0_{HA}}}$$ This is not a formula worth memorizing, it could be readily derived intuitively. A deprotonation percentage entails determining... well, what percentage of your initial acid exists as the deprotonated form ($\ce{[A-]}$). When it comes to any problem of equilibrium, especially in the realm of simple proton-transfer equilibria, one should write down balances: mass balances, charge balances and the expression of the acidity constant(s). $$ \ce{C^0_{HA} = [HA] + [A-]} \tag{1 - mass balance} $$ $$ \ce{[H3O+] = [HA-]} \tag{2 - charge balance} $$ $$ \mathrm{K_a} = \frac{\ce{[A-]\cdot [H3O+]}}{\ce{[HA]}} \tag{3 - the acidity constant} $$

3 unknowns, 3 equations! This should be more than possible to solve! But then, why are we given the $\mathrm{pH}$ supplementarily? If our system of equations is totally solvable, we should be able to calculate the $\mathrm{pH}$ on our own and, should it be different from the one in the problem statement, come up with an explanation, right?

Simple algebraic manipulation from the 3 equations alone leads to: $$ \ce{ C^0_{HA} = [HA] + [A-] = [A-] \left(1 + \frac{[H3O+]}{\mathrm{K_a}} \right) \Rightarrow [A-] = \frac{C^0_{HA}}{1+\frac{[H3O+]}{\mathrm{K_a}} } } $$

$$ \ce{ [H3O+] = [A-] \Rightarrow [H3O+] = \frac{ C^0_{HA} }{ 1+\frac{ [H3O+] }{ \mathrm{K_a} } } \Rightarrow C^0_{HA} = \frac{ [H3O+]^2 }{ \mathrm{K_a} } + [H3O+]} \Leftrightarrow $$ $$\ce{ \Leftrightarrow [H3O+]^2 + [H3O+]\cdot \mathrm{K_a} - C^0_{HA}\mathrm{K_a} = 0} $$ Upon solving this 2nd-degree polynomial, we arrive at a $\mathrm{pH}$ of $\sim \textbf{1.97}$. Not that far away from 2! But not exactly 2 either. By plugging in this value in the identities above we arrive at:

$$ \ce{ [H3O+] = [A-] = 1.073 \cdot 10^{-2} \mathrm{M} \Rightarrow p \approx \textbf{3.58}\% \\ [HA] = 2.893 \cdot 10^{-1} \mathrm{M} } $$

But this $\mathrm{pH}$ is still not 2. We could, for sure, accept the kindness of the authors of the problem of providing us with an approximation for the $\mathrm{pH}$, and then rejoice in our victory. Or, we could accept the $\mathrm{pH}$ value from the problem statement as a buffered one. On many occasions, analytical chemistry makes use of buffer solutions: mixtures of weak acid/conjugate base couples meant to keep the $\mathrm{pH}$ steady. In this sense, there is absolutely nothing stopping us from attaining a $\mathrm{pH}$ of absolute 2, with the assumptions made above still in place.

However, expect something to change! Well, the acidity constant won't change: it's constant! The mass balance won't change: after all, all of your mandelic/mandelate species come from the initial small quantity of mandelic acid we put in - $\ce{C^0_{HA}}$ - there's no mandelate or acid popping up from the ether. It's the charge balance that changes, as adding a buffer will add a new charged species into the equation (in this case, we want to make the $\mathrm{pH}$ more basic, from 1.97 to 2, we could add small quantities of $\ce{NH3/NH4Cl}$ which will add $\ce{NH4+}$ into the balance).

If you want your answer to be precise, as long as your $\mathrm{pH}$ is different from the 1.97 value we calculated earlier, you can no longer assume that pristine charge balance: your balance will be spoilt by the addition of whatever changes it from 1.97 to 2. Since a buffer is meant to keep the $\mathrm{pH}$ steady, we could determine $\ce{[A-]}$ by using the exact same equation as above:

$$ \ce{ [A-] = \frac{C^0_{HA}}{ 1+\frac{ [H3O+] }{ \mathrm{K_a} } } } = \frac{0.3}{ 1+\frac{ 10^{-2} }{ 10^{-3.4 } } } = 1.149 \cdot 10^{-2}\ \mathrm{M} $$

Finally, $$ p = 100 \cdot \ce{ \frac{ [A-] }{ C^0_{HA} }} = 3.83\% $$

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    $\begingroup$ Wouah, it makes much more sense after reading this. Thanks a lot for the thorough explanation ! $\endgroup$
    – wengen
    Dec 18, 2022 at 0:15

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