# How to find the deprotonation percentage of an acid?

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

• @wengen . The following information is absurd :$\ce{10^{−2}=[H+] → [H+]/0.3=3.33 }$ Dec 17, 2022 at 9:55
• Is there any difference between "deprotonation percentage" and "dissociation degree $\ce{\alpha}$" ? Dec 17, 2022 at 16:22
• @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. Dec 17, 2022 at 18:19
• 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. Dec 17, 2022 at 19:23
• $$K_\mathrm a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{(c \alpha)(c \alpha)}{c(1-\alpha)} = c \frac{\alpha^2 }{ 1-\alpha}$$ Dec 17, 2022 at 22:15

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

• Wouah, it makes much more sense after reading this. Thanks a lot for the thorough explanation ! Dec 18, 2022 at 0:15