Greetings dear chemists!

I got a nice exam problem on wich i am stuck for hours, well maybe the problem is with me. I know the problem can be solved. So here we go.

We got a weak monoprotic acid, we don't know its concentration, dissociation degree nor the dissociation constant. If we dilute it 3 times its volume, then the dissociated ions grow with an unknown constant, lets call it $y$. If we dilute the diluted solution again by 5 times its volume, then the dissociated ions grow 4 times. Whats the dissociation degree of the undiluted solution?

Yeah woah that looks tough. So lets give it my try.

Lets call the concentrations in order of dilution by $c_0, c_1, c_2$ and the dissociation degree by $\alpha_0,\alpha_1,\alpha_2$, and the dissociation constant of course $K_s$.

I managed to get three equations for this problem, but i feel i missed something or messed up my algebra someway i dont understand.

$$c_0=?,\alpha_0=?,K_s=?$$ $$\alpha_0=\frac{[A^-]}{c_0}$$ $$HA=H^++A^-$$ $$c_1=\frac{c_0}{3},\qquad c_2=\frac{c_1}{5} \to c_2=\frac{c_0}{15}$$ $$\alpha_1=y\alpha_0,\qquad \alpha_2=4\alpha_1 \to \alpha_2=4y\alpha_0$$ We call forth Ostwald; $$K_s=\frac{c_0\alpha_0^2}{1-\alpha_0}\qquad \qquad \qquad \qquad (1)$$ $$K_s=\frac{c_1\alpha_1^2}{1-\alpha_1}=\frac{c_0y^2\alpha_0^2}{3(1-y\alpha_0)}\qquad \qquad(2)$$ $$K_s=\frac{c_2\alpha_2^2}{1-\alpha_2}=\frac{c_016y^2\alpha_0^2}{15(1-4y\alpha_0)}\qquad \qquad(3)$$ Equating the first and the second equation, i could get an expression for $y$. By setting $\beta=\frac{3}{1-\alpha_0}$ and ignoring the negative root $$y^2+y\beta\alpha_0-\beta=0$$ $$y=\frac{-\beta\alpha_0+\sqrt{\beta^2\alpha^2+4\beta}}{2}$$ My problem starts when i equate the second and the third equation; $$\frac{c_0y^2\alpha_0^2}{3(1-y\alpha_0)}=\frac{c_016y^2\alpha_0^2}{15(1-4y\alpha_0)}$$ $$5-20y\alpha_0=16-16y\alpha_0$$ $$11+4y\alpha_0=0$$ From wich i clearly see that im on the wrong track, since $\alpha_0$ and $y$ cant be negative, please help me on solving this nightmare. Maybe i missed something obvious, i dont really know that im on the right method to solve this either.

  • $\begingroup$ Is the original problem in English? Can we get it exactly as stated? $\endgroup$ – MaxW Mar 28 '19 at 17:39
  • $\begingroup$ No sorry its in hungarian, these kind of problems pop out from my professors head, i wrote it from my memory, we talked about it a lot after the exam so im pretty sure its correct. $\endgroup$ – Andrew Kovács Mar 28 '19 at 17:41
  • $\begingroup$ If you do ask the prof, and he clarifies the problem, I'd sure like to know what he proposed as the answer. $\endgroup$ – MaxW Mar 29 '19 at 17:33

You are on the right track according to the directions given to you ( = the problem statement), but the directions themselves are wrong.

Say you had an equilibrium with $\ce{{[H^+][A^-]\over[HA]}=K}$. You dilute it five-fold and tell your particles to wait a bit with dissociation until you are done with the math. That makes all concentrations 5 times smaller, and consequently, the LHS is now $1\over5$ of what it used to be. Now you tell them to increase their dissociation by a factor of 4. In doing so, each of $\ce{[H+]}$ and $\ce{[A-]}$ increases four-fold, thus growing the LHS to $16\over5$ of its initial value, or more than 3 times greater than $K$, and they do that at the expense of the remaining $\ce{[HA]}$, decreasing the denominator and increasing the LHS further yet. Just how is it going to be the same $\ce{K}$ again?

I suppose it is some subtlety of wording that we are misinterpreting here. It might have been lost in translation, but can't we at least make a reasonable guess? Could it be that the 4-fold increase is compared to the initial state, that is, to $\alpha_0$ and not to $\alpha_1$? No, that would not suffice. Still, there might be other possibilities...

  • $\begingroup$ Thank you for your answer and time reading my post, your reasoning seems correct, obviously i misunderstood something or it might be but its more likely not that the problem itself i was doing is wrong. Feel free to delete my post, but i think your thoughts are a nice example for learning so i'll leave it. $\endgroup$ – Andrew Kovács Mar 28 '19 at 17:54
  • $\begingroup$ @IvanNeretin - The degree of dissociation is relative not absolute. So for the diluted acid a greater fraction ionizes, but the concentrations are all smaller. $\endgroup$ – MaxW Mar 28 '19 at 17:58
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    $\begingroup$ @MaxW They are all smaller in my answer. [H+] and [A-] are 4/5 of what they were before the dilution, and [HA] is less than 1/5 of its former value. $\endgroup$ – Ivan Neretin Mar 28 '19 at 19:00

$K_a = \dfrac{\ce{[H^+][A^-]}}{\ce{[HA]}}$

Let's assume a weak acid so that we'll assume a negligible fraction ionizes. So for all concentrations we get

$\ce{[H^+] = [A^-]} = \sqrt{(K_a)\ce{([HA])}}$

and the dissociation degree (fraction) is:

$\dfrac{\ce{[A^-]}}{\ce{[HA]}} = \sqrt{\dfrac{K_a}{\ce{[HA]}}}$

So if the concentration of HA is reduced by a factor of 5, then the dissociation degree should only be $\sqrt{5} \approx 2.24$ greater, not 4 times greater.

Now to get a greater degree of ionization we could have a very dilute solution of acid, and get into a situation where the autodissociation of water is important.

  • $\begingroup$ Thank you for all your efforts, nice answer though i have tried it too with this shortcut but didnt gave me results i wanted, I'll look into it tomorrow, i didnt want to bother my prof with my stupid questions since he is the dean, but it looks like i have no other choice. $\endgroup$ – Andrew Kovács Mar 28 '19 at 20:40
  • $\begingroup$ @AndrewKovács - As I said, the only way that I can think of to have a five-fold dilution get a four fold increase in ionization is to have the ionization of water come into play. $\endgroup$ – MaxW Mar 28 '19 at 20:44

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