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For example, if a question asks for $[\ce{H+}]$ of a $0.100~\mathrm{M}$ solution of picric acid (monoprotic), given that its $K_\mathrm{a}$ at the system temperature is $4.2 \times 10^{-1},$ following my teacher's instructions I would use the ICE table, which looks like this (for a different example).

$$\begin{array}{cccccccc} & \ce{CO (g)} & + & \ce{H2O (g)} & \ce{<=>} & \ce{CO2 (g)} & + & \ce{H2 (g)} \\ \text{Initial} & 0.100~\mathrm{M} & & 0.100~\mathrm{M} & & 0 & & 0 \\ \text{Change} & -x~\mathrm{M} & & -x~\mathrm{M} & & x~\mathrm{M} & & x~\mathrm{M} \\ \text{Equilibrium} & (0.100-x)~\mathrm{M} & & (0.100-x)~\mathrm{M} & & x~\mathrm{M} & & x~\mathrm{M} \\ \end{array}$$

I would then solve for $x$ by using the expression for $K_\mathrm{a}$. However, since $0.100/K_\mathrm{a} < 100$ (lowest initial concentration divided by $K_\mathrm{a}$ is less than 100), $x$ is significant and cannot be omitted from the $0.100-x$. Thus, you need to use the quadratic equation, which yields two solutions for my example above: $x=0.0834, 0.503$. The latter is obviously incorrect because it is greater than the initial concentration of the undissociated compound.

However, my question is: Is it possible that both solutions to a given reaction will be nontrivial, that is, is it possible that both answers appear to work at a glance? (No conflicts with the initial concentration, etc.) This question refers not specifically to the example in question, but rather looks at if it is possible at all (is there a reaction and starting concentration for which both answers would seem correct). In addition, if it is possible, how would one determine which answer is nontrivially impossible?

Whenever I get into a new subject I'm not able to "do" it until I work out some small details that stick with me. It would be best if the answer had some logical explanation, or even a proof as to why the two answers will never be both possible at a glance, or if the answer shows a way to determine without fail which answer is correct. I don't see an obvious reason why both solutions resulting from the quadratic equation will always have one blatantly incorrect result.

Please ask for clarifications if necessary.

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  • $\begingroup$ This has no real bearing on the question itself, but please pay attention to your units. Prior to my edit, the way you earlier defined $x$ in the ICE table, $x$ needed to have units of concentration. It cannot be a pure number because then $0.100~\mathrm{M} - x$ makes no sense. You either define $x$ as a dimensionless quantity, or define $x$ as a concentration and then tack the units along with it. Either way is fine but just be consistent. $\endgroup$ – orthocresol Nov 22 '16 at 11:28
  • $\begingroup$ With more complicated systems, you might end up with a cubic or even a higher order equation. Still, the nature guarantees the existence of a unique physically meaningful solution, no matter what. $\endgroup$ – Ivan Neretin Nov 22 '16 at 14:04
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What you have got here is a strong acid ($\mathrm{p}K_\mathrm{a} \approx 0.38$) that is not diluted, whereof the latter tells you that you can safely ignore the autoprotolysis of water.

The pH approximation to choose is therefor: $$[\ce{H+}] = \frac{[\ce{HA}]_0 K_\mathrm{a}}{[\ce{H+}] + K_\mathrm{a}}$$

or in a quadratic form: $$[\ce{H+}]^2 + [\ce{H+}] K_\mathrm{a} - [\ce{HA}]_0 K_\mathrm{a} = 0$$

which has two solutions: $$[\ce{H+}] = -\frac{K_\mathrm{a}}{2} \pm \sqrt{\left(\frac{K_\mathrm{a}}{2}\right)^2 + [\ce{HA}]_0 K_\mathrm{a}}$$

From those two solutions, only one is chemically meaningful, which is the positive combination which yields one of your solutions ... $[\ce{H+}] = 0.0834$.

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    $\begingroup$ Just to make it crystal clear for OP: the other solution is meaningless because it's negative. $\endgroup$ – orthocresol Nov 22 '16 at 11:20
  • $\begingroup$ The question is asking in general, not specifically referring to the example. I've edited the question to make it more clear. $\endgroup$ – Max Li Nov 23 '16 at 0:30
  • $\begingroup$ It's impossible to have two meaningful results for a pH calculation as a solution simply can't have two pH values. $\endgroup$ – pH13 - Yet another Philipp Nov 23 '16 at 19:19

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