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The reported solubility product of $\ce{HgS}$ is $4 \times 10^{-54}$. How is this possible? The minimum concentration that is possible is 1/avogadro's constant which shall be about $10^{-23}\ \mathrm{M}$. Hence the minimum value of solubility product for a diatomic specie like $\ce{HgS}$ would be about $10^{-46}$. What am I missing?

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    $\begingroup$ True, homeopathic concentrations like that can't be measured directly. I guess the solubility product is calculated from thermodynamic data. $\endgroup$ – Ivan Neretin Nov 14 '16 at 11:23
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    $\begingroup$ Can you provide a source for the "minimum concentration that is possible" being $1/N_A$? I was unaware of such a limitation. $\endgroup$ – Ben Norris Nov 14 '16 at 12:42
  • $\begingroup$ It's just an implication. The minimum amount of a specie you can have in any solution is one molecule of it. Hence the inverse of Avogadro's number. $\endgroup$ – Swaroop Chandra Nov 14 '16 at 14:44
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    $\begingroup$ Can't you have ten times the Avogadro's number of water per one dissolved molecule? $\endgroup$ – Ivan Neretin Nov 14 '16 at 14:52
  • $\begingroup$ Please read my comment to the answer below. $\endgroup$ – Swaroop Chandra Nov 14 '16 at 14:53
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Imagine you have one googolplex ($10^{10^{100}}$) liters of water. Now let's dissolve one mole of sugar in your water. Setting aside where would you find so much water, the concentration would be way lower than the inverse of Avogadro's constant.

Long story short, there's no lower limit to the concentration of a substance. You can always add more solvent to the mixture thus making the concentration smaller.

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  • $\begingroup$ Aren't all the results and laws that we apply, statistical i.e., a collective picture and hence even if you keep adding the solvent there is only one molecule in there and hence won't the statistical compilation be pointless. $\endgroup$ – Swaroop Chandra Nov 14 '16 at 14:48
  • $\begingroup$ you say the concentration does not have a lower limit but isn't it a little illogical to consider one molecule to be distributed throughout the solvent uniformly. $\endgroup$ – Swaroop Chandra Nov 14 '16 at 14:52
  • $\begingroup$ Swaroop - it isn't necessarily illogical. It depends what you are trying to describe about the system. And no one (except you) said "one molecule". Variax said "one mole". $\endgroup$ – Curt F. Nov 14 '16 at 14:57
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    $\begingroup$ I still think my reasoning is pretty solid. Dissolve one molecule of substance in one liter of water and you get a molarity of 1/NA . Add an extra liter of water and you halve that concentration. Granted, a concentration so enormously low may not be evenly distributed throughout the whole volume of the solution, but that's not the point at all. We're describing a macroscopic system, not its microscopic state. $\endgroup$ – Variax Nov 14 '16 at 15:32
  • $\begingroup$ The microscopic states statistically give the macroscopic system . Here the microscopic states being are so limited that taking a statistical conclusion is pointless. $\endgroup$ – Swaroop Chandra Nov 14 '16 at 17:08
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Variax's answer addresses the minimum concentration misconception. Perhaps you are also wondering how we could conceivably measure concentrations that small.

If:

$$\ce{HgS(s) <=> Hg^2+(aq) + S^2-(aq)}\ \ \ \ K_{sp}=4\times 10^{-54}$$

then, in a saturated solution prepared by adding solid $\ce{HgS}$ to ultrapure deionized water:

$$[\ce{Hg^2+}]=[\ce{S^2-}]=\sqrt{4\times 10^{-54}}=2\times 10^{-27}$$

As an aside, actually $[\ce{S^2-}]\approx 0$ because $\ce{S^2-}$ is a strong base with a very large $K_B$. So in fact, the reaction is

$$\ce{HgS(s) +H2O(\ell) <=> Hg^2+ (aq) + SH^- (aq) + OH-(aq)}\ \ \ \ K_{sp}=4\times 10^{-54}$$

However, let's stick with the simpler case. If we can measure $[\ce{Hg^2+}]$ or $[\ce{S^2-}]$, we can compute the value of $K_{sp}$. According to this pdf presentation from Perkin Elmer, their IPC-MS system can detect mercury at 0.016 ppb and sulfur at 28 ppb ppb limit of quantitation will be a little higher). 0.016 ppb is approximately $0.016\ \mathrm{\mu g \ per\ L}$:

$$ 0.016\ \mathrm{\mu g/L} = 1.6\times 10^{-8}\ \mathrm{g/L}\\ 1.6\times 10^{-8}\ \mathrm{g/L\times \dfrac{mol\ Hg}{200.59\ g\ Hg} = 7.98\times 10^{-11}\ M} $$

Clearly, this instrument is not sensitive enough.

One way the might work is by using a concentration cell. Because of Le Chatelier's principle, an electrochemical cell containing the same ions at different concentrations will produce a voltage. The larger the concentration difference, the larger the potential difference:

$$E_{cell}= E^\circ -\dfrac{RT}{nF}\ln{\dfrac{[\ce{Hg^2+}]_{dilute}}{[\ce{Hg^2+}]_{concentrated}}}$$

If we take our saturated $\ce{HgS}$ solution and couple it to a $1\ \mathrm{M}\ \ce{Hg(NO3)2}$ solution, we get $(E^\circ = 0)$:

$$E_{cell}= \mathrm{-\dfrac{(8.314\ J/mol)(298\ K)}{2(9.65\times 10^4\ C/mol}\ln{\dfrac{2\times 10^{-27}\ M}{1\ M}}=0.781\ V}$$

As mentioned by Ivan in the comments, this $K_{sp}$ value could also be estimated from thermodynamic data using

$$\Delta G^\circ = -RT\ln K$$

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  • $\begingroup$ Determining the solubility product is accepted. My doubt is at the validity of the assigning such a value to such extreme solutions. $\endgroup$ – Swaroop Chandra Nov 16 '16 at 2:12

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