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This question concerns a textbook problem* on precipitation. Briefly, there are $2\cdot10^{-4}$ of Mn$^{2+}$ and Cu$^{2+}$ ions in a liter of solution with $0.003$ M HClO$_4$. This solution is then saturated with H$_2$S. The solubility of H$_2$S is given as 0.1 mol/L and assumed independent of the presence of other substances in solution. K$_{sp}$ for MnS is $3\cdot 10^{-14}$, K$_{sp}$ for CuS is $8\cdot 10^{-37}.$ K$_1$ and K$_2$ for H$_2$S are given as $1.0\cdot 10^{-7}$ and $1.2\cdot 10^{-13}$ respectively.

In part one of the question it is found that Mn remains in solution and Cu$^{2+}$ will precipitate because the product [Cu][S] = $2.6\cdot 10^{-20}$ far exceeds K$_{sp}$ of CuS. In part two the problem is to determine how much Cu remains in solution.

The solution begins by stating that "most of the Cu$^{2+}$ will precipitate" and explains that a corresponding increment of [H$^+$] will be added to the solution from the H$_2$S which has lost S to precipitate, requiring a correction which leads to $8\cdot 10^{-21}$ moles Cu remaining in solution vs. $6\cdot 10^{-21}$ moles without the correction.

To make the correction the text assumes all the Cu precipitates, so here is my question.

The question being asked is: How much Cu remains in solution? To answer the question, it is assumed that none remains in solution...(!) This leads to a slight correction (from $3\cdot 10^{-21}$ % to $4\cdot 10^{-21}$ %) Cu remaining. The text does ignore the small change in [S] that accompanies the change in [H$^+$].

If there is a situation in which this very small difference matters, what is the right way to find it?

My guess is that there would be a simple ODE governing this problem--but I'm not sure.

The details above are included for completeness but the answer I think does not depend on the numbers here.

*The source of this problem is Shaum's College Chem. (10 ed.), 18.13-14.

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  • $\begingroup$ Please correct your text - does the problem refer to Mn or Mg? You have both in the text. $\endgroup$ – Waylander Oct 25 '18 at 12:39
  • $\begingroup$ I'm afraid your K2 of H2S is many order of magnitude too high, common thing in books using outdated data. $\endgroup$ – Mithoron Oct 25 '18 at 16:13
  • $\begingroup$ @Mithoron - The OP must use whatever constants are given in the problem, regardless if they are outdated values. However having a value that is "off" should get the OP to double check that the values were copied correctly from the problem. $\endgroup$ – MaxW Oct 25 '18 at 16:44
  • $\begingroup$ @daniel - Could you add a picture of the problem and solution as given, or rewrite the problem and answer exactly as given? $\endgroup$ – MaxW Oct 25 '18 at 17:01
  • $\begingroup$ @MaxW: I will post more of this when I get a moment, yes. $\endgroup$ – daniel Oct 25 '18 at 19:03
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Ok let's rewrite the problem for four significant figures.

There are $2.000\cdot10^{-4}$ of Mn$^{2+}$ and Cu$^{2+}$ ions in a liter of solution with $0.003000$ M HClO$_4$. This solution is then saturated with H$_2$S. The solubility of H$_2$S is given as 0.1000 mol/L and assumed independent of the presence of other substances in solution. K$_{sp}$ for MnS is $3.000\cdot 10^{-14}$, K$_{sp}$ for CuS is $8.000\cdot 10^{-37}.$ K$_1$ and K$_2$ for H$_2$S are given as $1.000\cdot 10^{-7}$ and $1.200\cdot 10^{-13}$ respectively.

Now let's solve this using significant figures in a reasonable fashion.

Carry six significant figure in intermediate calculations to try to avoid rounding errors and then round the final result to four significant figures.

First let's assume that the $\ce{[H^+] = [ClO4-]}$, in other words the perchloric acid completely dissociates, and all the acid comes from the perchloric acid.

  • For $\ce{H2S}$ we can multiply $K_{a1}\times K_{a2}$ to get an overall ionization product $$K_t = K_{a1}\times K_{a2} = 1.20000\times10^{-20} = \dfrac{\ce{[H+]_f[HS-]_f}}{\ce{[H2S]_f}}\times\dfrac{\ce{[H+]_f[S^{2-}]_f}}{\ce{[HS-]_f}} = \dfrac{\ce{[H+]_f^2[S^{2-}]_f}}{\ce{[H2S]_f}}$$ also that
    $$\ce{[S^{2-}]_f} = \dfrac{K_t\ce{[H2S]_f}}{\ce{[H^+]_f^2}}= \dfrac{(1.20000\cdot10^{-20})(0.10)}{(3.000\cdot10^{-3})^2} = 1.33333\cdot10^{-16}$$

Now using the two $K_{sp}$ values we can calculate the final concentrations of the $\ce{Cu^{2+}}$ and $\ce{Mn^{2+}}$ cations

$$\ce{[Cu^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{8.000\cdot10^{-37}}{1.33333\cdot10^{-16}} = 6.00002\cdot10^{-21}$$

$$\ce{[Mn^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{3.000\cdot10^{-14}}{1.333\cdot10^{-16}} = 225.001$$

So virtually all of the $\ce{Cu^{2+}}$ will ppt, but none of the $\ce{Mn^{2+}}$ will ppt.

Now knowing that the $\ce{Cu^{2+}}$ combining with $\ce{S^{2-}}$ frees $\ce{H+}$ according to the overall reaction: $$\ce{Cu^{2+} + H2S -> CuS + 2H+}$$ a pH correction is needed. We can ignore $\ce{H+}$ contributions of $\ce{[HS-]_f}$ and $\ce{[S^{2-}]_f}$ as being insignificant. Likewise the $\ce{OH-}$ from the ionization of water ($10^{-7} - \ce{[OH-]}$) that the acid neutralizes is insignificant. So

$$\ce{[H]_f = [ClO4-] + 2[Cu^{2+}]} = 3.40000\cdot10^{-3}$$ and $$\ce{[S^{2-}]_f} = \dfrac{K_t\ce{[H2S]_f}}{\ce{[H^+]_f^2}}= \dfrac{(1.200\cdot10^{-20})(0.1000)}{(3.40000\cdot10^{-3})^2} = 1.03806\cdot10^{-16}$$ $$\ce{[Cu^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{8.000\cdot10^{-37}}{1.03806\cdot10^{-16}} =7.70668\cdot10^{-21} \ce{->[rounding]} = 7.707\cdot10^{-21}$$

$\text{* * * * * * * * * * * * }$
$\text{* *}\quad\text{CHECKS}\quad\text{* *}$
$\text{* * * * * * * * * * * * }$

  • $\ce{[HS-]_f \ll [H+]_f}$ $$\ce{[HS-]} = \dfrac{K_{a1}\ce{[H2S]_f}}{\ce{[H]_f}} = \dfrac{(1.000\cdot10^{-7})(0.1000)}{3.40000\cdot10^{-3}} = 2.94118\cdot10^{-6}$$ $$0.00294\cdot10^{-3} \ll 3.40000\cdot10^{-3}$$
    THIS FAILS

  • $\ce{[S^{2-}]_f \ll [H+]_f}$ $$\ce{[S^{2-}]_f} = 1.333\cdot10^{-16} \ll 3.40000\cdot10^{-3}$$

  • $\ce{10^{-7} - [OH-]_f \ll [H+]_f}$ $$\ce{10^{-7} - [OH-]_f} = 1.0\cdot10^{-7} - \dfrac{K_w}{\ce{[H+]_f}} =1.0\cdot10^{-7} - \dfrac{1\cdot10^{-14}}{3.40000\cdot10^{-3}} = 9.99971\cdot10^{-8}$$ $$9.99971\cdot10^{-8} \ll 3.40000\cdot10^{-3}$$

This fails in the last of the six digits being used for intermediate calculations. Since we are going to round to four significant figures this should be OK, but it is right on the ragged edge of being acceptable.

$\text{* * * * * * * * * * * * * * * * * *}$
$\text{* *}\quad\text{RECALCULATION}\quad\text{* *}$
$\text{* * * * * * * * * * * * * * * * * *}$

The pH correction made was not quite correct. We can ignore $\ce{H+}$ contributions of $\ce{[S^{2-}]_f}$ and likewise the $\ce{H+}$ consummed in the neutralization $\ce{OH-}$ from the ionization of water ($10^{-7} - \ce{[OH-]}$). However the contribution of $\ce{[HS-]_f}$ is not insignificant. So

$$\ce{[H]_f = [ClO4-]_f + 2[Cu^{2+}]_i + [HS-]_f}$$

Now the correction for $\ce{[HS-]_f}$ is small. So we can probably just calculate the correction needed form the $K_{a1}$ equation using a guess that the $\ce{H+}=3.40294\cdot10^{-3}$. So let's check that for consistency.

$$\ce{[HS-]} = \dfrac{K_{a1}\ce{[H2S]_f}}{\ce{[H]_f}} = \dfrac{(1.000\cdot10^{-7})(0.1000)}{3.40294\cdot10^{-3}} = 2.93864\cdot10^{-6} = 0.00294\cdot10^{-3}$$

We don't need to match all six figures for the $\ce{[HS-]}$, but just the first three. So the small correction works.

Recalculating $\ce{[S^{2-}]}$ :

$$\ce{[S^{2-}]_f} = \dfrac{K_t\ce{[H2S]_f}}{\ce{[H^+]_f^2}}= \dfrac{(1.20000\cdot10^{-20})(0.1000)}{(3.40294\cdot10^{-3})^2} = 1.03627\cdot10^{-16}$$

Recalculating $\ce{[CuS^{2+}]}$ :

$$\ce{[Cu^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{8.000\cdot10^{-37}}{1.03627\cdot10^{-16}} = 7.72000\cdot10^{-21} \ce{->[Rounding]} 7.720\cdot10^{-21}$$

Note that we still haven't even needed to use a quadratic equation for the solution.

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  • $\begingroup$ Safe to say this would never show up on a gen. chem. test but very satisfying to see the calculation. We didn't use a quadratic... but the mental process of adjusting the pH correction is not trivial. $\endgroup$ – daniel Oct 28 '18 at 7:22
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    $\begingroup$ The point of course is to use significant figures to your advantage by simplifying the problem. The calc for $\ce{HS-}$ could easily be setup as a quadratic to get a "better" solution, but why go to the effort when a quick and dirty trick will work? $\endgroup$ – MaxW Oct 28 '18 at 7:27
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Ok let's rewrite the problem for two significant figures.

There are $2.0\cdot10^{-4}$ of Mn$^{2+}$ and Cu$^{2+}$ ions in a liter of solution with $0.0030$ M HClO$_4$. This solution is then saturated with H$_2$S. The solubility of H$_2$S is given as 0.10 mol/L and assumed independent of the presence of other substances in solution. K$_{sp}$ for MnS is $3.0\cdot 10^{-14}$, K$_{sp}$ for CuS is $8.0\cdot 10^{-37}.$ K$_1$ and K$_2$ for H$_2$S are given as $1.0\cdot 10^{-7}$ and $1.2\cdot 10^{-13}$ respectively.

Now let's solve this in a reasonable fashion.

First let's assume that the $\ce{[H^+] = [HClO4]}$

  • For $\ce{H2S}$ we can multiply $K_{a1}\times K_{a2}$ to get an overall ionization product $$K_t = K_{a1}\times K_{a2} = 1.2\times10^{-20} = \dfrac{\ce{[H+]_f[HS-]_f}}{\ce{[H2S]_f}}\times\dfrac{\ce{[H+]_f[S^{2-}]_f}}{\ce{[HS-]_f}} = \dfrac{\ce{[H+]_f^2[S^{2-}]_f}}{\ce{[H2S]_f}}$$ also that
    $$\ce{[S^{2-}]_f} = \dfrac{K_t\ce{[H2S]_f}}{\ce{[H^+]_f^2}}= \dfrac{(1.2\cdot10^{-20})(0.10)}{(3\cdot10^{-3})^2} = 1.333\cdot10^{-16}$$ carrying two extra digits in intermediate calcs

Now using the two $K_{sp}$ values we can calculate the final concentrations of the $\ce{Cu^{2+}}$ and $\ce{Mn^{2+}}$ cations

$$\ce{[Cu^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{8.0\cdot10^{-37}}{1.333\cdot10^{-16}} = 6.002\cdot10^{-21}$$

$$\ce{[Mn^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{3.0\cdot10^{-14}}{1.333\cdot10^{-16}} = 225$$

So virtually all of the $\ce{Cu^{2+}}$ will ppt, but none of the $\ce{Mn^{2+}}$ will ppt.

Now knowing that the $\ce{Cu^{2+}}$ combining with $\ce{S^{2-}}$ frees $\ce{H+}$ according to the overall reaction: $$\ce{Cu^{2+} + H2S -> CuS + 2H+}$$ a pH correction is needed. We can ignore $\ce{H+}$ contributions of $\ce{[HS-]_f}$ and $\ce{[S^{2-}]_f}$ as being insignificant. Likewise the $\ce{OH-}$ from the ionization of water ($10^{-7} - \ce{[OH-]}$) that the acid neutralizes is insignificant. So

$$\ce{[H]_f = [HClO4] + 2[Cu^{2+}]} = 3.4\cdot10^{-3}$$ and $$\ce{[S^{2-}]_f} = \dfrac{K_t\ce{[H2S]_f}}{\ce{[H^+]_f^2}}= \dfrac{(1.2\cdot10^{-20})(0.10)}{(3.4\cdot10^{-3})^2} = 1.038\cdot10^{-16}$$ $$\ce{[Cu^{2+}]_f} = \dfrac{K_{sp}}{\ce{[S^{2-}]}} = \dfrac{8.0\cdot10^{-37}}{1.038\cdot10^{-16}} =7.707\cdot10^{-21} \ce{->[rounding]} = 7.7\cdot10^{-21}$$

$\text{* * * * * * * * * * * * }$
$\text{* *}\quad\text{CHECKS}\quad\text{* *}$
$\text{* * * * * * * * * * * * }$

  • $\ce{[HS-]_f \ll [H+]_f}$ $$\ce{[HS-]} = \dfrac{K_{a1}\ce{[H2S]_f}}{\ce{[H]_f}} = \dfrac{(1.0\cdot10^{-7})(0.10)}{3.4\cdot10^{-3}} = 2.94\cdot10^{-6} \ll 3.4\cdot10^{-3}$$

  • $\ce{[S^{2-}]_f \ll [H+]_f}$ $$\ce{[S^{2-}]_f} = 1.333\cdot10^{-16} \ll 3.4\cdot10^{-3}$$

  • $\ce{10^{-7} - [OH-]_f \ll [H+]_f}$ $$\ce{10^{-7} - [OH-]_f} = 1.0\cdot10^{-7} - \dfrac{K_w}{\ce{[H+]_f}} =1.0\cdot10^{-7} - \dfrac{1\cdot10^{-14}}{3.4\cdot10^{-3}} = 9.99971\cdot10^{-8}$$ $$9.99971\cdot10^{-8} \ll 3.4\cdot10^{-3}$$

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  • $\begingroup$ The point here being that $$2.0\cdot10^{-4} - 7.7\cdot10^{-21} = 1.999999999999999923\cdot10^{-4} \ce{->[rounding]} = 2.0\cdot10^{-4}$$ $\endgroup$ – MaxW Oct 26 '18 at 18:34
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Ok, let's do the "complete" and "exact" problem.

First, I'm just going to ignore the $\ce{Mn^{2+}}$. Using a first pass with the simplifying assumptions would be enough to show that MnS will not ppt.

Second the problem statement notes that:

This solution is then saturated with H2S. The solubility of $\ce{H2S}$ is given as 0.1 mol/L and assumed independent of the presence of other substances in solution.

Chemically $\ce{H2S}$ would be bubbled through the solution to saturate it. While the $\ce{H2S}$ was bubbling $\ce{CuS}$ would be ppt'ing. So when the final concentration of $\ce{H2S}$ is 0.1 molar the ppt of $\ce{CuS}$ is complete.

Let:

  • The final $\ce{[H2S]_f = 0.1\bar0}$ molar

  • This means that the final concentration of all the sulfur species ($\ce{[H2S]_f, [HS-]_f,}$ and $\ce{[S^{2-}]_f}$) will be a bit more than $0.1\bar0$ molar.

  • The initial and final $\ce{[HClO4]} = 3.\bar 0\times10^{-3}$ molar

  • From $K_{sp}$ for CuS we know that: $$K_{sp} = [Cu^{2+}]_f[S^{2-}]_f$$ or $$\ce{[S^{2-}]_f} = \dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}}$$

  • For $\ce{H2S}$ we can multiply $K_{a1}\times K_{a2}$ to get an overall ionization product $$K_t = K_{a1}\times K_{a2} = 1.2\bar0\times10^{-20} = \dfrac{\ce{[H+]_f[HS-]_f}}{\ce{[H2S]_f}}\times\dfrac{\ce{[H+]_f[S^{2-}]_f}}{\ce{[HS-]_f}} = \dfrac{\ce{[H+]_f^2[S^{2-}]_f}}{\ce{[H2S]_f}}$$ also that
    $$\ce{[H+]_f} = \sqrt{\dfrac{K_t\ce{[H2S]_f}}{\ce{[S^{2-}]_f}}} = \sqrt{\dfrac{K_t\ce{[H2S]_f[Cu^{2+}]_f}}{K_{sp}}}$$ and
    $$\ce{[HS-]_f} = \dfrac{K_{a1}\ce{[H2S]_f}}{\ce{[H+]_f}} = K_{a1}\ce{[H2S]_f}\sqrt{\dfrac{\ce{[S^{2-}]_f}}{K_t\ce{[H2S]_f}}}=K_{a1}\sqrt{\dfrac{\ce{[H2S]_f[S^{2-}]_f}}{K_t}}= K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}}$$

  • $\ce{[Cu^{2+}]_f}$ be the final concentration of copper.

  • The initial $\ce{[Cu^{2+}]_i}=2\times10^{-4}$ molar which we will assume to be $2.\bar0\times10^{-4}$ molar. Thus the amount of copper that is removed from solution is $2.\bar0\times10^{-4} - \ce{[Cu^{2+}]_f}$

  • For the final $\ce{[H+]}$ we know that $$\ce{[H+]_f = [HClO4] + [HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - \ce{[Cu^{2+}]_f}) - (10^{-7} -[OH-])}$$ or $$ 0 = \ce{[HClO4] + [HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - [Cu^{2+}]_f) - (10^{-7} -\dfrac{K_w}{\ce{[H+]}}) - [H+]}$$ substituting for $\ce{[HClO4]}$ $$0 = 3.0000\times10^{-3} + \ce{[HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - [Cu^{2+}]_f) - (10^{-7} -\dfrac{K_w}{\ce{[H+]}}) - [H+]}$$ and collecting terms $$0 = 3.3999\times10^{-3} + \ce{[HS^-]_f + 2[S^{2-}]_f - 2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$ substituting for $\ce{[S^{2-}]_f}$ $$0 = 3.3999\times10^{-3} + \ce{[HS^-]_f} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$ substituting for $\ce{[HS^-]_f}$ $$0 = 3.3999\times10^{-3} + K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$ and finally substituting for $\ce{[H+]_f}$ $$0 = 3.3999\times10^{-3} + K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +K_w\sqrt{\dfrac{K_{sp}}{K_t\ce{[H2S]_f[Cu^{2+}]_f}}} - \sqrt{\dfrac{K_t\ce{[H2S]_f[Cu^{2+}]_f}}{K_{sp}}}}$$

Since $\ce{[H2S]_f = 0.10}$ per problem statement, the only variable left is $\ce{[Cu^{2+}]_f}$. I'll leave it to the reader to solve this mess for $\ce{[Cu^{2+}]_f}$, but to 2 significant figures the answer should be $7.7\cdot10^{-21}$.

The whole point in using significant figures to simply the problem is not to get the "wrong" answer, but to avoid a computational mess that doesn't yield any better answer.

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  • $\begingroup$ As I begin to look at this, I notice that of the six terms on the r.h.s. of the final equation, just taking [H2S] as 1 and using [Cu] from the previous answer (which is surely close), all the terms are < 9x10[-6] but the sixth term is about -.0107... $\endgroup$ – daniel Oct 27 '18 at 7:24
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    $\begingroup$ If the answer changes it will only be because I made a mistake. // $\ce{[H2S]_f = 0.10}$ as noted in specification of the problem. So the only variable left is $\ce{[Cu^{2+}]_f}$ // Notice in equations above the following equation was derived $$\ce{[H+]_f} = \sqrt{\dfrac{K_t\ce{[H2S]_f[Cu^{2+}]}}{K_{sp}}}$$ $\endgroup$ – MaxW Oct 27 '18 at 7:55
  • $\begingroup$ Bother--my mistake. I think this is correct and will be looking at it more carefully. $\endgroup$ – daniel Oct 27 '18 at 8:12
  • $\begingroup$ Totally agree with your last point. One lingering confusion--solving your last eq. for [Cu] gives about 7.71954x10[-21]. This corresponds to [H] =0.0034029 and I think we can't exceed 0.0034. I can't find any problem with your reasoning, so I may have made an error calculating--but I am prepared to call it a day and accept your upvoted answer--with much appreciation for the help . $\endgroup$ – daniel Oct 27 '18 at 19:43
  • $\begingroup$ Well to 2 significant figures $7.71954\cdot10^{-21} = 7.7\cdot10^{-21}$ and $0.0034029 = 0.0034$. That is the whole point. So do you want to do the "easy" calc and get $7.707\cdot10^{-21}$ which rounds to $7.7\cdot10^{-21}$, or do you want to do the "hard" calc which yields $7.71954\cdot10^{-21}$ and rounds to $7.7\cdot10^{-21}$? $\endgroup$ – MaxW Oct 27 '18 at 22:05
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It is just an algebra problem, but I'm lost as to exactly how the significant figures should be handled. The gist is that significant figures are used to simply the problem by assuming that certain aspects can be ignored. Without the simplifications you'll end up with a cubic or higher equation that can't be easily solved by hand.

For example, the perchloric acid will completely dissociate so the $\ce{[H+] = 3\times10^{−3}}$. The $\ce{Cu^{2+}=2\times10^{−4}}$, so if "all" the $\ce{Cu^{2+}}$ ppts, then the boost to the $\ce{[H+] = 4\times10^{−4}}$. So is the final $\ce{[H+]}$ supposed to be $3\times10^{−3}$ or $3.4\times10^{−3}$?

RANT - If the problem wanted the perchloric acid to be 0.00300 molar, then the !@#$%^& problem should state that.

The point here is that if we start adding extra significant figures, then how many should be added, and to which species? 1? 2? 3? 4? ...

So if we get slightly different answers assuming that:

  • The perchloric acid is 0.003 molar, the copper is $2\times10^{-4}$ molar and the hydrogen sulfide is 0.1 molar

  • The perchloric acid is 0.00300 molar, the copper is $2\times10^{-4}$ molar and the hydrogen sulfide is 0.10 molar

  • The perchloric acid is 0.00300 molar, the copper is $2.00\times10^{-4}$ molar and the hydrogen sulfide is 0.100 molar

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  • $\begingroup$ I did double-check $K_1,K_2$ and they are as given. And yes, the final [H] is assumed 3.4 x 10$^-3$ in the text. But we know this is approximate because all the Cu does not precipitate. I am interested in how to find the answer without assuming an artificially high boost of [H]. I will post more of the original question when I get a moment but you touch on what interests me in your first point--how do we formulate the "not easily solved by hand" version-- $\endgroup$ – daniel Oct 25 '18 at 19:03
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    $\begingroup$ How crazy do you want to get? You gave the initial concentration of copper as $2\times10^{-4}$ but the final is about $10^{-21}$. So do you want to work out problem "exactly" to say 18 significant figures? $\endgroup$ – MaxW Oct 25 '18 at 20:12
  • $\begingroup$ Let me put it this way--I put in a lot of time trying to understand a solution that was essentially an exercise in tail-chasing (almost all of the Cu precipitates... first we assume that all the the Cu precipitates...). So if you get crazy I will put in a lot more work to make sure I understand it. It's obviously a more subtle point than the authors wanted to address. $\endgroup$ – daniel Oct 25 '18 at 21:45
  • $\begingroup$ The reason I thought ODE was that this is not instantaneous and it seems like a "coroner's equation" equilibrium situation with d[Cu]/dt=k f ([Cu]) and after integrating w'd get a function of t that will reach some eq. point. The [H] will fall somewhere between 0.003 and 0.00034. But I am not sure. $\endgroup$ – daniel Oct 26 '18 at 6:01
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    $\begingroup$ A ODE would be useful to examine the kinetics. You could use kinetics and extrapolate to infinity time, but that is the very hard way to solve the equation. The problem also doesn't give any data for forward and backward reactions so there really isn't way to solve this problem using kinetics. $\endgroup$ – MaxW Oct 26 '18 at 6:11

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