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What is the maximum number of silver(Ⅰ) ions that can be present dissolved in one litre of a $\pu{0.003 M}$ $\ce{Na2S}$ solution?

According to my book, silver(I) reacts with sulfide producing $\ce{Ag2S}$ with the solubility product constant

$$K_\mathrm{sp}(\ce{Ag2S}) = [\ce{Ag+}]^2\,[\ce{S^2-}] = \pu{8E-51}.$$

Now here is what I don't understand. I have seen several questions where they ask to solve for the maximum solubility of a substance in water, in this case $\ce{Ag2S}$. The usual thing they do is set up an equation with $x$, like $K_\mathrm{sp} = (2x)^2\,x.$

But in our case, we are given an initial concentration $c_0(\ce{Na2S}) = \pu{0.003 M}$, from which we know that the concentration $c_0(\ce{S}) = \pu{0.003 M}$. We also know the $K_\mathrm{sp}$ value, and they ask specifically for the silver(I) ions, not silver(I) sulfide.

So I am not sure if I need to proceed like above, i.e. set up an equation with $x$ and solve for it, because it seems to me that would give the maximum solubility of $\ce{Ag2S}$, not the maximum number of silver ions.

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    $\begingroup$ Try setting up an ICE table, and be aware about the concentration of $\ce{S2-}$ in the row with Initial concentration. $\endgroup$
    – M.L
    Dec 25, 2022 at 19:23
  • $\begingroup$ @M.L You mean $\ce{Ag_2}$ instead of $\ce{S_2}^-$ right ? There is only one $\ce{S}$ for every two $\ce{Ag_2}$ or $\ce{Na_2}$ as far as I know $\endgroup$
    – wengen
    Dec 25, 2022 at 19:27
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    $\begingroup$ I meant $\ce{S^{2-}}$ $\endgroup$
    – M.L
    Dec 25, 2022 at 19:39
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    $\begingroup$ If $\ce{[Ag+]}$ = x, and if $\ce{[S^{2-}] = 0.003}$, it is possible to write $\ce{K_{sp} = x^2 ·0.003 = 8 10^{-51}}$. So the solubility of silver ions is $$\ce{x = \sqrt(8 10^{-51}/0.003) = 1.63 10^{-24}}$$ and this result is nearly $x = 1/N_A$ where $N_A$ is the Avogadro number. It means that the concentration of the silver ion is about $1$ ion per liter. $\endgroup$
    – Maurice
    Dec 25, 2022 at 19:52
  • $\begingroup$ @Maurice That makes sense, but why is there no "2" in front of the x in this case ? For every S in $Ag_2S$, there are two Ag, so shouldn't there be a "2" like $(2x)^2 \cdot 0.003 = 8 \cdot 10^{-51} $ ? $\endgroup$
    – wengen
    Dec 25, 2022 at 22:49

1 Answer 1

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Product solubility problems with non-zero initial concentrations can be generalized for a dissociation reaction of the form:

$$\ce{A_aB_b(s)<=>aA^{b+}(aq) + bB^{a-}(aq)}$$

With the condition: ($a≠b$) or ($a=b=1)$, and $K_{sp}>Q_{sp}$

And a general equilibrium expression:

$$K_{sp}=[A^{b+}]^a\;[B^{a-}]^b$$

Since initial concentrations are non-zero:

$$[A^{b+}]=[A^{b+}]_o + ax$$

$$[B^{a-}]=[B^{a-}]_o + bx$$

The resulting expression can be solved for x:

$$K_{sp}=\left([A^{b+}]_o + ax\right)^a\;\left([B^{a-}]_o + bx\right)^b$$

In our particular case, the dissociation reaction is:

$$\ce{Ag2S(aq)<=>2Ag+(aq) + S^2-(aq)}$$

So we can define $a$ and $b$ from the stoichiometric coefficients:

$$a=2$$

$$b=1$$

Initially, no $\ce{Ag+}$ is present, but $\ce{S-}$ is, so in terms of initial concentrations we have:

$$[A^{b+}]_o=[\ce{Ag+}]_o=0$$

$$[B^{a-}]_o=[\ce{S^{2-}}]_o=\pu{0.003M}$$

So the resulting equilibrium expression would be:

$$K_{sp}=\left(2x\right)^2\;\left(0.003+x\right)=8\cdot10^{-51}$$

Solving for $x$:

$$x=\pu{8.2\cdot10^{-25}M}$$

Calculating the resulting concentration of $\ce{Ag+}$ at equilibrium:

$$[\ce{Ag+}]=2x=\pu{1.64\cdot10^{-24}M}$$

Finally, the number of silver ions can be calculated using Avogadro's constant and the volume of the solution given:

$$N_{\ce{Ag+}}=[\ce{Ag+}]\;V\;L=(\pu{1.64\cdot10^{-24}mol/L})(\pu{1L})(\pu{6.022\cdot10^{23}ions/mol})≈\pu{1 ion}$$

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  • $\begingroup$ Thank you for your thorough answer ! I have a last question, I’m a bit confused when we need to set up an equation with x and 2x, and when we need to solve for [A+] directly like [Ag+], in other word what does the x and 2x actually represent compared to [Ag+] and [S-] ? Don’t both x and [Ag+] represent some form of concentration ? What’s the difference between them ? Are those different approaches ? At the end, you say that [Ag+] is equal to two x, because for every [S-], there are two [Ag+] ? But then why do we get the same result than if we simply did [Ag+]=sqrt( K_sp / [S-]) like @Maurice ? $\endgroup$
    – wengen
    Dec 26, 2022 at 12:28
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    $\begingroup$ You are correct in acknowledging the stoichiometric ratio between species, but that's already accounted for when including the stoichiometric coefficients ($a,b$) when calculating the equilibrium concentrations. $x$ is arbitrary, so you could associate $[\ce{Ag+}]$ with $x$ or $2x$, as long as its proportion with $\ce{S^{2-}}$ is preserved. If you choose $x$ for $[\ce{Ag+}]$, then the corresponding factor for $[\ce{S^{2-}}]$ would be $\frac{x}{2}$. Conversely, if you choose $2x$ for $[\ce{Ag+}]$, then the corresponding factor for $[\ce{S^{2-}}]$ would be $x$. $\endgroup$
    – Sam202
    Dec 26, 2022 at 14:13
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    $\begingroup$ I prefer the second option, since it guarantees all factors involving $x$ will be whole numbers. Note that the reason why i had to multiply $x$ by two near the end of my answer is because I defined the concentration of $\ce{Ag+}$ at equilibrium as $\ce{[Ag+]=[Ag+]_o + ax}$, where a=2 since silver's molar coefficient in the reaction is 2. If I had left it as $x$, then that would only represent half of the equilibrium concentration of silver. Simply put, $x$ is the change in concentration per molar coefficient of a particular species. $\endgroup$
    – Sam202
    Dec 26, 2022 at 14:18

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