What is the maximum number of silver(Ⅰ) ions in one litre of a 0.003 M sodium sulfide solution?


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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?
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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.




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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