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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ I meant $\ce{S^{2-}}$ $\endgroup$
    – M.L
    Dec 25, 2022 at 19:39
  • 1
    $\begingroup$ 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


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:


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:



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



So the resulting equilibrium expression would be:


Solving for $x$:


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


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

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.