Related to my previous question: Is solubility coefficient affected if ion data is given in Ksp?

$200\,\mathrm{mL}$ solution of $0.02\,\mathrm{M}\,\ce{AgNO3}$ is added to $200\,\mathrm{mL}$ $\ce{CrO4^{2-}}$ and $\ce{PO4^3-}$ ions. Find out both of $Q_{\mathrm{sp}}$.

Actually the question is asking "Will it precipitate", but I skip it.

Here is my half approach:

$$200\,\mathrm{mL}=2\times 10^{-1}\,\mathrm{L}$$ $$V_{\mathrm{total}}=400\,\mathrm{mL}=4\times 10^{-1}\,\mathrm{L}$$ $$[\ce{Ag+}]=[\ce{CrO4^2-}]=[\ce{PO4^3-}]=\frac{0.02\,\mathrm{M}\cdot 2\times 10^{-1}\,\mathrm{L}}{4\times 10^{-1}\,\mathrm{L}}=0.01\,\mathrm{M}$$ $$\ce{Ag2CrO4 -> 2Ag^{+} +CrO4^{2-}}$$ $$Q_{\mathrm{sp}}(\ce{Ag2CrO4})=[\ce{Ag+}]^{2} \cdot [\ce{CrO4^2-}]$$ $$Q_{\mathrm{sp}}(\ce{Ag2CrO4})=[2s]^{2}[0.01\,\mathrm{M}]$$

I know that to find $K_{\mathrm{sp}}$, the $s$ needs to be multiplied (and to the power) with the coefficient from the reaction, but how about $Q_{\mathrm{sp}}$? Does the $s$ also need to be multiplied?

I asked this because my teacher said that only to the power is affecting the $Q_{\mathrm{sp}}$.


1 Answer 1


You don't need to multiply s, since $Q_\text{sp}$ is the result of multiplication of ions concentration (and you don't need to multiply with the coefficient in any $Q$). So the right way to answer $Q_\text{sp}$ is

$$Q_{\text{sp}}\:(\ce{Ag_{2}CrO_{4}})=[\ce{Ag^{+}}]^{2}[\ce{CrO_{4}^{2-}}]$$ $$Q_{\text{sp}}\:(\ce{Ag_{2}CrO_{4}})=[0.01]^{2}[0.01] = 10^{-6}$$

  • $\begingroup$ But Ksp is also the result of multiplication of ions concentration. What's the different between them? $\endgroup$
    – làntèrn
    May 14, 2015 at 15:18
  • $\begingroup$ K_sp and Q_sp differ in that K_sp is the solubility product at equilibrium (and Q_sp = K_sp at equilibrium). Q_sp is the solubility product when the system is not at equilibrium. $\endgroup$
    – imaginov
    May 14, 2015 at 15:34

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