Is solubility in Qsp affected by coefficient?

Related to my previous question: [Is solubility coefficient affected if ion data is given in Ksp?](https://chemistry.stackexchange.com/questions/31135/is-solubility-coefficient-affected-if-ion-data-is-given-in-ksp)



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> $200\,\mathrm{mL}$ 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 $Q\_{\mathrm{sp}}$.
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*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 $Q\_{\mathrm{sp}}$.*




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 


$$Q\_{\text{sp}}\:(\ce{Ag\_{2}CrO\_{4}})=[\ce{Ag^{+}}]^{2}[\ce{CrO\_{4}^{2-}}]$$
$$Q\_{\text{sp}}\:(\ce{Ag\_{2}CrO\_{4}})=[0.01]^{2}[0.01] = 10^{-6}$$


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