Is solubility in Qsp affected by coefficient? 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:

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

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 

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