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In my textbook written that:

Given that $K_\mathrm{sp}(\ce{Ag2CrO4})=9.0\times 10^{-12}$, consider the solubility of $\ce{Ag2CrO4}$ in a $0.100\:\mathrm{M}$ solution of $\ce{AgNO3}$.

Initial concentrations (before any $\ce{Ag2CrO4}$ dissolves) are, $[\ce{Ag+}]_0=0.100\:\mathrm{M}$ (from $\ce{AgNO3})$ and $[\ce{CrO4^{2-}}]_0=0\:\mathrm{M}$

The system comes to equilibrium as $\ce{Ag2CrO4}$ dissolves according to the reaction: $$\ce{Ag2CrO4(s)\rightleftharpoons 2Ag+(aq) + CrO4^{2-}(aq)}$$ $$K_\mathrm{sp}=[\ce{Ag+}]^{2}[\ce{CrO4^{2-}}]=9.0\times 10^{-12}$$

We assume that $x\:\mathrm{M}$ of $\ce{Ag2CrO4}$ dissolves to reach equilibrium, which means that: $$x\:\mathrm{M\:\ce{Ag2CrO4}(s)}\rightarrow 2x\:\mathrm{M\:\ce{Ag+}(aq)}+x\:\mathrm{M\:\ce{CrO4^{2-}}\:(aq)}$$

Now we can specify the equilibrium concentrations in terms of $x$: $$[\ce{Ag+}]=[\ce{Ag+}]_{0}+\Delta[\ce{Ag+}]=(0.100+2x)\:\mathrm{M}$$ $$[\ce{CrO4^2-}]=[\ce{CrO4^2-}]_{0}+\Delta[\ce{CrO4^2-}]=(0+x)\:\mathrm{M}$$

Giving: $$9.0\times 10^{-12}=[0.100+2x]^{2} \cdot [x]$$

Also, here is line I want to discuss (again, taken from the textbook):

Since the $K_{\mathrm{sp}}$ value for $\ce{Ag2CrO4}$ is small (the position of the equilibrium lies far to the left), $x$ is expected to be small compared with $0.100\:\mathrm{M}$. Therefore: $${9.0\times 10^{-12}=[0.100+2x]^{2}[x] \approx [0.100}]^{2}[x]$$

When can this approximation be made and when can it not be made?

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  • $\begingroup$ I edited this post so it contains the correct formatting. Please use this formatting in future posts, as it makes it easier for everyone to understand your question and thereby help you:) $\endgroup$ – ringo May 17 '15 at 17:14
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Really, your question seems to be a math question.

You need to solve the cubic equation:

$$\mathrm{9.0\cdot10^{-12}=[0.100+}2x]^{2}[x]$$

Cubic equations are exactly solvable; however, it is common to use an approximate method.

Whether or not it is appropriate to approximate 0.100 +2x as 0.100 depends upon how small x is and how precise an answer you need.

Once you get an approximate solution for x, you could recalculate 0.100 +2x and use that value to re-solve the problem. By repeating this process you can get arbitrary close to the correct value by iteration.

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  • $\begingroup$ Really, I didn't realize that. Should I just delete this question? $\endgroup$ – làntèrn May 17 '15 at 13:29
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    $\begingroup$ @Ridho Don't delete it. It might be useful for other people in the future. $\endgroup$ – Binary Geek May 17 '15 at 13:35

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