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the question is, Lead metal is added to $0.100\ \text{M} \ \ce{Cr^{+3}(aq)}$. What are $[\ce{Pb^{+2}}]$, $[\ce{Cr^{+2}}]$, and $[\ce{Cr^{+3}}]$ when equilibrium is established in the reaction? $$K_c = 3.2\times 10^{-10}$$

Given this equilibrium equation:

$\ce{Pb(s) + 2Cr^{+3}(aq) <=> Pb^{+2}(aq) + 2Cr^{+2}(aq)}$

          0.100 M          0              0

          0.100-2x        +x            +2x

          0.100-2x         x             2x

$$K_c = \dfrac{[\ce{Pb^{+2}}][\ce{Cr^{+2}}]^2}{[\ce{Cr^{+3}}]^2}$$

(We do not include [Pb] because it is a solid) $$3.2\times 10^{-10} = \dfrac{[x][2x]^2}{[0.100-2x]^2}$$ Since $2x<<0.100$

$$3.2\times 10^{-10} = \dfrac{4x^3}{0.0100}$$

$$3.2\times 10^{-12} = 4x^3$$

$$8\times 10^{-13} = x^3$$

$$\sqrt[3]{8\times 10^{-13}}=x$$

$$x = 2.7\times 10^{-6}$$

$$[\ce{Pb^{+2}}] = 2.7\times 10^{-6} \ \text{M}$$

$$[\ce{Cr^{+2}}] = 2.7\times 10^{-6} \ \text{M}$$

$$[\ce{Cr^{+3}}] = 0.100 \text{ M} - = 2.7\times 10^{-6} \ \text{M} = 0.0999973 \text{ M} = 0.1 \text{ M}$$

Did I do my calculation for x correctly? If the question asks for $[\ce{Cr^{+3}}]$ and $[\ce{Cr^{+2}}]$, do I still have to multiply $x$ by 2 at the end as their coefficient is 2 in the equilibrium constant equation?

Please point out my mistakes! Thank you very much!

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Yes, you need to use $2x$ when you calculate the concentrates of the chromium ions. The stoichiometry of the reaction is still important.

I also see one big math error - when you calculate the cube root of $8\times 10{-13}$, you seem to have interpreted cube root as three square roots.

$$\sqrt[3]{A}\ne 3\sqrt{A}$$

As a simple example, take $x^3 = 8$ $$x=\sqrt[3]{8} = 2$$

$$3\sqrt{8} =8.485...$$

To double check, we can do the inverse process:

$$2^3=8$$

$$(8.485...)^3=610.94...$$

Thus, $$\sqrt[3]{8\times 10^{-13}}\ne 2.7\times 10^{-6}$$

$$\sqrt[3]{8\times 10^{-13}}= (8\times 10^{-13})^{1/3}=9.28\times 10^{-5}$$

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    $\begingroup$ Thank you very much for pointing out my mistakes and giving me a clear explanation! I was confused at how to solve the cube root part and wasnt sure if I got the answer for x correctly. $\endgroup$ – Jesse Mar 2 '14 at 23:04

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