2
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The reaction is $$\ce{ CoCl_{2(s)} + 6 H_{2}O_{(g)} <=> [Co(H2O)6]Cl_{2(s)}}$$

Note that the reaction is at equilibrium. I have determined the value of $K_\text{p}=5.5\cdot 10^{12}$. The next step in the solution manual is isolating for $P_{\ce{H2O}}$: $$K_\text{p}=\dfrac{1}{(P_{\ce{H2O}})^6}=5.5\cdot 10^{12}$$

How are we justified in making the above expression? Do we make the numerator 1 because the equation is at equilibrium?

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1
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How are we justified in making the above expression. Do we make the numerator 1 because the equation is at equilibrium?

No. The activity of pure solids and liquids is idealized as one. Hence the one in the numerator.

In the case of solutes, their concentrations are generally approximately equal to their activities at low concentrations (below 0.1 M).

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  • $\begingroup$ In this case, the water is a vapor. $\endgroup$ – LDC3 Aug 17 '14 at 14:33
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    $\begingroup$ I was talking about the numerator or where the concentration (or activity) of the product would go. The product is a solid. $\endgroup$ – Dissenter Aug 17 '14 at 14:55

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