Using the appropriate values of #K_(sp)# and #K _f#, what is the equilibrium constant for the following reaction: #PbCl_2(s) + 3OH^(-)(aq) rightleftharpoons Pb (OH)_3^(-)(aq) + 2Cl^(-)(aq)# ?
Now, the idea here is that lead(II) chloride is considered Insoluble in aqueous solution, which means that a dissociation equilibrium is established when you place this salt in water
The thing to notice here is that adding hydroxide anions, to the aqueous solution of lead(II) chloride will consume the lead(II) cations dissolved in solution.
This means that you'll have
If you were to add these two reactions, you would get
This is equivalent to
Plug in your values to find
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[ K_{\text{eq}} = \frac{[\text{Pb(OH)}3^-][\text{Cl}^-]^2}{[\text{PbCl}2]} ] [ K{\text{eq}} = \frac{K{\text{sp}} \cdot K_{\text{f}}^3}{1} ]
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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