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Andrei [34K]
3 years ago
10

A solution has a Pb2+ concentration of 1.9 x 10-3 M, and an F-1 concentration of 3.8 x 10-3 M. The value of Ksp for PbF2 at room

temperature is 4 x 10-8. Will this solution form a precipitate?
The solubility equilibrium equation is as follows.



.

Yes

No
Chemistry
2 answers:
horrorfan [7]3 years ago
6 0

A solution has a Pb2+ concentration of 1.9 x 10^-3, and an F-1 concentration of 3.8 x 10^-3 M. The value of Mps for Pbf2 at room temperature is 4 x 10^-8. This solution will NOT form a precipitate.
The solubility equilibrium equation is as follows.
PbF2 (s) <--> Pb2 (aq) + 2F- (aq)

valentina_108 [34]3 years ago
5 0
We are going to get the value of Qsp and compare it with the Ksp value:

when Qsp is the concentration of the initial products:

Qsp = [Pb2+][F-]^2

when the [Pb2+] = 1.9 x 10^-3

and [F-] = 3.8 x 10^-3

by substitution:

∴Qsp = (1.9 x 10^-3) * (3.8 x 10^-3)^2

           = 2.7 x 10^-8 

when Ksp = 4 x 10^-8

so Qsp < Ksp

∴ precipitate will not be formed 
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The mixture would contain

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Construct a RICE table for the hydrolysis of \text{Ac}^{-} under a basic aqueous environment (with a negligible hydronium concentration.)

\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}

The question supplied the <em>acid</em> dissociation constant pK_afor acetic acid \text{HAc}; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant pK_b for its conjugate base, \text{Ac}^{-}. The following relationship relates the two quantities:

pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})

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[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b =  10^{-pK_{b}}

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