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The Ksp of MC12 is 1.5 x 107 at 25 °C.(M is a fictitious metal cation with +2 charge) A 1.0 L aqueous solution contains 0.0025 M M(NO3)2. If 0.10 g of NaCl is added, will a precipitate form? Assume no change in volume.

The Ksp of MC12 is 1.5 x 107 at 25 °C.(M is a fictitious metal cation with +2 charge) A 1.0 L aqueous solution contains 0.0025 M M(NO3)2. If 0.10 g of NaCl is added, will a precipitate form? Assume no change in volume.

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Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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The Ksp of MC12 is 1.5 x 107 at 25 °C.(M is a fictitious metal cation with +2 charge) A 1.0 L aqueous solution contains 0.0025 M M(NO3)2. If 0.10 g of NaCl is added, will a precipitate form? Assume no change in volume.
Transcribed Image Text:The Ksp of MC12 is 1.5 x 107 at 25 °C.(M is a fictitious metal cation with +2 charge) A 1.0 L aqueous solution contains 0.0025 M M(NO3)2. If 0.10 g of NaCl is added, will a precipitate form? Assume no change in volume.
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Step 1

Since the solution is having M(NO3)2 which has M2+ ion 

Hence the concentration of M2+ ion = [ M2+ ] = concentration of  M(NO3)2 = 0.0025 M

Because we have 1 M in 1 M(NO3)2 

And since the solubility reaction of MCl2 is

=> MCl2 (s) -------> M2+ (aq) + 2 Cl- (aq) 

Hence Ksp = [ M2+ ]  X [Cl-]2 

Hence substituting the values we get 

1.5 X 10-7 = 0.0025 X [Cl-]2 

=> [Cl-]= concentration of Cl- ion required for precipitation = 7.75 X 10-3 M 

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