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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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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.
Expert Solution
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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