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.

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter4: Stoichiometry: Quantitative Information About Chemical Reactions

Section4.9: Spectrophotometry

Problem 2.1ACP: Excess KI is added to a 100.0-mL sample of a soft drink that had been contaminated with bleach,...

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Question

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 K_sp = [ 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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