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8.Lead (II) iodide, Pb2, is an ionic compound with asolubility product constant Ksp of 7.9 x10-9. Calculate the solubility of this compoundin

  1. a. pure water.

b. 0.050 mol L-1 KI solution.

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Beverley Smith
Beverley SmithLv2
30 Sep 2019

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Related questions

Please help, and show how you did it! Excuse the numbers1-5.

#1: Calculate the number ofgrams of solid Mg(OH)2 that will dissolve in 650 mL of water. ForMg(OH)2, Ksp = 1.5e-11. Assume no volume change for the solutionwhen the solid is added.

  1. A. 0.0092
  2. B. 0.0059
  3. C. 0.012
  4. D. 0.019
  5. E. 0.060

#2: Which of the followingbest describes what happens to the solubility of lead (II) iodidewhen it is dissolved in a 0.10 M NaI solution, compared with purewater? For lead (II) iodide, Ksp = 6.5e-9.

  1. A. Lead (II) iodide is more solublein 0.10 M Nal solution than in pure water.
  2. B. Lead (II) iodide is less solublein 0.10 M Nal solution than in pure water.
  3. C. The solubility of lead (II)iodide is the same in both a 0.10 M Nal solution and in purewater.

a) Write the dissolution reaction for lead(II) iodide. How many grams of lead(II) iodide will dissolve in 250.0 mL of water? What are the molarities of lead(II) ion and iodide ion? Neglect any volume change due to the lead(II) iodide. Ksp of lead(II) iodide = 1.4 x 10-8

b) How many grams of lead(II) iodide will dissolve in 250.0 mL of 0.35 M HI?

c) Will the solubility of lead(II) iodide in 250.0 mL of 0.35 M HNO3 be the same as (a) or (b) ?

d ) How would the answers to (a) and (c) have compared if the salt under consideration were lead(II) fluoride, rather than lead(II) iodide? Ksp PbF2 = 7.1 x 10-7 KaHF = 7.2 x 10-4

e ) Lead(II) ion forms a 1:1 complex with a particular ligand, L (forms PbL2+) . Write the reaction for formation of the complex ion using รขย€ยœLรขย€ย as the ligand. If the formation constant for this complex is 2.9 x 104 , calculate the solubility of lead(II) iodide in a 1.5 M solution of L.

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