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FSU - CHM 1046 - Class Notes - Week 11

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FSU - CHM 1046 - Class Notes - Week 11

School: Florida State University
Department: Chemistry
Course: General Chemistry II
Professor: Robert Cameron
Term: Spring 2018
Tags: Calculus
Name: Week 11 Notes
Description: These notes cover lecture material from the week of 3/26 - 3/30.
Uploaded: 03/30/2018
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Join more than 18,000+ college students at Florida State University who use StudySoup to get ahead
School: Florida State University
Department: Chemistry
Course: General Chemistry II
Professor: Robert Cameron
Term: Spring 2018
Tags: Calculus
Name: Week 11 Notes
Description: These notes cover lecture material from the week of 3/26 - 3/30.
Uploaded: 03/30/2018
2 Pages 30 Views 24 Unlocks
  • Better Grades Guarantee
  • 24/7 Homework help
  • Notes, Study Guides, Flashcards + More!

Unformatted text preview: 2 mmol Immol 3/28/18 HA + NaOH NaA + H20 Immol Immo pH = pka kako po pra 10X10 I I Chapter 16 Caf, 2 Catfreuerse nxn oculos concr - dissolution of ions increase - precipitation Kspja (caF]3 k solubility product constant malar solubilia malori cono, of salt in a Saturated sol in gram solubility (slomi table of Kse values. o wcite the express for the Ksp foethe disso.of each salt od potassium permanganat b) sodium dichro mate ke [No Lir2017 commonium phosphateau so (NHVDPO upa orogen phosphate P Ksp = K (HPO KHOy2 Copper ( bromide has a molari solf 2 G mol/L 2c.calec sp Cu Br2 2 600 2310 2,0x10 molse Cu (Bad RSPRONOS 040(042 ..Ksp F 13.2 x 10 3 Calc. the Ksect b'smuth sulfide which has Bi2S22B3+ 354 k sp = (Bit(s2 ] By Kse 20 ore 230 MORE Ksp XVO 3 The ksp for copper iodate is 192 Calea sol. allo 25 C) ac mol/L +9 100ml. CU (103), Cut + 210, 2x 1.359* 0.149 IOON 4071 me silver iodide se - 1.SY 15) Choose the most soluble soit capoe co odide kse 5.0 um sultatele 610 TorgestChoose mgist soluble salt and a) copper (IDsulfide sp8. SKI-45 b) silversulfide Ksp GTX10-49 c.) bismuth (un sulfide kspc0 -23 Cu$ = Cu? + S molar solubility 8.6 x 10-45 - 9.2 x 10-23 Ag 2 S 2 2Ag + S 1.60X10:49 - H X 3.xos Bi2S: 22B3 35? Ix013 = (2x72 (3X3 1x1035 = (4x19x 17X1078= 3xS ->X 101018

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