chem lab 19 - 8 pKa=5.66 Molar mass =101.3 g/mol Name=...

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Chem 404 section 7 4-10-08 Weak Acid Titration 1. 2M NaOH x XL= .05M x .2L XL= .01ML/2M X=.005L 2. KHP 1: .304g . . / 304 g KHP204 23 g mol = . 40 2mL Titrant1000 mL x XM Molarity = .0370 KHP 2: .302g . . / 302 g KHP204 23 g mol = . 26 9mL Titrant1000 mL x XM Molarity = .0549 KHP 3: .301g . . / 301 g KHP204 23 g mol = . 26 8mL Titrant1000 mL x XM Molarity = .0549 3. See Appendix A for Graph labeled “Acid-Base Titration” 4. The equivalence point and the volume of NaOH solution that is required to reach the equivalence point is 30mL. 5. The PKa of the unknown acid is 5.66 and the pH at the half equivalence point is 5.66 6. Moles to reach equivalence point = .03L X 1.563 M= .04689 moles NaOH 7. molar mass= .152g/ .0015 moles= 101.3 g/mol
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Unformatted text preview: 8. pKa=5.66 Molar mass =101.3 g/mol Name= potassium hydrogen phthalate 9. See Appendix A; page two for Tables 1 and 2 10. % error molar mass = 101.3-204.1/101.3 x 100= 101.4% % error pka= 5.66-5.51/ 5.51 x 100= 2.72% 11. pH= -log (5.2X10-6 )=5.28 12. Before pH at ½ equivalence point : Lots of HA little A-½ equivalence point: A little more HA than A-After ½ equivalence point but before equivalence point: About same HA as A-Equivalence point: Same HA as A-After equivalence point: More A-than HA 13. pH= 8.79 Sheldon Pope Individual Lab C=pK b A: [B-]> >[HB] H + <<OH-B: [B-]>[HB] H + < OH-C: [B-]>[HB] H + =OH-D: [B-]=[HB] H + > OH-E: [B-]<[HB] H + >OH-F: [B-]<<[HB] H + >>OH-...
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This lab report was uploaded on 04/18/2008 for the course CHEM 404 taught by Professor Bauer during the Spring '08 term at New Hampshire.

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chem lab 19 - 8 pKa=5.66 Molar mass =101.3 g/mol Name=...

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