Prelab for Phun with Phosphate Buffers

Prelab for Phun with Phosphate Buffers - Prelab for Phun...

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Prelab for Phun with Phosphate Buffers Chemistry 112 Lab Section 005 November 5, 2014 1. Describe how you would prepare 1.50 L of a 0.5 M phosphate buffer with a pH of 7.50. Be sure to show all your work; no credit will be given otherwise. pH = p K a + log [ base ] [ acid ] P t = molacid + molbase 7.50 = 7.21 + log [ base ] [ acid ] 0.5 M 1.50 L = mol acid + molbase 0.29 = log [ base ] [ acid ] 0.75 mol = molacide + mol base 10 0.29 = 10 log [ base ] [ acid ] mole/molarity ratio are equal, so I will substitute M for mol 1.9498446 = [ base ] [ acid ] 0.75 mol = molacid + 1.95 mol acid 1.9498446 [ acid ] =[ base ] 0.75 mol = 2.95 molacid 0.25425 mol = molacid 0.75 mol = 0.25 + molbase 0.49575 mol = molbase I would put 0.5 moles of my base and 0.3 (0.2 54) moles of my acid in a beaker and then fill it up
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Unformatted text preview: to 1.50 L. 2. If 5.0 mL of 0.15 M HCl is added to 750mL of your buFer from (1), what will be the new pH? 0.005 L ∗ 0.15 M = 7.5 ∗ 10 − 4 mol 7.5 ∗ 10 − 4 mol + 1 2 ( 0.25425 mol ) = 0.127875 mol log 1 2 ( 0.49575 ) 0.127875 + p K a = pH pH = 7.497445326 pH = 7.5 3. If 10.0 mL of 1.5 M NaOH is added to 750 mL of your buFer from (1), what will be the new pH? 0.01 L ∗ 1.5 M = 0.015 mol 0.015 mol + 1 2 ( 0.49575 mol ) = 0.262875 mol log 0.262875 1 2 ( 0.25425 ) + p K a = pH pH = 7.525516593 pH = 7.5...
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  • Spring '08
  • LEMASTER
  • Chemistry, pH, mol, buffer solution, Lab section, Phun

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