HW 9-UV visible spectroscopy Key

# HW 9-UV visible spectroscopy Key -...

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Unformatted text preview: HW 9: UV‐Visible Spectroscopy  (1)  Express the following absorbances as % transmiBance: 0.692, 0.038.  0.20; 0.92  (2) Convert the following transmiBance data to absorbances: 15.8%, 0.085.  0.74; 1.1  (3) A soluSon containing 6.23 ppm KMnO4 had a transmiBance of 0.195 in a 1.00 cm  cell at 520 nm. Calculate the molar absorpSvity (exSncSon coeﬃcient) at 520 nm.  A = 0.71; Molar absorpSvity = 0.11 ppm‐1cm‐1; AlternaSvely, ppm could be converted  to molarity to give an absorpSvity with units of M‐1cm‐1.  (4)  A  25.0  mL  aliquot  of  an  aqueous  quinine  soluSon  was  diluted  to  50.0  mL  and  found to have an absorbance of 0.656 at 348 nm when measured in a 2.50 cm  cell. A second 25.0  mL aliquot was mixed with 10.00  mL of a soluSon containing  25.7  ppm of quinine; a[er diluSon to 50  mL, this soluSon had an absorbance of  0.976 (2.50 cm cell). Calculate the concentraSon of quinine in ppm in the sample.  The  problem  is  idenScal  to  the  one  assigned  with  the  calibraSon  curve  HW.  It  is  a  standard addiSon problem.  The unknown concentraSon is 5.2 ppm.  (5) The acid‐base indicator HIn undergoes the following reacSon in dilute aqueous soluSon:               HIn  H+ + In‐  The following absorbance data were obtained for a 5.00 x 10‐4 M soluSon of HIn in 0.1 M NaOH  and 0.1 M HCl. Measurements were made at wavelengths of 485 nm and 625 nm with 1.00  cm cells.  0.1 M NaOH A485 = 0.075; A625 = 0.904  0.1 M HCl A485 = 0.487; A625 = 0.181  In the  NaOH soluSon essenSally all of the indicator is present as In‐; in the acidic soluSon, it is  essenSally all in the form of HIn.  (a)  Calculate molar absorpSviSes for In‐ and HIn at 485 and 625 nm.  In‐: 150 and 1808 M‐1 cm‐1 at 485 nm and 625 nm  Hin: 974 and 362 M‐1 cm‐1 at 485 nm and 625 nm  (b)  Calculate  the  acid  dissociaSon  constant  for  the  indicator  if  a  pH  5.00  buﬀer  containing  a  small  amount  of  the  indicator  exhibits  an  absorbance  of  0.567  at  485  nm  and  0.395  and  625 nm (1.00 cm cells).  Solve  for  [In]  and  [HIn]  by  treaSng  this  as  a  2‐component  Beer’s  law  problem.  Then  use  the  Henderson‐Haaselbalch equaSon to solve for the pKa.  [In] = 1.05 x 10‐4 M; [HIn] = 5.66 x 10‐4 M; pKa = 5.73  (c)  What  is  the  pH  of  a  soluSon  containing  a  small  amount  of  the  indicator  that  exhibits  an  absorbance of 0.492 at 485 nm and 0.245 at 635 nm?   Same thing. Find the concentraSons and use the HH equaSon with the pKa determined above.  ...
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## This note was uploaded on 11/16/2011 for the course CHEM 4011 at Colorado.

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