lab 17.docx - Lab 18 Buffer Lab Purpose The purpose of this...

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Lab 18: Buffer LabPurpose: The purpose of this experiment is to prepare a buffer solution at a specific pH then to putstress on the system with the addition of an acid or base and re-take the pH. Students will thenuse the Henderson-Hasselbach to determine the theoretical pH as well as the percent error. Data Analysis I: Determining Chemical Equations for Buffer Solutions and the Added StressPart I: Buffer Equilibrium Equation:HC2H3O2(aq) + H2O(l)H3O++ C2H3O2-(s) Addition of the Base, NaOH, to the Buffer SystemNaOH(aq)+ HC2H3O2(aq)H2O(l) + NaC2H3O2(aq)Addition of the Acid, HCl, to the Buffer SystemHCl(aq)+ NaC2H3O2(aq) NaCl(aq) + HC2H3O2(aq) Part II:Buffer Equilibrium Equation:H2PO4-(s) + H2O(l) H3O+ + HPO42-(s)Addition of the Base, NaOH, to the Buffer SystemNaOH(aq)+ NaH2PO4(s)Na2HPO4(aq) + H2O(l)Addition of the Acid, HCl, to the Buffer System
HCl(aq)+ Na2HPO4(s) NaCl(aq)+ NaH2PO4(aq)Data Analysis II:Calculating Theoretical pH and DeterminingPercent ErrorHenderson-Hasselbach equation: pH = pKa+ log ( [A-] / [HA] )Theoretical pH and Percent Error of the Buffer Solution:Moles of HC2H3O2: (0.100M HC2H3O2)(0.050L) = 0.005moles HC2H3O2Moles of NaC2H3O2: 0.076g NaC2H3O2 / 82.0337g NaC2H3O2 = 9.26E-4 moles NaC2H3O2pH = -log(1.8E-5) + log(9.26E-4 / 0.005)pH = 4.01Percent Error = [(Calculated pH – Actual pH) / Calculated pH] x 100Actual pH = 3.88% Error = [(4.01 – 3.88) / 4.01] x 100 = 3.24%Theoretical pH and Percent Error of the Buffer Solution Stressed with a Base: NaOH(aq)+ HC2H3O2(aq)H2O(l) + NaC2H3O2(aq)
Moles of HC2H3O2:(0.100M HC2H3O2)(0.050L) = 0.005 moles HC2H3O2Moles of NaC2H3O2:0.076g NaC2H3O2 / 82.0337g NaC2H3O2 = 9.26E-4 moles NaC2H3O2Concentration of HC2H3O2:0.005 moles HC2H3O2/ 0.050L = 0.1M HC2H3O2Concentration of NaC2H3O2:9.26E-4 moles NaC2H3O2 / 0.050L = 0.02M NaC2H3O2Moles of HC2H3O2 after NaOH was Added to 10mL of Buffer:(0.1M HC2H3O2)(0.01L) = 0.001 moles HC2H3O2Moles of NaC2H3O2 after NaOH was Added to 10mL of Buffer:(0.02M NaC2H3O2)(0.01L) = 2E-4 moles NaC2H3O2Moles of NaOH:(0.02M NaOH)(0.002L) = 4E-5 moles NaOHExcess HC2H3O2:0.001 moles HC2H3O2 - 4E-5 moles NaOH = 9.6E-4 moles HC2H3O2Excess NaC2H3O2:2E-4 moles NaC2H3O2+ 4E-5 moles NaOH = 2.4E-4 moles NaC2H3O2
pH = -log(1.8E-5) + log(2.4E-4 / 9.6E-4) = 4.14Percent Error = [(Calculated pH – Actual pH) / Calculated pH] x 100Actual pH = 4.05% Error = [(4.14 – 4.05) / 4.14] x 100 = 2.24%Theoretical pH and Percent Error of the Buffer Solution Stressed with an Acid: HCl(aq)+ NaC2H3O2(aq) → NaCl(aq) + HC2H3O2(aq)Moles of HC2H3O2:(0.100M HC2H3O2)(0.050L) = 0.005 moles HC2H3O2Moles of NaC2H3O2:

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