Part A What volume of the 11.6 M HCl must the assistant add to a one liter volum
ID: 627774 • Letter: P
Question
Part A What volume of the 11.6 M HCl must the assistant add to a one liter volumetric flask so that the concentration of the stock solution will be 0.64 M after dilution to the mark with water? Part B What is the pH of the stock solution? (Remember the significant figure rules for logarithms!) You will now use this 0.56 M HCl solution to carry out a series of dilutions, measuring the pH of each diluted sample. Sample A will be produced by pipetting 1.00 mL of the stock solution into a 25 mL volumetric, and diluting to volume with distilled water. Sample B will be produced by pipetting 1.00 mL of sample A into a 25 mL volumetric, and diluting to volume with distilled water. Sample C will be produced by pipetting 1.00 mL of sample B into a 25 mL volumetric, and diluting to volume with distilled water. Sample D will be produced by pipetting 1.00 mL of sample C into a 25 mL volumetric, and diluting to volume with distilled water. Sample E will be produced by pipetting 1.00 mL of sample D into a 25 mL volumetric, and diluting to volume with distilled water. Sample F will be produced by pipetting 1.00 mL of sample E into a 25 mL volumetric, and diluting to volume with distilled water. Part C Calculate theCl- concentration in sample A. Part D Calculate H+ and the pH for sample A. Part E Calculate the Cl- concentration in sample B. Part F Calculate H+ and the pH for sample B. part g Calculate Cl- the concentration in sample C. Part H Calculate H+ and the pH for sample C. Part I Calculate Cl- the concentration in sample D. Part J Calculate H+ and the pH for sample D. Part K Calculate the Cl-concentration in sample E. Part L Calculate H+ and the pH for sample E. Part M Calculate the Cl-concentration in sample F. Part N Calculate H+ and the pH for sample F.Explanation / Answer
part A ) volume to be added = 55.17 mL part B) pH = 0.19 part c) Cl- concentration in A= 0.0224M part D) [H+] = .0224 M pH= 1.65 part E) [Cl-] = 8.96*10^-4M part f) [H+] = 8.96*10^-4M pH = 3.05 part G) [Cl-]= 3.584*10^-5M part H) [H+] = 3.584*10^-5M pH = 4.45 part I) [Cl-] = 1.434*10^-6 M part j) [H+] = 1.434*10^-6 M pH = 5.84