Part A What volume of the 11.6 M HCl must the assistant add to a one liter volum

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 0.56 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

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