Show all work for full credit. 1) Constant boiling HCI has a concentration of 11

Question

Show all work for full credit. 1) Constant boiling HCI has a concentration of 11.6 M. Your laboratory assistant is going to dilute this acid to create a stock solution of HCI for you to use in this experiment. What volume of the 11.6 M HCI must the assistant add to water in one-liter metric flask so that the concentration of the stock solution will be 0.62 Mafter dilution to the mark with water? What is the pH of this stock solution? (Remember the significant figure rules for logarithms) You will now use this 0.62 M HCl solution to camy out a series of dilutions, measuring the pH of each diluted sample. Calculate the [Cl and [H concentrations and the pH of each sample (Remember the significant figure rules for logarithms!) Sample A will be produced by pipetting 1.00 mL off the stock solution into a 25 mL volumetric flask, and diluting to volume with distilled water. [H] pH Sample B will be produced by pipetting 1.00 mL of sample A into a 25 mL volumetric flask, and diluting to volume with distilled water. [H Sample C will be produced by pipetting 1.00 mL of sample B into a 25 mL volumetric flask, and diluting to volume with distilled water. [H pH Sample D will be produced by pipetting 1.00 mL of sample C into a 25 mL volumetric flask, and diluting to volume with distilled water. 120

Explanation / Answer

Given

Molarity of conc. acid = 11.6 M ( or mol/L)

Molarity of stock solution = 0.62 M

Volume of stock solution = 1 L

Volume of Conc. acid = 0.62 mol/L * 1 L / 11.6 mol/L = 0.05345 L = 53.45 ml

so he should take 53.45 ml of given conc. acid and make it upto 1 L of solution

molarity of HCl = 0.62 mol/l

one mole HCl will yield one mole of H+ and one mole of Cl-

[H+] = 0.62 mol/L

pH = - log([H+]) = 0.208 Answer

similarly we have to calculate molarity of new solution of HCl each time and use the same molarity as [H+] and [Cl-] and calculate pH using the above formula

Stock solution

Molarity M1 = 0.62 M

Volume V1 = 1ml =0.001 L

dilute solution

volume V2 = 25 ml = 0.025 L

Molarity M2 = M1 * V1 / V2 = 0.62 mol/L * 0.001 L / 0.025 L = 0.0248 mol/L

[H+] = 0.0248 mol/L

[Cl-] = 0.0248 mol/L

pH = - log ([H+]) = - log ( 0.0248) = 1.6 Answer

similarly you can do for rest

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---