Question

Constant boiling HCl has a concentration of 11.6 M. Your laboratory assistant is going to dilute this acid for you to create a stock solution of HCl for you to use in this experiment. What volume of the 11.6 M HCl must the assistant add to one liter volumetric flask so that the concentration of the stock solution will be 0.62 M after dilution to the mark the water?

What is the pH of this stock solution?

You will now use this 0.62 M HCl solution to carry out a series of dilutions, measuring the pH of each diluted sample. Calculate the [Cl^-] and [H+] concentrations and the pH of each sample.

Sample A will be produced bby pipetting 1.00 mL of the stock solution into a 25 mL volumetric, and diluting to the volume with distilled water.

[Cl^-]=

[H+]=

pH=

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.

[Cl^-]=

[H+]=

pH=

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.

[Cl^-]=

[H+]=

pH=

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

[Cl^-]=

[H+]=

pH=

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

[Cl^-]=

[H+]=

pH=

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

[Cl^-]=

[H+]=

pH=

Answer 1

HCl concentration M1 = 11.6

volume of HCl V1 = ??

concentration of stock solution M2 = 0.62

volume of stock solution V2 = 1 lit

M1 V1 = M2 V2

11.6 x V1 = 0.62 x 1
V1 = 0.0534 L = 53.4 mL

volume of HCl = 53.4 mL

pH = - log 0.62 = 0.21

pH of stock solution = 0.21

a) sample A

moles HCl = 0.62 M x 1.00 x 10^-3 L = 0.00062

volume = 25 mL

concentration = moles / volume

new concentration = 0.00062 / 0.025 L

                              = 0.025 M
[H+] =0.025 M

[Cl-]= 0.025 M

pH = 1.6

pH = -log[H+]

pH = -log (0.025) = 1.6

b) Sample B

moles HCl = 0.025 M x 1.0 x 10^-3 L = 0.000025

new concentration = 0.000025 / 0.025 L = 0.0010 M

[H+] = 0.0010 M

[Cl-] = 0.0010 M

pH = 3.0

c) sample C

moles HCl = 0.0010 M x 1.0 x 10^-3 L=1.0 x 10^-6

new concentration = 1.0 x 10^-6 / 0.025 L = 0.000040 M

[H+] = 0.000040 M

[Cl-] = 0.000040 M

pH = 4.4

d) sample D

moles HCl = 0.000040 x 1.0 x 10^-3 =4.0 x 10^-8

new concentration = 4.0 x 10^-8/ 0.025 L=1.6 x 10^-6 M

[H+] = 1.6 x 10^-6 M

[Cl-] = 1.6 x 10^-6 M

pH = 5.8

e) sample E

moles HCl = 1.6 x 10^-6 x 1.0 x 10^-3 L = 1.6 x 10^-9

new concentration = 1.6 x 10^-9 / 0.025 L = 6.4 x 10^-8 M

now we can not neglet the concentration of H+ by the autoionization of the water :[H+]= 1.0 x 10^-7

[H+] = 1.0 x 10^-7 + 6.4 x 10^-8

        =1.6 x 10^-7 M

[H+] =1.6 x 10^-7 M

pH = 6.8

[Cl-]= 6.4 x 10^-8 M