Constant boiling HCl has a concentration of 11.6 M. Your laboratory assistant is
ID: 903949 • Letter: C
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=
Explanation / Answer
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