I. Canstant boiling HCI has a concentration of 11.6 ML Your labaratory assistant
ID: 712043 • Letter: I
Question
I. Canstant boiling HCI has a concentration of 11.6 ML Your labaratory assistant is going to dilute this acid for you to create a stock sohution of HCI to use in this experiment What volene of the 116 MHCI must fhe assistant add to one-liter volumetric flask so that the concentration of the stock solation will be 0.73 M after dilution to the mark with water? What is the pH of this stock solution? (Remember the sigaificant figure rules for You will now use thi‘ 0.78 M HCl solution to carry out a series of dilutions meas nngthe pH of each diluted sample. Calculate the (Cll and [H J concestrations and the pli of each sample. (Reamember the significant figuze rules for logarithams) Sample A will be produced by pipetting 1.00 mL of the stock solution into a 25 mL volumetric and diluting to voume with distilled water. . [CI pH distilled water. Sample B will be produced by pipetting 1.00 mL of sample A into a 25 ml vohumetric and diluting to volume with ICTH Sample C will be prodaced by pipetting 1.00 mL of sample B into a 25 mL volumetric and diluting to volume with distilled water. [CI [H+] = pH= . Sample D will be produced by pipetting 1.00 ml of sample C into a 25 ml rohumetric and diluting to vohume with distilled water. pH= Sample E will be produced by pipetting 1.00 mL of sample D into a 25 ml volunetric and diluting to volume with distilled water. [CIjz [H+] = Sample F wll be produced by pipetting 1.00 ml of sample E into a 25 ml volumetic and diluting to volune with distilled uater. [CI-1= .Explanation / Answer
Q1)
the pH of 0.78M HCl solution
pH = -log [H+]
= - log 0.78
=0.11
Q2) Sample A
1mL x 0.78M = 25 XMa
M of Sample A = 0.0312 M
HCL --------------> H+ + Cl- (strong acid , dissociates completely)
Thus [H+] = 0.0312M
[Cl-] 0.0312M
and pH = -log 0.0312
=1.506
Q3) sample B
1mL x 0.0312 M = 25mL x Mb
M of solution B = 0.00125 M
thus [H+] = 1.25x10-3 M
[Cl-] = 1.25x10-3M
and pH = -log 1.25x10-3M
=2.903
Q4) Sample C
1mL x 1.25x10-3M = 25 x Mc
molarity of sample C = 4.992x10-5 M
Thus
[H+] =4.992x10-5 M
[Cl-] = 4.992x10-5 M
and
pH = -log 4.992x10-5 M
=4.3017
Q5) sample D
1mL x 4.992x10-5 M = 25mL x Md
molarity of sample D = 1.997x 10-6 M
Thus
[H+] = 1.997x 10-6 M
[Cl-] = 1.997x 10-6 M
and
pH = -log 1.997x 10-6 M
=5.699
Q6) Sample E
1mL x 1.997x 10-6 M = 25mL x Me
molarity of sample E = 7.99x10-8 M
Now the solution is so dilute that the [H+] is less than the [H=] fro autoionisation of water.
Thus the [H+] in solution = 7.99x10-8 M + 1.0x10-7M
= 1.799 x10-7 M
and [Cl-] = 7.99x10-8 M
pH = -log 1.799x10-7 M
=6.7449