Bonus: Students complained that a professor\'s tests were much too difficult. Th
ID: 580564 • Letter: B
Question
Bonus: Students complained that a professor's tests were much too difficult. The professor then agreed to give a one question test. Correct responses would earn a100; incorrect ones, a 0. The professor made the first entry; the students made the others. One student made the following entries, explain why this student received a [HCI] in mole/L,S 1.0 x 10-2 1.0 x 10- 1.0 x 10-6 1.0 x 10-8 1.0 x 10-10 2.00 (professor's entry) 4.00 (student's entry) 6.00 (student's entry) 8.00 (student's entry) 10.00 (student's entry)Explanation / Answer
Solution:- Last two entries are not correct since HCl is an acid and the pH of the acid can't be equal to or greater than 7. For the last two entries student has written the pH as 8.00 and 10.00 and these two are incorrect as these must be less than 7.
Even if the acid is stronger but it's concentration is too low and you think it would give you pH equal to or greater than 7 then always included the dissociation of water also along with the acid dissociation. So, total H+ would be the sum of the H+ concentrations that we get from the acid as well as water.
When HCl concentration is 1.0 x 10-8 M. For H2O. H+ is 1.0 x 10-7 M.
So, Total [H+] = 1.0 x 10-8 + 1.0 x 10-7
for making the exponents equal, It could be written as..
Total [H+] = 1.0 x 10-8 + 10.0 x 10-8
Total [H+] = 10-8[ 1.0 + 10.0] = 11 x 10-8 = 1.1 x 10-7
So, pH = - log[H+]
pH = - log 1.1 x 10-7
pH = 7log10 - log1.1
pH = 7(1) - 0.04
pH = 7 - 0.04
pH = 6.96
Similarly, when HCl concentration is 1.0 x 10-10 M then...
Total [H+] = 1.0 x 10-10 + 1.0 x 10-7
Total [H+] = 1.0 x 10-10 + 1000 x 10-10
Total [H+] = 10-10[ 1.0 + 1000]
Total [H+] = 1001 x 10-10 = 1.001 x 10-7
pH = - log 1.001 x 10-7
pH = 7log10 - log 1.001
pH = 7(1) - 0.00043
pH = 7 - 0.00043
pH = 6.99957