Final [ENDORSED]

[Golbarg Rahimi 3k]

Hi,
Didn’t Dr lavelle mention that salts were not going to be included in our final due to the fire? Question 8a was asking for the PH of a solution containing both NaOH and HCl which produced nacl salt. Were we supposed to know how to solve this problem without the knowledge on salts? If so how could we solve it?
Thank you

[Ozhen Atoyan 1F]

Yes, Dr. Lavelle said that salts were not supposed to be on the final. I don't exactly remember the problem, but maybe you could have solved this problem by finding the molarities and then doing -log of the concentration. What I'm drawing from is pH=-log[concentration]. In this case we would disregard that they are salts.

[Chris Pleman 3E]

What I did was find the molarity of both the H+ and OH- ion in the 1.00 L solution, and then took the difference in molarities between the two ions, because I wanted to find the net concentration of a certain ion. I chose to find the net concentration, because I believed that the H+ and OH- ion would neutralize each other on a 1:1 ratio (making H20), therefore, with whatever was left you'd take the negative log of that concentration. I found that my net concentration was .0498 M H+, and took the negative log to get a pH of 1.30. I'm not sure if this is the correct approach, but it seemed to make the most sense to me at the time.

[Golbarg Rahimi 3k]

Couldn’t we just say that since the products would be salt and water and salt is solid then the ph would equal the ph of water that is 7?

[Chem_Mod]

What I did was find the molarity of both the H+ and OH- ion in the 1.00 L solution, and then took the difference in molarities between the two ions, because I wanted to find the net concentration of a certain ion. I chose to find the net concentration, because I believed that the H+ and OH- ion would neutralize each other on a 1:1 ratio (making H20), therefore, with whatever was left you'd take the negative log of that concentration. I found that my net concentration was .0498 M H+, and took the negative log to get a pH of 1.30. I'm not sure if this is the correct approach, but it seemed to make the most sense to me at the time.

Well stated Chris. This is the correct answer.

Of the exams graded so far students are doing well with a class average of 7/10 on this question. We always give partial credit.
Over 100 students already have a perfect score (10/10) on this question.

The above answer shows this question (and answer) has nothing to do with salts or titrations. For comments on strong acid and strong base in the same solution see Section J and my lecture notes. This was a more straight forward calculation as no equilibrium table was needed.