6B. 3  [ENDORSED]

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Gisela F Ramirez 2H
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Joined: Fri Sep 28, 2018 12:27 am

6B. 3

Postby Gisela F Ramirez 2H » Thu Dec 06, 2018 1:02 am

A careless lab tech wants to prepare 200.00 mL of a 0.025M HCl solution but uses a volumetric flask of volume 250.0mL.
a.) What would be the pH of the desired solution?
b.) What will be the actual pH of the prepared solution?

How do you find the pH? I know pH= -log[H3O+] but how do we find the molarity of H3O when we are given the molarity of HCl?

Michelle Wang 4I
Posts: 63
Joined: Fri Sep 28, 2018 12:27 am

Re: 6B. 3

Postby Michelle Wang 4I » Thu Dec 06, 2018 1:07 am

Since HCl is a strong acid, we can assume that there is 100% dissociation. Since there is .025 M HCl, then there is .025 moles of HCl. When HCl dissociates, there would be .025 moles of H3O+. You would then divide that by the volumes to get the molarity, and use that to solve for the two pH levels.

Henri_de_Guzman_3L
Posts: 88
Joined: Fri Sep 28, 2018 12:25 am

Re: 6B. 3

Postby Henri_de_Guzman_3L » Fri Dec 07, 2018 12:31 am

I'm stuck on this too. I can get the right answer for part A but part B doesn't make any sense. Wouldn't the pH of either volume be the same??

Henri_de_Guzman_3L
Posts: 88
Joined: Fri Sep 28, 2018 12:25 am

Re: 6B. 3  [ENDORSED]

Postby Henri_de_Guzman_3L » Fri Dec 07, 2018 12:40 am

This is how I did it and I got the correct answers. The phrasing of the question is terrible. You have to assume that the solution of 250 ml is DILUTED with water

20181207_003620.jpg


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