Bookwork 6B #3

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Bailey Giovanoli 1L
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Bookwork 6B #3

Postby Bailey Giovanoli 1L » Mon Dec 07, 2020 4:16 pm

I've always struggled a little with molarity calculations, so this question confuses me a little:

A careless laboratory technician wants to prepare 200.0 mL of a 0.025 M HCl(aq) solution but uses a volumetric flask of volume 250.0 mL by mistake. (a) What would the pH of the desired solution have been? (b) What will be the actual pH of the solution as prepared?

I know it will involve the M1V1 = M2V2, but I don't understand the explanation of the solution that
-log((0.200 x .025)/.250)= 1.7
-log(0.025)= 1.6
Is this because the volume of water is being used as a proportional constant to show that the solution with a higher volume is more diluted? If someone could please explain this to me, I would appreciate it.

Chem_Mod
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Re: Bookwork 6B #3

Postby Chem_Mod » Mon Dec 07, 2020 4:42 pm

For the pH of the desired solution, you just use the concentration of HCl, which is desired to be 0.025M. 0.025 M of HCl is 0.025 M of H+ since HCl is a strong acid and completely dissociates into H+ and Cl-. pH= -log[H+], therefore, the pH of the desired solution is -log(0.025). For the second part of the problem which asks for the pH of the solution that was actually made, use M1V1= M2V2 for the molarity of H+, solve for M2, the molarity of H+ in the actual solution. M2= M1V1/V2 = (0.025 M)(.200 L)/.250 L. Then plug this M2 into the pH formula

Bailey Giovanoli 1L
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Re: Bookwork 6B #3

Postby Bailey Giovanoli 1L » Mon Dec 07, 2020 4:45 pm

Okay, this makes much better sense now. Thank you:)

Marco Morales 2G
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Re: Bookwork 6B #3

Postby Marco Morales 2G » Fri Dec 11, 2020 12:45 am

Chem_Mod wrote:For the pH of the desired solution, you just use the concentration of HCl, which is desired to be 0.025M. 0.025 M of HCl is 0.025 M of H+ since HCl is a strong acid and completely dissociates into H+ and Cl-. pH= -log[H+], therefore, the pH of the desired solution is -log(0.025). For the second part of the problem which asks for the pH of the solution that was actually made, use M1V1= M2V2 for the molarity of H+, solve for M2, the molarity of H+ in the actual solution. M2= M1V1/V2 = (0.025 M)(.200 L)/.250 L. Then plug this M2 into the pH formula


for the second part, how do you know that you have to convert mL to L ?

Marco Morales 2G
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Re: Bookwork 6B #3

Postby Marco Morales 2G » Fri Dec 11, 2020 1:01 am

Also, how do you know V1=.200L ? and V2=.250L ?

Is it because we initially "think" we used .200 L at first?

RitaThomas_3G
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Re: Bookwork 6B #3

Postby RitaThomas_3G » Fri Dec 11, 2020 10:55 pm

Answering the question above ^, I believe that could be the case. However, what I believe is more important than classifying as V1 or V2 is making sure that you put the 200 mL and 0.025 M on the same side. This way, even if you incorrectly classify it as V1 or V2, you would still get the right answer.

Emma Strassner 1J
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Re: Bookwork 6B #3

Postby Emma Strassner 1J » Sat Dec 12, 2020 12:26 pm

Hi, in general for these calculations you should convert from mL to L because Liters is the unit used in the formula itself.


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