6B.3

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ayushibanerjee06
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6B.3

Postby ayushibanerjee06 » Sun Dec 01, 2019 5:18 pm

Can someone explain how to do 6B.3?
The question is: 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 started doing part a) by multiplying 0.2 L * 0.025 M and got 0.005 mol HCl. I am not sure if I can do -log(0.005 mol) so I am kind of stuck.

Harry Zhang 1B
Posts: 101
Joined: Sat Sep 14, 2019 12:16 am

Re: 6B.3

Postby Harry Zhang 1B » Sun Dec 01, 2019 5:28 pm

We must input the concentration(mol*L^-1) when using the -log formula to calculate pH level, therefore, you will need to divide the result you got from your first calculation by 0.250L and then input the result into -log to calculate the pH level.

ayushibanerjee06
Posts: 177
Joined: Thu Jul 11, 2019 12:16 am
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Re: 6B.3

Postby ayushibanerjee06 » Sun Dec 01, 2019 5:38 pm

but part a) asks for the desired solution. 0.005 mol/0.2L is 0.025 M which is what you start with. same thing when you multiply 0.025 M * 0.25 L.


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